Answer:
$1.9006
Step-by-step explanation:
The expected value out of an event with value v and the probability p of that event to happen is the product of the value v and the probability p itself.
Therefore, the expected value of winning $215000 with probability of 0.00000884 bottle purchase is
E = pv = 215000 * 0.00000884 = $1.9006
The expected value, multiplied by the chance of winning the Lamborghini, is approximately $1.9004. This value means that if you could repeat this contest over and over, on average, you would gain about $1.90 per bottle purchased.
Explanation:The subject of this question is about the expected value in probability. In the given scenario, when a Pepsi or Mountain Dew bottle is purchased, there's a probability of .00000884 of winning a Lamborghini sports car worth $215,000.
The expected value is calculated as the product of the value of the prize and the probability of winning it. Therefore, the expected value for this case would be (215000*.00000884) which equals to $1.9004, rounded to 4 decimal places. However, in the previous calculation, there was an error, resulting in a negative value which should not be the case in this context.
Expected value offers a predicted value of a variable, calculated as the sum of all possible values each multiplied by the probability of its occurrence. It is widely used in Probability Theory and Statistics.
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For each call to the following method, indicate what console output is produced:public void mysteryXY(int x, int y) { if (y == 1) { System.out.print(x); } else { System.out.print(x * y + ", "); mysteryXY(x, y - 1); System.out.print(", " + x * y); }} mysteryXY(4, 1); mysteryXY(4, 2); mysteryXY(8, 2); mysteryXY(4, 3); mysteryXY(3, 4);
The mysteryXY method on Java demonstrates recursion, and calls to this method generate a sequence of multiplications based on the values of x and y, with the results printed in a particular format.
Explanation:The console output for each call to the mysteryXY method can be determined by following the recursion and evaluating the output statements. The method prints a series of multiplications, interspersed with commas, by recursively calling itself with a decremented y until y reaches 1.
Calling mysteryXY(4, 1) will output 4 since y is 1.Calling mysteryXY(4, 2) will output 8, 4, 4 because it prints 4 × 2, then calls itself with y-1 (which is 1, printing 4), and then prints 4 × 2 again after the recursive call.Calling mysteryXY(8, 2) will output 16, 8, 8 for the same reasons as the previous call but with x equal to 8.Calling mysteryXY(4, 3) will output 12, 8, 4, 4, 8, 12 which includes two recursive calls, printing 4 × 3, 4 × 2, then 4 × 1, and then 4 × 2 and 4 × 3 after the reversals of the calls.Finally, calling mysteryXY(3, 4) will output 12, 9, 6, 3, 3, 6, 9, 12 with three recursive calls and their reversals.A survey of magazine subscribers showed that 45.2% rented a car during the past 12 months for business reasons, 56% rented a car during the past 12 months for personal reasons, and 32% rented a car during the past 12 months for both business and personal reasons. (a) What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons? (b) What is the probability that a subscriber did not rent a car during the past 12 months for either business or personal rea
Answer:
a. 0.692 or 69.2%; b. 0.308 or 30.8%.
Step-by-step explanation:
This is the case of the probability of the sum of two events, which is defined by the formula:
[tex] \\ P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex] (1)
Where [tex] \\ P(A \cup B)[/tex] represents the probability of the union of both events, that is, the probability of event A plus the probability of event B.
On the other hand, [tex] \\ P(A \cap B)[/tex] represents the probability that both events happen at once or the probability of event A times the probability of event B (if both events are independent).
Notice the negative symbol for the last probability. The reason behind it is that we have to subtract those common results from event A and event B to avoid count them twice when calculating [tex] \\ P(A \cup B)[/tex].
We have to remember that a sample space (sometimes denoted as S) is the set of the all possible results for a random experiment.
Calculation of the probabilitiesFrom the question, we have two events:
Event A: event subscribers rented a car during the past 12 months for business reasons.
Event B: event subscribers rented a car during the past 12 months for personal reasons.
[tex] \\ P(A) = 45.2\%\;or\;0.452[/tex]
[tex] \\ P(B) = 56\%\;or\;0.56[/tex]
[tex] \\ P(A \cap B) = 32\%\;or\;0.32[/tex]
With all this information, we can proceed as follows in the next lines.
The probability that a subscriber rented a car during the past 12 months for business or personal reasons.
We have to use here the formula (1) because of the sum of two probabilities, one for event A and the other for event B.
Then
[tex] \\ P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]
[tex] \\ P(A \cup B) = 0.452 + 0.56 - 0.32[/tex]
[tex] \\ P(A \cup B) = 0.692\;or\;69.2\%[/tex]
Thus, the probability that a subscriber rented a car during the past 12 months for business or personal reasons is 0.692 or 69.2%.
The probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons.
As we can notice, this is the probability for the complement event that a subscriber did not rent a car during the past 12 months, that is, the probability of the events that remain in the sample space. In this way, the sum of the probability for the event that a subscriber rented a car plus the event that a subscriber did not rent a car equals 1, or mathematically:
[tex] \\ P(\overline{A \cup B}) + P(A \cup B)= 1[/tex]
[tex] \\ P(\overline{A \cup B}) = 1 - P(A \cup B)[/tex]
[tex] \\ P(\overline{A \cup B}) = 1 - 0.692[/tex]
[tex] \\ P(\overline{A \cup B}) = 0.308\;or\;30.8\%[/tex]
As a result, the requested probability for a subscriber that did not rent a car during the past 12 months for either business or personal reasons is 0.308 or 30.8%.
We can also find the same result if we determine the complement for each probability in formula (1), or:
[tex] \\ P(\overline{A}) = 1 - P(A) = 1 - 0.452 = 0.548[/tex]
[tex] \\ P(\overline{B}) = 1 - P(B) = 1 - 0.56 = 0.44[/tex]
[tex] \\ P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - 0.32 = 0.68[/tex]
Then
[tex] \\ P(\overline{A \cup B}) = P(\overline{A}) + P(\overline{B}) - P(\overline{A\cap B})[/tex]
[tex] \\ P(\overline{A \cup B}) = 0.548 + 0.44 - 0.68[/tex]
[tex] \\ P(\overline{A \cup B}) = 0.308[/tex]
Final answer:
The probability of a magazine subscriber renting a car in the past 12 months for business or personal reasons is 69.2%, and the probability of not renting for either reason is 30.8%.
Explanation:
To solve the question: What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons? and What is the probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons?, we can use the principle of inclusion-exclusion in probability.
(a) The probability of renting a car for either reason can be calculated by adding the probabilities of renting for each reason and then subtracting the probability of renting for both reasons (to avoid double counting). Hence, the formula is: P(Business or Personal) = P(Business) + P(Personal) - P(Both).
Substituting the given values: P(Business or Personal) = 45.2% + 56% - 32% = 69.2%.
(b) The probability that a subscriber did not rent a car for either business or personal reasons is the complement of the probability calculated in part (a). Hence, it is calculated as 100% - 69.2% = 30.8%.
