Answer:
(a) 0.023835
(b) 0.6281
Step-by-step explanation:
(a) The chance someone from this population will test positive is given by the percentage of people who have the virus multiplied by the change of testing positive (1 - false-negative rate) added to the percentage of people who do not have the virus multiplied by the change of testing positive (false-positive rate)
[tex]P(+) = 0.015*(1-0.002)+(1-0.015)*0.009\\P(+) = 0.023835[/tex]
(b) The probability that someone actually has the virus given that they have tested positive is determined as the probability of having the virus and testing positive divided by the probability of testing positive:
[tex]P(V|+) = \frac{ 0.015*(1-0.002)}{0.023835}\\P(V|+) = 0.6281[/tex]
Write a function rule for "The output is four more than the input." Let x represent the input and let y represent the output.
Final answer:
A function rule that states "The output is four more than the input" is expressed as y = x + 4, where x is the input and y is the output.
Explanation:
To write a function rule that describes "The output is four more than the input," we let x represent the input and y represent the output. According to the statement, for any given value of x, the value of y will always be 4 units larger. Therefore, the function rule can be written as y = x + 4.
This means that if you have an input value, simply add 4 to it to get the output value. For example, if the input, x, is 5, the output, y, would be 5 + 4, which equals 9.
How many quarts of water must be added to 3 gallons of soup that is 60% chicken broth to make the soup 40% chicken broth
Answer:
6 quarts
Step-by-step explanation:
60% 3 gallons = 1.8 gallons of broth
water = 0% broth
1.8=1.2+0.4x
0.6=0.4x
x=1.5
1.5 gallons = 6 quarts
The amount of water must be added to 3 gallons of soup which is 60% chicken broth to make the soup 40% chicken broth is 6 quartz.
What is equation?In other terms, it is a mathematical statement stating that "this is equivalent to that." It appears to be a mathematical expression on the left, an equal sign in the center, and a mathematical expression on the right.
Given:
There are 3 gallons of soup that is 60% chicken broth to make the soup 40% chicken broth,
Assume the number of gallons is x then write the equation as shown below,
60% 3 gallons = 1.8 gallons of broth
water = 0% broth
1.8 = 1.2 + 0.4x
0.6 = 0.4x
x = 1.5
As we know that 1 gallon = 4 quartz,
1.5 gallons = 6 quarts
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A distribution for a set of wrist circumferences (measured in centimeters) taken from the right wrist of a random sample of newborn female infants is represented by:______
Answer:
A Histogram will be used to represent the size of right wrist of the random sample of newborn infants.
Step-by-step explanation:
A histogram is the graphical representation of the frequency distribution in the given sample. As the value of circumference can be a positive real number, therefore a Histogram with class boundaries can be formed such that the overall frequency of a wrist size is also visible in the graph.
Also as the distribution will be of continuous nature thus a histogram is a more suitable option as compared to a bar or stem and leaf graph.
A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centimeters and variance 9. Calculate the probability that a component is at least 12 centimeters long.
Final answer:
The probability that a component is at least 12 centimeters long, given that the lengths follow a normal distribution with mean 14 cm and variance 9, is approximately 74.86%.
Explanation:
To calculate the probability that a component is at least 12 centimeters long given that X (the length of a component) follows a normal distribution with mean 14 centimeters and variance 9, we first need to standardize the random variable X to convert it to the standard normal distribution Z.
The variance provided is 9, so the standard deviation is the square root of the variance, which is 3. We standardize using the formula
Z = (X - µ) / σ,
where µ is the mean and
σ is the standard deviation.
For X = 12 centimeters, Z = (12 - 14) / 3 = -2 / 3 ≈ -0.67.
Now, we look up the value of -0.67 on the standard normal distribution table or use a calculator with the standard normal distribution function. Let's denote this value as P(Z < -0.67).
Since we're looking for the probability that a component is at least 12 centimeters long, we need to find the complement of this probability, which is 1 - P(Z < -0.67).
Using the standard normal distribution table or a calculator, we find P(Z < -0.67) ≈ 0.2514.
Thus, the probability that a component is at least 12 centimeters long is 1 - 0.2514 ≈ 0.7486, or approximately 74.86%.
The union of two events A and B is the event that: a) The intersection of A and B does not occur. b) Both A and B occur. c) Either A or B or both occur. d) Either A or B, but not both occur. e) A and B occur at the same time. f) None of the above
Answer:
c) Either A or B or both occur.
Step-by-step explanation:
Suppose that we have two events
Event A
Event B
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a a happens and b does not and [tex]A \cap B[/tex] is the probability that aboth events happen
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
The union of events A and B is:
[tex](A \cup B) = a + b + (A \cap B)[/tex]
Which includes either one of them or both.
So the correct answer is:
c) Either A or B or both occur.
Nicole deposited $4400 in a savings account earning 6% compounded
monthly. If she makes no other deposits or withdrawals, how much will
she have in her account in two years?
$4959.50
$4928.00
$9342.76
$9328.00
Answer:
$4928.00
Step-by-step explanation:
This question is solved by the compound interest formula:
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
In which A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.
In this problem, we have that:
Nicole deposited $4400, so [tex]P = 4400[/tex]
6% compounded monthly, which means that [tex]r = 0.06, n = 12[/tex]
How much will she have in her account in two years?
This is A when [tex]t = 2[/tex].
So
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
[tex]A = 4400(1 + \frac{0.06}{12})^{12*2}[/tex]
[tex]A = 4959.50[/tex]
So the correct answer is:
$4928.00
Historical data for a local manufacturing company show that the average number of defects per product produced is 2. In addition, the number of defects per unit is distributed according to a Poisson distribution. What is the probability that there will be a total of 7 defects on four units
Answer:
The probability that there will be a total of 7 defects on four units is 0.14.
Step-by-step explanation:
A Poisson distribution describes the probability distribution of number of success in a specified time interval.
The probability distribution function for a Poisson distribution is:
[tex]P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...[/tex]
Let X = number of defects in a unit produced.
It is provided that there are, on average, 2 defects per unit produced.
Then in 4 units the number of defects is, [tex](2\times4)=8[/tex].
Compute the probability of exactly 7 defects in 4 units as follows:
[tex]P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14[/tex]
Thus, the probability of exactly 7 defects in 4 units is 0.14.
The number of bats in a colony is growing exponentially. After 2 years, there were 180 bats. After 5 years, there were 1440 bats. If the colony continues to grow at the same rate, how many bats are expected to be in the colony after 9 years
Answer:
23040 bats
Step-by-step explanation:
Let N(t) be the number of bats at time t
We know that exponential function
[tex]y=ab^t[/tex]
According to question
[tex]N(t)=ab^t[/tex]
Where t (in years)
Substitute t=2 and N(2)=180
[tex]180=ab^2[/tex]...(1)
Substitute t=5 and N(5)=1440
[tex]1440=ab^5[/tex]...(2)
Equation (1) divided by equation (2)
[tex]\frac{180}{1440}=\frac{ab^2}{ab^5}=\frac{1}{b^{5-2}}[/tex]
By using the property [tex]a^x\div a^y=a^{x-y}[/tex]
[tex]\frac{1}{8}=\frac{1}{b^3}[/tex]
[tex]b^3=8=2\times 2\times 2=2^3[/tex]
[tex]b=2[/tex]
Substitute the values of b in equation (1)
[tex]180=a(2)^2=4a[/tex]
[tex]a=\frac{180}{4}=45[/tex]
Substitute t=9
[tex]N(9)=45(2)^9=23040 bats[/tex]
Hence, after 9 years the expected bats in the colony=23040 bats
To find the number of bats expected to be in the colony after 9 years, we can use the equation for exponential growth. By plugging in the given population values and solving for the growth rate, we can then calculate the population after 9 years.
Explanation:To find the number of bats expected to be in the colony after 9 years, we need to determine the growth rate. Let's use the equation for exponential growth: N = P * e^(kt), where N is the final population, P is the initial population, e is the base of the natural logarithm, k is the growth rate, and t is the time.
We are given the population after 2 years (P = 180) and after 5 years (P = 1440). Plugging these values into the equation, we can solve for k:
180 = P * e^(2k) and 1440 = P * e^(5k).
Dividing the second equation by the first equation, we can eliminate P and solve for e^(3k): 8 = e^(3k).
Taking the natural logarithm of both sides, we get: ln(8) = 3k.
Finally, solving for k, we have: k = ln(8) / 3.
Now, we can use the calculated value of k to find the population after 9 years:
N = P * e^(9k).
Plugging in the value of P and k, we get: N = 180 * e^(9 * ln(8) / 3). Calculating this expression gives us the expected number of bats in the colony after 9 years.