The Bruin Stock Fund sells Class A shares that have a front-end load of 4.8 percent, a 12b-1 fee of 0.42 percent, and other fees of 1.3 percent. There are also Class B shares with a 5 percent CDSC that declines 1 percent per year, a 12b-1 fee of 1.95 percent, and other fees of 1.3 percent. Assume the portfolio return is 11 percent per year. What is the value of $1 invested in each share class if your investment horizon is 3 years? Class A $ Class B $ What if your investment horizon is 20 years?
Answer:
Step-by-step explanation:
Investment Amount Net of Front end Load (1 - 0.048) = $0.952
The value of $1 invested in each share class if investment horizon is 3 years
After 3 years: (For every dollar invested)Class A: 0.952 x (1+0.11-0.0042-0.013)^3 = $1.24Class B: (1 x (1+0.11-0.0195-0.013)^3) x (1-0.02) = $1.23SIMILARLY,
The value of $1 invested in each share class if investment horizon is 20 years
After 20 years: (For every dollar invested)Class A: 0.952 x (1+0.11-0.0042-0.013)^20 = $5.62Class B: 1 x (1+0.11-0.0195-0.013)^20 = $4.45Answer:
Part 1 :
The total investment for,
Class A = $1.24
Class B = $ 1.23
Part 2:
The total investment for,
Class A = $ 5.62
Class B = $ 4.45
Step-by-step explanation:
The amount of investment Net of Front end Load,
[tex]= 1 - 0.048[/tex]
[tex]= $0.952[/tex]
Part 1:
If the investment horizon is 3 years, the value of $1 invested in each share class for Class A become,
[tex]= (0.952)(1 + 0.11 - 0.0042 - 0.013)^{3}[/tex]
[tex]= $1.24[/tex]
The value of $1 invested in each share class for Class A is $1.24.
If the investment horizon is 3 years, the value of $1 invested in each share class for Class B become,
[tex]= ((1)(1 + 0.11 - 0.0195 - 0.013)^{3} )(1-0.02)[/tex]
[tex]= $1.23[/tex]
The value of $1 invested in each share class for Class B is $1.23.
Part 2:
If the investment horizon is 20 years, the value of $1 invested in each share class for Class A become,
[tex]= (0.952)(1 + 0.11 - 0.0042 - 0.013)^{20}[/tex]
[tex]= 5.62[/tex]
The value of $1 invested in each share class for Class A is $5.62.
If the investment horizon is 20 years, the value of $1 invested in each share class for Class B become,
[tex]= (1)(1 + 0.11 - 0.0195 - 0.013)^{20}[/tex]
[tex]= $4.45[/tex]
The value of $1 invested in each share class for Class B is $4.45.
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to cone back to the skipped part. The half-life of cesium-137 is 30 years. Suppose we have a 30 mg sample. Exercise
(a) Find the mass that remains after t years. Step 1 Let y(t) be the mass (in mg) remaining after t years. Then we know the following Stop 2 Since the half-life is 30 years, then y(30) - More Information m Your answer cannot be understood or graded Submit Skie (you cannot come back) Exercise
(b) How much of the sample remains after 20 years? Step 1 After 20 years we have the following (20) 30 mg (Round your answer to two decimal places.)
Answer:
(a)y(t)=30exp(-0.0231t)
(b)y(20)=18.9mg
Step-by-step explanation:
At a particular time t, the mass of a radioactive substance like Cesium-137 is governed by the equation:
N=N₀e⁻ᵏᵗ where k=ln 2/half life
(a)Mass that remains after t years
Half Life= 30 years
k= ln2/30=0.0231
Initial Mass, N₀=30mg
Therefore the mass N that remains at time t
N=N₀e⁻ᵏᵗ
N=30exp(-0.0231t)
y(t)=30exp(-0.0231t)
(b)We want to determine how much of the sample remains after 20 years.
At t=20 years
y(t)=30exp(-0.0231t)
y(20)=30exp(-0.0231X20)
=30 X 0.63
y(20)=18.9mg
Final answer:
23.81 mg of the original 30 mg sample remaining, rounded to two decimal places.
Explanation:
The half-life of a radioactive isotope like cesium-137 is time it takes for half of the original amount of the substance to decay. For cesium-137, this period is 30 years. To find out how much of a sample remains after a specific amount of time, such as 20 years in the given question, we use the decay formula.
Step 1: Let y(t) be the mass remaining after t years.
Step 2: The decay formula is y(t) = y(0) · (1/2)^(t/half-life), where y(0) is the initial mass and t is the time in years.
For cesium-137 after 20 years: y(20) = 30 mg · (1/2)^(20/30).
Calculating the remaining mass: y(20) = 30 mg · (1/2)^(2/3) ≈ 30 mg · 0.7937 ≈ 23.81 mg.
3. At the Art Studio 3/8 were red and
# 4/8 were blue. The rest of the colors
were orange. What fraction of the colors
at the Art Studio were Orange?
Answer:
⅛
Step-by-step explanation:
Let the total be X
4/8 = 1/2
X - ⅜X - ½X = ⅛X are orange
Fraction ⅛X ÷ X = ⅛
Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficients. If possible, factor L. (Use D for the differential operator.) y(4) + 8y' = 6
Answer:
The differential equation will be like the one shown below
Step-by-step explanation:
Data:
Let the equation be given as:
y(4) + 8y' = 6
The equation will be expressed linearly as follows:
y(4) + 8[tex]\frac{dy}{dx} = 6[/tex]
[tex]8\frac{dy}{dx}+ 4y = 6[/tex]
This is the linear form of the differential equation.
Answer:
Step-by-step explanation:
Given the differential equation
y^(4) + 8y' = 6
We want to write this in the form
L(y) = g(x)
Where L is a linear differential operator with constant coefficient.
A linear differential operator of the nth order is a linear combination of derivative operators up to n.
L = D^n + a_1D^(n-1) + a_2D^(n-2) + ... + a_n,
defined by
Ly = y^n + a_1y^(n-1) + a_2y^(n-2) + ... + a_(n-1)y' + a_ny
Where a_i are continuous functions of x.
Now, we have
y^(4) + 8y' = 6
Let d/dx = D
Then
D^4 y + 8Dy = 6
D(D³ + 8)y = 6
Consider
D(D³ + 8)y = 0
The auxiliary equation is
m(m³ + 8) = 0
m = 0
Or
m³ + 8 = 0
=> m³ = -8
=> m = -2
The complimentary solution is
y = C1 + (C2 + C3x + C4x²)e^(-2x)
The particular integral is
y_p = Ax
y' = A
y'' = y''' = y^(4) = 0
Using these
0 + 8A = 6
A = 6/8 = 3/4
So
y = C1 + (C2 + C3x + C4x²)e^(-2x) + 3/4
Every month, there are 1000 independent TIE ghter ights, and each TIE ghter ight crashes with a probability of 0.0035. (a) What is the probability that at least 2 crashes occur in the next month
Answer:
86.46% probability that at least 2 crashes occur in the next month.
Step-by-step explanation:
For each TIE ghter ights, there are only two possible outcomes. Either it crashes, or it does not. The probability of a TIE ghter ights crashing is independent from other TIE ghter ights. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Every month, there are 1000 independent TIE ghter ights, and each TIE ghter ight crashes with a probability of 0.0035.
This means that [tex]n = 1000, p = 0.0035[/tex]
(a) What is the probability that at least 2 crashes occur in the next month?
Either less than 2 crashes occur, or at least 2 do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{1000,0}.(0.0035)^{0}.(0.9965)^{1000} = 0.0300[/tex]
[tex]P(X = 1) = C_{1000,1}.(0.0035)^{1}.(0.9965)^{999} = 0.1054[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0300 + 0.1054 = 0.1354[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1354 = 0.8646[/tex]
86.46% probability that at least 2 crashes occur in the next month.
Sara is the recipient of a trust that will pay her $500 on the first day of each month, starting immediately and continuing for 40 years. What is the value of this inheritance today if the applicable discount rate is 7.3 percent, compounded monthly
Answer:
The value of this inheritance is $78,192.28
Step-by-step explanation:
The monthly payments Sara will receive starting today for next 40 years is $500.
Annual Interest Rate = 7.3%
Monthly Interest Rate = Annual Interest Rate/12
=7.3/12
Monthly Interest Rate = 0.6083%
Present Value = $500 + $500/1.006083 + $500/1.006083^2 + $500/1.006083^3 + ... + $500/1.006083^479
Present Value = $500 * 1.006083 * (1 - (1/1.006083)^480) / 0.006083
Present Value = $500 * 156.39156
Present Value = $78,192.28
Thus, the value of this inheritance is $78,192.28
Anna goes to a frozen yogurt shop. She can choose from any of the following toppings: peanuts, caramel sauce, butterscotch chips, strawberries, and cookie dough bits. How many different variations of yogurt and toppings can be made?
Answer:
32 variations
Step-by-step explanation:
Anna can choose any number from 0 to 5 different toppings, which is the sum of the combinations of 0 out of 5, 1 out of 5, 2 out of 5, 3 out of 5, 4 out of 5 and 5, out of 5. The number of different variations is:
[tex]n = \frac{5!}{(5-0)!0!} + \frac{5!}{(5-1)!1!}+ \frac{5!}{(5-2)!2!} +\frac{5!}{(5-3)!3!} +\frac{5!}{(5-4)!4!} +\frac{5!}{(5-5)!5!}\\ n=1+5+10+10+5+1\\n=32[/tex]
There are 32 different variations of yogurt and toppings.
The question involves combinatorics, a branch of Mathematics. For each topping, Anna can decide to either have it or not, generating 2^5 possible combinations. Subtracting the option with no toppings, there are 31 different yogurt and toppings combinations.
Explanation:When Anna goes to the frozen yogurt shop, she has five options for toppings: peanuts, caramel sauce, butterscotch chips, strawberries, and cookie dough bits. In this scenario, she could have her yogurt with any possible combination of these five toppings. We're entering the domain of combinatorics, a branch of Mathematics. For each topping, Anna can decide to either have it or not. Therefore, the problem is essentially a set of five binary decisions, which means there are 2^5 possible outcomes. However, one of these possibilities is that she chooses no toppings at all, leaving us with 2^5 - 1 = 31 different combinations of toppings that she can choose.
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If sin θ=− 8/17 and 270°<θ<360°, what is cos θ?
Final answer:
To determine cos θ when sin θ is -8/17 and θ is in the fourth quadrant, use the Pythagorean identity; the result is cos θ = 15/17.
Explanation:
Given the information that sin θ is -8/17 and the angle θ is between 270° and 360°, we need to determine the cos θ. In the given range, θ lies in the fourth quadrant where sine is negative and cosine is positive.
Using the Pythagorean identity sin2 θ + cos2 θ = 1, we can find cos θ. Since sin θ = -8/17, we have:
sin2 θ = (-8/17)2 = 64/289
cos2 θ = 1 - sin2 θ = 1 - 64/289 = 225/289
cos θ = √(225/289)
cos θ = 15/17 (since cosine is positive in the fourth quadrant)
Therefore, cos θ = 15/17.
The average age of CEOs is 56 years. Assume the variable is normally distributed. If the SD is four years, find the probability that the age of randomly selected CEO will be between 50 and 55 years old.
Answer: the probability that the age of randomly selected CEO will be between 50 and 55 years old is 0.334
Step-by-step explanation:
Assuming that the age of randomly selected CEOs is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = age of randomly selected CEOs.
µ = mean age
σ = standard deviation
From the information given,
µ = 56 years
σ = 4 years
We want to find the probability that the age of randomly selected CEO will be between 50 and 55 years old. It is expressed as
P(50 ≤ x ≤ 55)
For x = 50,
z = (50 - 56)/4 = - 1.5
Looking at the normal distribution table, the probability corresponding to the z score is 0.067
For x = 55,
z = (55 - 56)/4 = - 0.25
Looking at the normal distribution table, the probability corresponding to the z score is 0.401
Therefore,
P(50 ≤ x ≤ 55) = 0.401 - 0.067 = 0.334
B. Calculate the distance Kinsey jogs if
she jogs'mile each day for 9 days.
Answer:
9x miles
Step-by-step explanation:
Distance covered per day = x mile per day
Number of days = 9days
Unknown:
Distance covered for 9 days = ?
Solution:
This is a simple problem.
Since Kinsey covered a distance of x mile/day, in 9 days:
x mile is covered in 1 day;
1 day, Kinsey covers x mile
9 days, she will cover 9 x (x) miles = 9x miles
Final answer:
Kinsey has jogged 9 miles in total over the 9 days.
Explanation:
The question asks us to calculate the distance Kinsey jogs if she jogs 1 mile each day for 9 days.
To find the total distance, we simply need to multiply the daily distance by the number of days she jogs.
To calculate:
Identify the daily distance jogged: 1 mile.
Identify the number of days jogged: 9 days.
Multiply the daily distance by the number of days: 1 mile/day imes 9 days.
After performing the multiplication, we find that Kinsey has jogged 9 miles in total over the 9 days.
The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution: x 0 1 2 3 4 5 P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05 On average, how many accidents are there in the intersection in a week? a. 5.3 b. 2.5 c. 1.8 d. 0.30 e. 0.1667
The average number of accidents in the intersection in a week, given by the expected value of the random variable, is calculated by multiplying each possible value by its corresponding probability and summing these products. The resulting average is 1.8 accidents per week.
Explanation:This question is about calculating the expected value or mean of a random variable. For a discrete random variable, we calculate the expected value by summing the product of each possible value and its corresponding probability.
Here, you should multiply the number of accidents (x) by the corresponding probability (P(X = x)), then sum these products. So, the calculation becomes:
0*0.20 + 1*0.30 + 2*0.20 + 3*0.15 + 4*0.10 + 5*0.05 = 0 + 0.30 + 0.40 + 0.45 + 0.40 + 0.25 = 1.80.
Therefore, on average, there are 1.8 accidents in the intersection in a week, so the answer is (c).
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Biologists stocked a lake with 160 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 4,000. The number of fish tripled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years. P
Answer:
Therefore logistic equation is
[tex]P(t) = \frac{4000}{1+ 24(\frac{36}{11})^t}[/tex]
Step-by-step explanation:
The logistic equation is
[tex]\frac{dP}{dt}= kP(1-\frac{P}{k})[/tex]
The analytic solution is
[tex]P(t)= \frac{K}{1+Ae^{-kt}}[/tex] ; [tex]A=\frac{K-P_0}{P_0}[/tex]
P(t) = the population at time t.