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All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be proficient in reading, 78% were found to be proficient in mathematics, and 65% were found to be proficient in both reading and mathematics. A student is chosen at random.
What is the probability that the student is proficient in neither reading nor mathematics?
Answer:
There is a 2% probability that the student is proficient in neither reading nor mathematics.
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a student is proficient in reading
B is the probability that a student is proficient in mathematics.
C is the probability that a student is proficient in neither reading nor mathematics.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a student is proficient in reading but not mathematics and [tex]A \cap B[/tex] is the probability that a student is proficient in both reading and mathematics.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So
[tex](A \cup B) + C = 1[/tex]
In which
[tex](A \cup B) = a + b + (A \cap B)[/tex]
65% were found to be proficient in both reading and mathematics.
This means that [tex]A \cap B = 0.65[/tex]
78% were found to be proficient in mathematics
This means that [tex]B = 0.78[/tex]
[tex]B = b + (A \cap B)[/tex]
[tex]0.78 = b + 0.65[/tex]
[tex]b = 0.13[/tex]
85% of the students were found to be proficient in reading
This means that [tex]A = 0.85[/tex]
[tex]A = a + (A \cap B)[/tex]
[tex]0.85 = a + 0.65[/tex]
[tex]a = 0.20[/tex]
Proficient in at least one:
[tex](A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98[/tex]
What is the probability that the student is proficient in neither reading nor mathematics?
[tex](A \cup B) + C = 1[/tex]
[tex]C = 1 - (A \cup B) = 1 - 0.98 = 0.02[/tex]
There is a 2% probability that the student is proficient in neither reading nor mathematics.
Quantity A of an ideal gas is at absolute temperature TTT, and a second quantity B of the same gas is at absolute temperature 2T2T. Heat is added to each gas, and both gases are allowed to expand isothermally.
Answer:
The question is incomplete, here is the complete question ; Quantity A of an ideal gas is at absolute temperature T, and a second quantity B of the same gas is at absolute temperature 2T. Heat is added to each gas, and both gases are allowed to expand isothermally. If both gases undergo the same entropy change, is more heat added to gas A or gas B?
a. More heat is added to gas A
b. More heat is added to gas B
c.The same amount of heat is added to each gas
Option B is the correct answer = more heat is added to gas B
Step-by-step explanation:
Considering dQ = dS/T
dQ(A) = dS/T
dQ(B) = dS/2T
From this, it implies that dQ(B) = dQ(A)/2
and as such, more heat is added to gas B or gas B will undergo the greater entropy change
(1 point) For the equation given below, evaluate ′ at the point (−1,2). (5−)^4+4^3=2433. ′ at (−1,2) =
Answer:
[tex]\dfrac{343}{71}[/tex]
Step-by-step explanation:
Given the equation
[tex](5x-y)^4+4y^3=2433[/tex]
Find the derivative:
[tex]((5x-y)^4+4y^3)'=(2433)'\\ \\4(5x-y)^3\cdot (5x-y)'+4\cdot 3y^2\cdot y'=0\\ \\4(5x-y)^3\cdot (5-y')+12y^2y'=0[/tex]
Substitute
[tex]x=-1\\ \\y=2,[/tex]
then
[tex]4(5\cdot (-1)-2)^3\cdot (5-y')+12\cdot 2^2\cdot y'=0\\ \\4(-5-2)^3(5-y')+48y'=0\\ \\4\cdot (-7)^3\cdot (5-y')+48y'=0\\ \\-1,372(5-y')+48y'=0\\ \\-6,860+1,372y'+48y'=0\\ \\1,420y'=6,860\\ \\y'=\dfrac{6,860}{1,420}=\dfrac{686}{142}=\dfrac{343}{71}[/tex]
Multiple-choice questions each have fourfour possible answers left parenthesis a comma b comma c comma d right parenthesis(a, b, c, d), one of which is correct. Assume that you guess the answers to three such questions. Same question with multiplication rule to find P(WWC) with C as Correct and W as wrong__________.
Answer: 9/64
Step-by-step explanation:
Probability is a chance of prediction. It's a measure of how an event is likely to happen.
P(A) = Number of favorable outcome/Total Number of favorable outcome
Let's make W the correct answer and C the right answer.
The probability of choosing the correct answer from multiple choice question:
P(C) = 1/4
The probability of choosing the wrong answer from multiple choice question:
P(C) =1/4
P(W)= 1 - 1/4
P(W) = 3/4
Therefore, to find P(WWC)
P(WWC) = P(W) × P(W) × P(C)
P(WWC) = 3/4 ×3/4 × 1/4
P(WWC) = 9/64
The probability is 9/64.