K = the carrying capacity = 4,000
k= constant proportionality
t= time
[tex]P_0[/tex] = 160.
The number of fish tripled in the first year.
P(1) = 3×160 =480
[tex]A= \frac{4000-160}{160}[/tex] =24
[tex]\therefore P(t) =\frac{4000}{1+24 e^{-kt}}[/tex] .....(1)
When t= 1 , P(1)=480
[tex]480=\frac{4000}{1+24 e^{-k}}[/tex]
[tex]\Rightarrow 1+24e^{-k} =\frac{4000}{480}[/tex]
[tex]\Rightarrow 24e^{-k} =\frac{4000}{480} -1[/tex]
[tex]\Rightarrow 24e^{-k}= \frac{25}{3}-1[/tex]
[tex]\Rightarrow 24e^{-k}=\frac{22}{3}[/tex]
[tex]\Rightarrow e^{-k} = \frac{22}{3\times 24}[/tex]
[tex]\Rightarrow e^{-k} = \frac{11}{36}[/tex]
taking ln both sides
[tex]\Rightarrow ln e^{-k} =ln \frac{11}{36}[/tex]
[tex]\Rightarrow {-k} = ln\frac{11}{36}[/tex]
[tex]\Rightarrow k = -ln\frac{11}{36}[/tex]
Putting the value of k in equation (1)
[tex]P(t)=\frac{4000}{1+24e^{ln\frac{36}{11}t}}[/tex]
[tex]\Rightarrow P(t) = \frac{4000}{1+24 (e^{ln\frac{36}{11}})^t}[/tex]
[tex]\Rightarrow P(t) = \frac{4000}{1+ 24(\frac{36}{11})^t}[/tex]
Therefore logistic equation is
[tex]P(t) = \frac{4000}{1+ 24(\frac{36}{11})^t}[/tex]
2.10 Guessing on an exam: In a multiple choice exam, there are 6 questions and 4 choices for each question (a, b, c, d). Nancy has not studied for the exam at all and decides to randomly guess the answers. What is the probability that:
Answer:
a) [tex] p = (3/4)^5 *(1/4) =0.0593[/tex]
b) [tex] P(X=6) = (6C6) (0.25)^6 (1-0.25)^{6-6}= 0.000244[/tex]
c) [tex] P(X \geq 1)[/tex]
And we can use the complement rule like this:
[tex]P(X \geq 1) = 1-P(X<1) = 1-P(X=0) [/tex]
[tex] P(X=0) = (6C0) (0.25)^0 (1-0.25)^{6-0}= 0.17798[/tex]
And replacing we have:
[tex]P(X \geq 1) = 1-P(X<1) = 1-P(X=0) =1-0.17798 = 0.822[/tex]
Step-by-step explanation:
Previous concepts
A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]
We can model the number of correct questions answered with a binomial distribution [tex] X \sim Binom(n = 6, p = 1/4=0.25)[/tex]
Solution to the problem
Assuming the following questions:
a) the first question she gets right is the 6th question?
For this case we want the first 5 questions incorrect and the last one correct, assuming independence we have:
[tex] p = (3/4)^5 *(1/4) =0.0593[/tex]
(b) she gets all of the questions right?
For this case we want all the questions right so then we want this:
[tex] P(X=6) = (6C6) (0.25)^6 (1-0.25)^{6-6}= 0.000244[/tex]
(c) she gets at least one question right?
For this case we want this probability:
[tex] P(X \geq 1)[/tex]
And we can use the complement rule like this:
[tex]P(X \geq 1) = 1-P(X<1) = 1-P(X=0) [/tex]
[tex] P(X=0) = (6C0) (0.25)^0 (1-0.25)^{6-0}= 0.17798[/tex]
And replacing we have:
[tex]P(X \geq 1) = 1-P(X<1) = 1-P(X=0) =1-0.17798 = 0.822[/tex]
The probability of guessing all 6 answers correctly in a multiple choice exam is 1/4096.
Explanation:In mathematics, especially in the field of probability, the question describes a situation in which the answer to each of the questions on the exam is guessed at random. To calculate the probability of correctly guessing the answer to an individual question with 4 choices, we need to remember that probability is calculated as a ratio of the favorable outcomes to the total number of outcomes. In this case, each question has 1 correct answer and 3 incorrect answers, so the probability of correctly guessing an answer is 1/4 or 0.25 for each question.
However, if we were to calculate the probability of correctly guessing all the answers, such a probability would be significantly smaller. Because the answers to the questions are guessed independently, their probabilities multiply. Thus, the probability of correctly guessing the answer to all 6 questions is (1/4)^6 = 0.000244140625.
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For this assignment, your group will utilize the preliminary data collected in the Topic 2 assignment. Considering the specific requirements of your scenario, complete the following steps using Excel. The accuracy of formulas and calculations will be assessed.
Select the appropriate discrete probability distribution. If using a binomial distribution, use the constant probability from the collected data and assume a fixed number of events of 20. If using a Poisson distribution, use the applicable mean from the collected data.
Identify the following: the probability of 0 events occurring, the probability of <5 events occurring, and the probability of ≥10 events occurring.
Using the mean and standard deviation for the continuous data, identify the applicable values of X for the following: Identify the value of X of 20% of the data, identify the value of X for the top 10% of the data, and 95% of the data lies between two values of X.
How would I Calculate and do #3 and can it be explained step by step so I can better understnd?
Answer:
See the attached file.
Step-by-step explanation:
see the attached file for explanation.
The points (x, y) represented in the table below lie on a straight line. When the equation of this line is written in the form y = Ax + B, what is the value of A + B?
The equation y = Ax + B represents a straight line. In the given example, 'A' is the slope of the line which is 3, and 'B' is the y-intercept which is 9. Hence, the sum of 'A' and 'B', or A + B, equals 12.
Explanation:The student's question pertains to the equation of a straight line, which in slope-intercept form is y = mx + b, where 'm' represents the slope and 'b' represents the y-intercept. However, the equation mentioned in the question was written as y = Ax + B, where 'A' represents the slope and 'B' represents the y-intercept (equivalent to 'm' and 'b', respectively, in the more common form).
In the instance given, A (the slope) is 3 and B (the y-intercept) is 9. Therefore, when you add the values of A and B together as the question asked, the calculation is 3 + 9. So, A + B equals 12. In essence, the equation represents a line that, for every horizontal movement (x), the vertical movement (y) increases by 3 times the x value, and it intersects the y-axis at 9.
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The calculated value of A + B is -15/4 from the equation y = 3/4x - 9/2
How to determine the calculated value of A + BFrom the question, we have the following parameters that can be used in our computation:
The graph
A linear equation is represented as
y = mx + c
Where
c = y when x = 0
m = slope
Using the above as a guide, we have the following:
m = (-6 + 3)/(-2 - 2)
m = 3/4
So, we have
y = 3/4x + c
Using the points, we have
3/4 * 6 + c = 0
Evaluate
c = -9/2
So, we have
y = 3/4x - 9/2
So, we have
A + B = 3/4 - 9/2
Evaluate
A + B = -15/4
Hence, the value of A + B is -15/4
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Consider the following experiment. Pick a random integer from 1 to 1012 . (a) What is the probability that it is either a perfect square (1, 4, 9, 16, …) or a perfect cube (1, 8, 27, 64,…)? (b) What is the probability that it is either a perfect fourth power (1, 16, 81, 256, …) or a perfect sixth power (1, 64, 729, 4096,…)?