What is the mean? If the answer is a decimal, round it to the nearest tenth.
56 47 48 52 62 59 49 56 43 48
Answer:
The mean is 52.
Step-by-step explanation:
The mean is the sum of all elements divided by the number of elements.
In this problem, we have that:
Elements
56 47 48 52 62 59 49 56 43 48
Sum
[tex]56+47+48+52+62+59+49+56+43+48 = 520[/tex]
Number of Elements
10 elements
The mean is
[tex]M = \frac{520}{10} = 52[/tex]
write a function that represents the sequence 7, 14, 21, 28, ...
Answer:
a ₙ = 7n
Step-by-step explanation:
This is an arithmetic sequence, the common difference between each term is 14-7 = 21-14 = 28-21 = 7
to the previous term in the sequence addition of 7 gives the next term.
Arithmetic Sequence:
d = 7
This is the formula of an arithmetic sequence.
a ₙ = a₁ + d(n − 1)
Substitute in the values of
a₁ = 7 and d = 7
a ₙ = 7 + 7(n − 1)
a ₙ = 7 + 7n -7
a ₙ = 7 - 7 +7n = 7n
a ₙ = 7n
Answer:
Step-by-step explanation:
In an arithmetic sequence, consecutive terms differ by a common difference and it is always constant. Looking at the set of numbers,
14 - 7 = 21 - 14 = 28 - 21 = 7
Therefore, it is an arithmetic sequence with a common difference of 7.
The formula for determining the nth term of an arithmetic sequence is expressed as
Tn = a + (n - 1)d
Where
a represents the first term of the sequence.
d represents the common difference.
n represents the number of terms in the sequence.
From the information given,
a = 7
d = 7
The function that represents the sequence would be
Tn = 7 + (n - 1)7
Tn = 7 + 7n - 7
Tn = 7n
You get 3% commission on all sales. This month, you made a sale of $45,050 and a sale of $6,785.25. What is your commission for the month?
Answer:
Your commission for the month is $1,553.35.
Step-by-step explanation:
You made 2 sales.
In each you got a commission of 3%. Your total commission is the sum of both commisions. So
Sale of $45,050:
You got 3% of the sale. So
0.03*45050 = $1,351.5
Sale of $6,728.25:
You got 3% of this sale. So
0.03*6728.25 = $201.85
Total commision for the month:
$1,351.5 + $201.85 = $1,553.35.
Your commission for the month is $1,553.35.
Answer:
1,553.35
The commission for the month is a total of 1,553.35
Kevin Hall is considering an investment that pays 7.70 percent, compounded annually. How much will he have to invest today so that the investment will be worth $30,000 in six years
Answer:
He will have to invest $20,519.84 today.
Step-by-step explanation:
We can solve this question using the simple interest formula:
This is a simple interest problem.
The simple interest formula is given by:
[tex]E = P*I*t[/tex]
In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.
After t years, the total amount of money is:
[tex]T = E + P[/tex].
In this problem, we have that:
[tex]I = 0.077, t = 6, T = 30,000[/tex]
So
[tex]T = E + P[/tex].
[tex]E + P = 30000[/tex]
[tex]E = 30000 - P[/tex]
So
[tex]E = P*I*t[/tex]
[tex]30000 - P= P*0.077*6[/tex]
[tex]30000 - P = 0.462P[/tex]
[tex]1.462P = 30000[/tex]
[tex]P = \frac{30000}{1.462}[/tex]
[tex]P = 20519.84[/tex]
He will have to invest $20,519.84 today.
Kevin Hall needs to invest approximately $19,249.38 today to have $30,000 in six years at an annual interest rate of 7.70%, compounded annually.
To determine how much Kevin Hall should invest today to have $30,000 in six years with a 7.70% annual interest rate, we'll use the formula for present value (PV) of a future amount, which is:
PV = FV / (1 + r)ⁿ
Where:
FV = Future Value = $30,000r = annual interest rate = 7.70% or 0.077n = number of years = 6Plugging in the values:
PV = 30,000 / (1 + 0.077)⁶
Calculating the denominator:
(1 + 0.077)⁶ ≈ 1.5583
Thus,
PV = 30,000 / 1.5583 ≈ $19,249.38
Kevin Hall should invest approximately $19,249.38 today to have $30,000 in six years with a 7.70% annual interest rate, compounded annually.