Answer:
a.) 0.0402
b.) 0.00789
Step-by-step explanation:
In between the numbers 1 and 1012,
The number of perfect squares we have is 31, that is all numbers between 1 and 31 inclusive have their squares between the numbers 1 and 1012
The number of perfect cubes we have is 10, that is all numbers between 1 and 10 Inclusive have their perfect cube between the numbers 1 and 1012.
The number of perfect Fourth we have is 5, that is all numbers between 1 and 5 inclusive have their perfect fourth between the numbers 1 and 1012.
The number of perfect Sixth we have is 3, that is all numbers between 1 and 3 inclusive have their perfect sixth between the numbers 1 and 1012.
Hence,
a.) probability of choosing a perfect square = 31/1012
Probability of choosing a perfect cube = 10/1012
Probability of choosing a perfect square or a perfect cube = (31/1012) + (10/1012) - [(31/1012)*(10/1012)]
=41/1012 - 310/1024144
=0.0405 - 0.0003
=0.0402.
b.) Probability of choosing a perfect fourth or a perfect Sixth = (5/1012) + (3/1012) - [(5/1012)*(3/1012)]
= 8/1012 - 15/1024144
= 0.00789.
Mark each statement as True or False?
a) The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle.
b) The cosine rule is used when we are given either a) three sides or b) two sides and the included angle.
Answer:
a.True
b. True
Step-by-step explanation:
a. #Given two angles and one side
Let the given angles be A° and B° and side x, the angle and two remaining sides can be calculated as:
[tex]\frac{a}{sin \ A\textdegree}=\frac{b}{sin \ B\textdegree}=\frac{x}{sin \ (180-A-B)\textdegree}[/tex]
#Given two sides and one non-included angle.
Let the given angle A° and sides x and y, the angle remaining sideand angle can be calculated as:
[tex]\frac{x}{sin \ A\textdegree}=\frac{y}{sin \ B}=\frac{c}{sin \ C\textdegree}[/tex]
b.#Given three sides
Let the given sides be a, b and c:
The angles can be calculated as:
[tex]a^2=b^2+c^2-2bc\ cos \ A\\\\b^2=a^2+c^2-2ac\ cos \ C\\\\c^2=a^2+b^2-2ab\ cos \ B[/tex]
#Given two sides and the included angle
Let the sides given be c, b and the angle A, the remaining sides can be calculated by substituting values in the equation;
[tex]a^2=b^2+c^2-2bc\ cos \ A\\\\[/tex]
Eli, a pretty smart ape in the Salt Lake City zoo, has correctly picked the winner of the last 7 Super Bowls. Read about Eli's prowess at selecting winners here. Question. If Eli is guessing when he chooses a Super Bowl winner (that is, he mentally flips a fair coin to decide which team he chooses as the winner), what is the probability that Eli chooses 7 winners in a row
Answer:
0.0078125
Step-by-step explanation:
If the probability of flipping a fair coin (mentally) correctly for 1 row is 0.5, then the probability of flipping it correctly for 2 rows is
0.5 * 0.5 = 0.25
for 3 rows: 0.5*0.5*0.5 = 0.125
...
For 7 rows:
[tex]0.5*0.5*0.5*0.5*0.5*0.5*0.5 = 0.5^7 = 0.0078125[/tex]
n San Francisco, 30% of workers take public transportation daily. In a sample of 10 workers, what is the probability that exactly three workers take public transportation daily?
Answer:
[tex]X \sim Binom(n=10, p=0.30)[/tex]
And we want to find this probability:
[tex] P(X=3)[/tex]
And using the probability mass function we got:
[tex] P(X=3) = (10C3) (0.3)^3 (1-0.3)^{10-3]= 0.2668[/tex]
Step-by-step explanation:
Previous concepts
A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]
Solution to the problem
Let X the random variable of interest "workers that take public transportation daily", on this case we now that:
[tex]X \sim Binom(n=10, p=0.30)[/tex]
And we want to find this probability:
[tex] P(X=3)[/tex]
And using the probability mass function we got:
[tex] P(X=3) = (10C3) (0.3)^3 (1-0.3)^{10-3]= 0.2668[/tex]
Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following.
0 , x<0
f(x) = ((x^2)/4) , 0 <= x <= 2
1 , 2<= x
Use the cdf to obtain the following. (If necessary, round your answer to four decimal places.)
(a) P(X %u2264 1)
(b) P(0.5 %u2264 X %u2264 1)
(c) P(X > 1.5)
(d) The median checkout duration [solve 0.5 = F(mew)]
(e) Use F'(x) to obtain the density function f(x)
(f) Calculate E(X)
(g) Calculate V(X) and %u03C3x
(h) If the borrower is charged an amount h(X) = X2 when checkout duration is X, compute the expected charge
E[h(X)].
This is a set of solutions for a series of problems related to a given cumulative distribution function of a random variable X. The answers provide ways to calculate various probabilities, the median, the density function, the expected value, the variance and standard deviation, and the expected charge.
Explanation:Given that the cumulative distribution function (cdf) of the random variable X, which denotes the amount of time a book on two-hour reserve is checked out, is f(x) = (x^2)/4 for 0 <= x <= 2.
(a) P(X <= 1) can be obtained by substituting x = 1 into the cdf, which yields (1^2)/4 = 0.25.
(b) P(0.5 <= X <= 1) is the probability that x is between 0.5 and 1. This can be calculated by subtracting P(X <= 0.5) from P(X <= 1), giving (1^2)/4 - (0.5^2)/4 = 0.1875.
(c) P(X > 1.5) can be obtained by subtracting P(X <= 1.5) from 1, giving 1 - (1.5^2)/4 = 0.375.
(d) For the median, we need to solve (m^2)/4 = 0.5, yielding a median of sqrt(2).
(e) The density function f(x) is the derivative of the cdf, and so we need to calculate F'(x). This gives f(x) = x/2.
(f) The expected value E(X) can be calculated as ∫xf(x)dx from 0 to 2, yielding 4/3.
(g) The variance V(X) can be calculated as ∫(x-E(x))^2*f(x)dx from 0 to 2, yielding 4/45, and the standard deviation σx is the square root of the variance, which is 2sqrt(10)/15.
(h) If the borrower is charged an amount h(X) = X^2 when the checkout duration is X, the expected charge E[h(X)] can be calculated as ∫x^2f(x) dx from 0 to 2, yielding 4/5.
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Consider the following random sample from a normal population: 14, 10, 13, 16, 12, 18, 15, and 11. What is the 95% confidence interval for the population variance?
The 95% confidence interval for the population variance of the provided sample is between 2.34 and 24.87. This is calculated using the chi-square distribution, by finding the sample variance and then applying the chi-square formulas for each bound of the confidence interval.
Explanation:The question is asking for a 95% confidence interval for the variance of a normal population given a random sample. We would first utilize the chi-square distribution, specifically the chi-square statistic (χ^2), which measures the discrepancy between the observed data and what we would expect if the null hypothesis were true. The chi-square statistic is calculated as: χ^2 = (n-1) * (sample variance) / (population variance).
In this question, we are being asked to find the confidence interval for the population variance, but we are given a sample. From our sample we calculate the sample variance and then use the chi-square distribution to find the confidence interval for the population variance.
To calculate the sample variance: Add up the squared deviation of each number from the mean and divide by n-1 for the sample variance. For this sample, the variance is 7.67.
For the chi-square distribution, the degrees of freedom (df) will be n-1, in this case, 7. The confidence interval for the variance is (df * s^2 / χ^2_right, df * s^2 / χ^2_left), where χ^2_right and χ^2_left represent the right and left critical values for the chi-square distribution respectively. For a 95% confidence interval with 7 degrees of freedom, the right critical value is 2.167 and the left critical value is 16.013. Substituting in our values, we get the confidence interval to be (7.67*7/16.013, 7.67*7/2.167). This simplifies to (2.34, 24.87).
Therefore, we estimate with 95 percent confidence that the true value of the population variance is between 2.34 and 24.87.
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The 95% confidence interval for the population variance of the given sample is calculated using Chi-Square distribution. The steps include finding the sample mean, sample variance, and then applying the Chi-Square values to get the interval. The final interval is approximately (1.179, 8.703).
To calculate the 95% confidence interval for the population variance from the given sample data (14, 10, 13, 16, 12, 18, 15, and 11), we need to follow these steps:
Calculate the sample mean [tex]\bar{X}[/tex]= 13.625
Calculate the sample variance (s^2):≈ 2.696
Use Chi-Square distribution to find the interval:Lower limit = (df * s²) / χ² (0.975,7)
≈ (7 * 2.696) / 16.012
≈ 1.179
Upper limit = (df * s²) / χ² (0.025,7)
≈ (7 * 2.696) / 2.167
≈ 8.703
Thus, the 95% confidence interval for the population variance is approximately (1.179, 8.703).
Find the equilibrium point for the pair of demand and supply functions. Here q represents the number of units produced, in thousands, and x is the price, in dollars. Demand: qequals=59005900minus−6060x Supply: qequals=400400plus+5050x
Answer:
Q = 2,900 units
x = $50
Step-by-step explanation:
At the equilibrium point the quantity demanded and supplied at the equilibrium price are the same. The demand and supply functions are:
[tex]D: Q=590 - 60x\\S: Q=400+50x[/tex]
When the quantity is the same, the selling price x is:
[tex]5900 - 60x=400+50x\\x=\frac{5,500}{110}\\ x=\$50[/tex]
And the equilibrium quantity is:
[tex]Q = 400+50*50\\Q=2,900\ units[/tex]
According to an article, 47% of adults have experienced a breakup at least once during the last 10 years. Of 9 randomly selected adults, find the probability that the number, X, who have experienced a breakup at least once during the last 10 years is a
Answer:
a) [tex]P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228[/tex]
b) [tex]P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033[/tex]
And replacing we got:
[tex]P(X \geq 1)= 1-P(X<1) = 1-P(X=0)=1-0.0033= 0.9967[/tex]
c) [tex]P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257[/tex]
[tex]P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228[/tex]
[tex]P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135[/tex]
And adding we got:
[tex] P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620 [/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=9, p=0.47)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Assuming the following questions:
a. exactly five
For this case we can use the probability mass function and we got:
[tex]P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228[/tex]
b. at least one
For this case we want this probability:
[tex] P(X \geq 1)[/tex]
And we can use the complement rule and we got:
[tex]P(X \geq 1)= 1-P(X<1) = 1-P(X=0)[/tex]
[tex]P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033[/tex]
And replacing we got:
[tex]P(X \geq 1)= 1-P(X<1) = 1-P(X=0)=1-0.0033= 0.9967[/tex]
c. between four and six, inclusive.
For this case we want this probability:
[tex] P(4 \leq X \leq 6)[/tex]
[tex]P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257[/tex]
[tex]P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228[/tex]
[tex]P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135[/tex]
And adding we got:
[tex] P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620 [/tex]
It seems like your question was cut off at the end. You did not specify the exact number of adults out of the 9 randomly selected ones that have experienced a breakup. However, I can help you understand how to approach this kind of problem generally using the binomial probability formula.
When we randomly select adults and we are interested in those who have experienced a breakup, with a probability of 47% (or 0.47), this can be modeled as a binomial distribution because:
1. There are a fixed number of trials (n = 9).
2. Each trial has only two possible outcomes (experienced a breakup or did not experience a breakup).
3. The probability of experiencing a breakup is the same for each trial (p = 0.47).
4. Each trial is independent of the others.
The probability of exactly x adults (out of 9) having experienced a breakup can be calculated using the binomial probability formula:
P(X = x) = (n choose x) * p^x * (1 - p)^(n - x)
Where:
- "n choose x" is the binomial coefficient C(n, x) = n! / [x!(n - x)!],
- p is the probability of success on any given trial (in this case, 0.47),
- x is the number of successes out of n trials,
- n! denotes the factorial of n,
- x! denotes the factorial of x.
Because you didn't specify the value of x in your question, I can't give you the exact probability. However, if you give me a specific number of adults who have experienced a breakup (x), I can provide the calculation for that scenario.
For now, if you'd like to calculate this probability for a certain number of adults x, you would plug your values for x and n into the formula and calculate accordingly.
Let's say you wanted to find the probability that exactly 4 out of 9 adults have experienced a breakup, for example. You would calculate it as follows:
P(X = 4) = C(9, 4) * (0.47)^4 * (0.53)^(9 - 4)
= 126 * (0.47)^4 * (0.53)^5
C(9, 4) is the number of combinations of 9 things taken 4 at a time, which is 9! / (4!(9 - 4)!) = 126.
You would then calculate (0.47)^4, (0.53)^5, and multiply these by 126 to get the probability.
For other values of x, you would undergo a similar process, using the appropriate value of x in the formula.
If cos B = 3/5 , then what is the positive value of cos 1/2 B , in simplest radical form with a rational denominator?
To find cos(1/2 B) given cos B = 3/5, we can use the double angle identity for cosine. The positive value of cos(1/2 B) in simplest radical form with a rational denominator is (2√5) / 5.
Explanation:If cos B = 3/5, we are looking to find cos(1/2 B) in simplest radical form. To find this, we can use the double angle identity for cosine, which tells us that cos(2θ) = 2cos^2(θ) - 1. If we let θ = 1/2 B, then B = 2θ and the identity becomes cos(B) = 2cos^2(1/2 B) - 1. We can solve for cos(1/2 B) by rearranging into 2cos^2(1/2 B) = cos(B) + 1. Substituting the known value of cos(B) gives us:
2cos^2(1/2 B) = 3/5 + 1 = 8/5
So,
cos^2(1/2 B) = 8/10
cos^2(1/2 B) = 4/5
Taking the positive square root, since cosine is positive in the question:
cos(1/2 B) = √(4/5)
cos(1/2 B) = √4 / √5
cos(1/2 B) = 2 / √sqrt(5)
To rationalize the denominator, we multiply the numerator and denominator by √5:
cos(1/2 B) = (2√5) / 5
Therefore, the positive value of cos(1/2 B), in simplest radical form with a rational denominator, is (2√5) / 5.
The positive value of [tex]$\cos \frac{1}{2}B$[/tex] in simplest radical form with a rational denominator is [tex]$\frac{2}{\sqrt{5}}$[/tex].
To find the value of [tex]$\cos \frac{1}{2}[/tex][tex]B$ given $\cos B[/tex][tex]= \frac{3}{5}$[/tex], we can use the half-angle formula for cosine:
[tex]\[ \cos \frac{1}{2}B = \pm\sqrt{\frac{1 + \cos B}{2}} \][/tex]
Since we are looking for the positive value, we will use the positive square root. Plugging in the given value of [tex]$\cos B$[/tex]:
[tex]\[ \cos \frac{1}{2}B = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{5}{5} + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{8}{5} \cdot \frac{1}{2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} \][/tex]
Now, we can simplify the square root of a fraction by taking the square root of the numerator and the denominator separately:
[tex]\[ \cos \frac{1}{2}B = \frac{\sqrt{4}}{\sqrt{5}} = \frac{2}{\sqrt{5}} \][/tex]
This is the simplest radical form with a rational denominator. Note that we do not need to rationalize the denominator since the question only asks for a rational denominator, not necessarily an integer denominator. The answer[tex]$\frac{2}{\sqrt{5}}$[/tex] is already in the required form."
Suppose you like to keep a jar of change on your desk. Currently, the jar contains the following: 5 Pennies 26 Dimes 18 Nickels 12 Quarters What is the probability that you reach into the jar and randomly grab a quarter and then, without replacement, a penny? Express your answer as a fraction or a decimal number rounded to four decimal places.
Answer:
Step-by-step explanation:
number of pennies = 5
number of dimes = 26
number of nickels = 18
number of quarters = 12
Total = 5 + 26 + 18 + 12 = 61
Probability to take out a quarter = 12 / 61 = 0.1967
Probability to take out a penny without replacement = 5 / 60 = 0.0833
Total probability to take out a quarter and then penny without replacement
= 0.1967 x 0.0833 = 0.0164
The probability of pulling out a quarter and then a penny from the jar without replacement is 0.0131, calculated by multiplying the individual probabilities of each event.
To calculate the probability of pulling out a quarter and then a penny from the jar without replacement, we need to consider the total number of coins and the number of each kind of coin. Initially, the jar contains 5 pennies, 26 dimes, 18 nickels, and 12 quarters, for a total of 61 coins.
First, let's find the probability of picking a quarter. There are 12 quarters out of 61 coins, so the probability is 12/61. After picking a quarter, we do not replace it; thus, there are now 60 coins left in the jar, including only 4 pennies. Now, the probability of picking a penny is 4/60 or 1/15.
We find the combined probability of these two events occurring in sequence by multiplying their probabilities: (12/61) * (1/15).
Probability of picking a quarter then a penny without replacement = (12/61) * (1/15)
= 12/(61 * 15)
= 12/915
= 0.0131 (when rounded to four decimal places)
A poll conducted in 2013 found that 52% of U.S. adult Twitter user get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation justify each of your answers.
a) the data provide statistically significant evidence that more than half U.S. adult twitter users get some news throught Twitter. Use a significance level of alpha = 0.01
b) Since the standard error is 2.4%, we can conclude that 97.6% of all U.S. adult Twitter users were included in the study.
c) If we want to reduce the standard error of estimate, we should collect less data.
d) If we construct a 90% confidence interval for the perventage of U.S. adults Twitter suers who get some news through Twitter, this confidene interval will be wider than a corresponding 99% confidence interval.
Answer:
a) FALSE we obtain a NOT significant results after conduct the hypothesis tes.
b) FALSE, the standard error is not associated to a certain % of people included in the study
c) FALSE. if we want to reduce the standard error we need to increase the sample size or data
d) FALSE, always if we have a higher confidence level the confidence interval associated to this level would be wider than for a lower confidence interval
Step-by-step explanation:
Data given and notation n
n represent the random sample taken
[tex]\hat p=0.52[/tex] estimated proportion of U.S. adult Twitter user get at least some news on Twitter
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.99
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that more than half U.S. adult twitter users get some news throught Twitter:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
The standard error is given:
[tex] SE = \sqrt{\frac{p_o (1-p_o)}{n}}=0.024[/tex]
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.52 -0.5}{0.024}=0.833[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed is [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(z>0.833)=0.2024[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
Now let's idendity the statements
a) FALSE we obtain a NOT significant results after conduct the hypothesis tes.
b) FALSE, the standard error is not associated to a certain % of people included in the study
c) FALSE. if we want to reduce the standard error we need to increase the sample size or data
d) FALSE, always if we have a higher confidence level the confidence interval associated to this level would be wider than for a lower confidence interval
The statement a) is true because the data is statistically significant. Statements b), c) and d) are false because: standard error doesn't reflect the portion of the population included in the study; standard error reduces with more data; and a higher confidence level results in a wider interval, not narrower.
Explanation:a) This statement is true. The data can be considered statistically significant as long as the percentage is not within the standard error range below 50%. In this case, 52% minus 2.4% (standard error) = 49.6% which is below 50%. However, at a significance level of alpha = 0.01, it would be statistically significant since it is more than one standard error away from 50%.
b) This statement is false. The standard error is not an indication of what portion of the population is included in the study. Instead, it's a measure of statistical accuracy of the estimate.
c) This statement is false. The standard error of the estimate would typically decrease as we collect more data. As more data is collected, estimates are usually more accurate.
d) This statement is false. A 90% confidence interval is narrower than a 99% confidence interval. The larger the confidence level, the wider the interval to factor in more uncertainty.
Learn more about Statistical Significance here:https://brainly.com/question/33838241
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verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
I. x2 + xy - y2 = 1, (2,3) (10)
II x2 + y2 = 25, (3,-4)
III. x3y2 = 9, (-1,3)
IV. y2 – 2x – 4y - 1 = 0, (-2, 1)
For each curve, plug in the given point [tex](x,y)[/tex] and check if the equality holds. For example:
(I) (2, 3) does lie on [tex]x^2+xy-y^2=1[/tex] since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.
For part (a), compute the derivative [tex]\frac{\mathrm dy}{\mathrm dx}[/tex], and evaluate it for the given point [tex](x,y)[/tex]. This is the slope of the tangent line at the point. For example:
(I) The derivative is
[tex]x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}[/tex]
so the slope of the tangent at (2, 3) is
[tex]\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74[/tex]
and its equation is then
[tex]y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12[/tex]
For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, [tex]-\frac1{\frac{\mathrm dy}{\mathrm dx}}[/tex]. For example:
(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation
[tex]y-3=-\dfrac47(x-2)\implies y=-\dfrac47x+\dfrac{29}7[/tex]
The points (2,3) and (-1,3) are not on their respective curves, so there are no tangent or normal lines to find. The point (3,-4) lies on the curve. The tangent line at (3,-4) has the equation y = (3/4)x - (25/4) and the normal line has the equation y = (-4/3)x - (8/3). The point (-2,1) also lies on the curve. The tangent line at (-2,1) has the equation y = (-3/2)x - 2 and the normal line has the equation y = (2/3)x + 7/3.
Explanation:Question: Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
I. x2 + xy - y2 = 1, (2,3)
To verify if the given point (2,3) lies on the curve, substitute x=2 and y=3 into the equation:
22 + 2(2)(3) - 32 = 4 + 12 - 9 = 7 ≠ 1
Since the expression on the left side doesn't equal 1, the point (2,3) is not on the curve. Therefore, there are no tangent or normal lines to find.
II. x2 + y2 = 25, (3,-4)
To verify if the given point (3,-4) lies on the curve, substitute x=3 and y=-4 into the equation:
32 + (-4)2 = 9 + 16 = 25
Since the expression on the left side is equal to 25, the point (3,-4) lies on the curve. To find the tangent line, take the derivative of the equation and evaluate it at the point (3,-4). The derivative of the equation is:
2x + 2y(dy/dx) = 0
Substituting x=3 and y=-4 into the derivative equation:
2(3) + 2(-4)(dy/dx) = 0
6 - 8(dy/dx) = 0
-8(dy/dx) = -6
dy/dx = 6/8 = 3/4
The slope of the tangent line at the point (3,-4) is 3/4. Using the point-slope form of a line, the equation of the tangent line is:
y - (-4) = (3/4)(x - 3)
Simplifying the equation:
y + 4 = (3/4)x - (9/4)
y = (3/4)x - (9/4) - (16/4)
y = (3/4)x - (25/4)
To find the normal line, we need to find the negative reciprocal of the slope of the tangent line. The negative reciprocal of 3/4 is -4/3. Using the point-slope form of a line, the equation of the normal line is:
y - (-4) = (-4/3)(x - 3)
Simplifying the equation:
y + 4 = (-4/3)x + (12/3)
y = (-4/3)x + (4/3) - (12/3)
y = (-4/3)x - (8/3)
III. x3y2 = 9, (-1,3)
To verify if the given point (-1,3) lies on the curve, substitute x=-1 and y=3 into the equation:
(-1)3(3)2 = -1(9) = -9 ≠ 9
Since the expression on the left side doesn't equal 9, the point (-1,3) is not on the curve. Therefore, there are no tangent or normal lines to find.
IV. y2 – 2x – 4y - 1 = 0, (-2, 1)
To verify if the given point (-2,1) lies on the curve, substitute x=-2 and y=1 into the equation:
(1)2 – 2(-2) – 4(1) - 1 = 1 + 4 - 4 - 1 = 0
Since the expression on the left side is equal to 0, the point (-2,1) lies on the curve. To find the tangent line, take the derivative of the equation and evaluate it at the point (-2,1). The derivative of the equation is:
-2 - 4y(dy/dx) – 4 = 0
Substituting x=-2 and y=1 into the derivative equation:
-2 - 4(1)(dy/dx) – 4 = 0
-6 - 4(dy/dx) = 0
-4(dy/dx) = 6
dy/dx = 6/-4 = -3/2
The slope of the tangent line at the point (-2,1) is -3/2. Using the point-slope form of a line, the equation of the tangent line is:
y - 1 = (-3/2)(x + 2)
Simplifying the equation:
y - 1 = (-3/2)x - 3
y = (-3/2)x - 3 + 1
y = (-3/2)x - 2
To find the normal line, we need to find the negative reciprocal of the slope of the tangent line. The negative reciprocal of -3/2 is 2/3. Using the point-slope form of a line, the equation of the normal line is:
y - 1 = (2/3)(x + 2)
Simplifying the equation:
y - 1 = (2/3)x + 4/3
y = (2/3)x + 4/3 + 1
y = (2/3)x + 7/3
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A technician plans to test a certain type of resin developed in a laboratory to determine the nature of the time it takes before bonding takes place. It is known that the time to bonding is normally distributed with a mean of 4.5 hours and a standard deviation of 1.5 hours. It will be considered an undesirable product if the bonding time is either less than 2 hours or more than 6 hours. a. What is the probability that the bonding time will be less than 2 hours? b. What is the probability that the bonding time will be more than 6 hours? c. How often would the performance be considered undesirable (in a value of probability)? d. Between what two times (equally before the mean and equally after the mean) accounts for the a drying time of 95%?
Answer:
Step-by-step explanation:
Hello!
The objective is to test if the new type of resin has a bonding time between 2 and 6 hs, less than 2 hs or more than 6 hs makes it undesirable.
The bonding time of resin is normally distributed with mean μ= 4.5 hs and standard deviation δ= 1.5 hs.
a. What is the probability that the bonding time will be less than 2 hours?
Symbolically:
P(X<2)
Using thestandard normal distribution you have to standardize this value to reach the corresponding probability using Z= (X-μ)/δ ~N(0;1)
P(Z<(2-4.5)/1.5)= P(Z<-1.67)= 0.047
b. What is the probability that the bonding time will be more than 6 hours?
P(X>6)= 1 - P(X≤6)
1 - P(Z≤(6-4.5)/1.5)= 1 - P(Z≤1)= 1 - 0.841= 0.159
c. How often would the performance be considered undesirable (in a value of probability)?
The performance will be considered undesirable when the bonding time is less than 2 hs or when the time is higher than 6 hs,
P(X<2) + P(X>6)= 0.047 + 0.159= 0.206
20.6% of the new resin will be considered undesirable.
d. Between what two times (equally before the mean and equally after the mean) accounts for the a drying time of 95%?
I'll start working with the standard normal distribution and then transform the values to X.
Considering that the mean if the standard normal distribution is zero and that the distribution is symmertical regarding it's mean.
You need to look for two values of Z equidistant from the mean that surround 1-α: 0.95 of the distribution, leaving the remaining α: 0.05 outside, equally distributed in two tails α/2: 0.025 (See attachment)
These two value would be the same but with different sign, symbolically:
P(-d≤Z≤d)= 0.95
We can say that the accumulated probability until "d" is
P(Z≤d)= (1-α)+(α/2)
P(Z≤d)= 0.95+0.025= 0.975
And the accumulated until "-d" is:
P(Z≤-d)= α/2
P(Z≤-d)= 0.025
Now you look in the table of the standard normal distribution for the corresponding Z values:
P(Z≤d)= 0.975 ⇒ d= 1.965
P(Z≤-d)= 0.025 ⇒ -d= -1.965
Now for each value you have to use the variable information to reverse the standardization and reach the corresponding bonds of the interval:
Upper bond:
Z= (d-μ)/δ
d= (Z*δ)+μ
d= (1.965*1.5)+4.5
d= 7.45
Lower bond
Z= (-d-μ)/δ
-d= (Z*δ)+μ
-d= (-1.965*1.5)+4.5
-d= 1.55
95% of the bonding time will be between 1.55 and 7.45 hs
I hope it helps!