Among 27 external speakers, there are three defectives. An inspector examines 7 of these speakers.
Find the probability that there are at least 2 defective speakers among the 7
(round off to second decimal place).
The probability of randomly selecting atleast 2 defective speakers from 7 trials is 0.18
The probability of randomly selecting a defective speaker can be calculated thus :
P(defective) = number of defective speakers / total speakers P(defective) = 3 / 27 = 0.1111Using the binomial probability relation :
P(x = x) = nCx * p^x * q^(n-x) Probability of success, p = 0.1111n = number of trials = 7x ≥ 2 q = 1 - p = 1 - 0.1111 = 0.889P(x ≥ 2 ) = P(x = 2)+P(x = 3)+P(x = 4)+P(x = 5)+P(x =6)+P(x = 7)
Using a binomial probability calculator to save time :
P(x ≥ 2 ) = 0.17785
P(x ≥ 2 ) = 0.18 ( 2 decimal places)
Therefore, the probability of selecting atleast 2 defective speakers from 7 is 0.18
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Suppose that if θ = 1, then y has a normal distribution with mean 1 and standard deviation σ, and if θ = 2, then y has a normal distribution with mean 2 and standard deviation σ. Also, suppose Pr(θ = 1) = 0.5 and Pr(θ = 2) = 0.5.
Step-by-step explanation:
We have two cases for Ф,
1. Ф=1; it implies that Pr(Ф=1)=0.5, while y~N(1,α²)
2. Ф=2; it implies that Pr(Ф=2)=0.5, while y~N(2,α²)
Now,
For 1st case of α=2,
We have marginal probability density formula
p(y)=∑p(yIФ)p(Ф)
=p(yIФ=1)p(Ф=1)+p(yIФ=2)p(Ф=2)
=N(yI1,2²)(1/2)+N(yI2,2²)(1/2)
=(1/2)[N(yI1,2²)+N(yI2,2²)]
Now.
For Pr(Ф=1Iy=1) at α=2
We have,
=p(Ф=1Iy=1)
=[p(y=1,Ф=1)]/[p(y=1)]
=[p(y=1IФ=1)p(Ф=1)]/[p(y=1)]
={(1/[tex]\sqrt{2x-2}[/tex])exp[(-1/(2*2²))(1-1)²(1/2)]}/{(1/[tex]\sqrt{2x-2}[/tex])(1/2)[exp[(-1/(2*2²))(1-1)²]+exp[(-1/(2*2²))(1-2)²]}
=0.53 Answer
Now, to describe the changes in shape of Ф when α is increased and decreased:
The formula for posterioir density is p(ФIy)=p(yIФ)p(Ф)/p(y)
=exp[(-1/(2α²)(y-Ф)²]/{exp[(-1/(2α²)(y-1)²]+exp[(-1/(2α²))(y-2)²]}
Now at Ф=1 and solving the equation, we get
p(Ф=1Iy)=1 / {1+exp[(2y-3)/2α²]}
Similarly at Ф=1 and solving the equation, we get
p(Ф=2Iy)=1 / {1+exp[(2y-3)/2α²]}
Conclusion:
α² → ∞ ⇒p(ФIy) → p(Ф) = 1/2
α² → 0 ⇒ two cases
y > 3/2, α² → 0 ⇒p(Ф=2Iy) → 1
y < 3/2, α² → 0 ⇒p(Ф=1Iy) → 1
The value of θ determines the mean of the normal distribution for y, while σ remains constant. The probabilities of θ being 1 or 2 are both 0.5.
The given information states that if θ = 1, then y has a normal distribution with a mean of 1 and standard deviation σ, and if θ = 2, then y has a normal distribution with a mean of 2 and standard deviation σ.
The probabilities of θ being 1 or 2 are both 0.5.
This means that there is a 50% chance of θ being 1, and a 50% chance of θ being 2.
This information allows us to understand how the value of θ affects the distribution of y. When θ is 1, y follows a normal distribution with mean 1 and standard deviation σ.
When θ is 2, y follows a normal distribution with mean 2 and standard deviation σ. The probabilities of these scenarios happening are equal.
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Scores for a common standardized college aptitude test are normally distributed with a mean of 512 and a standard deviation of 106. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the test has no effect
If 1 of the men is randomly selected, find the probability that his score is at least 559.5.
P(X > 559.5) =
If 18 of the men are randomly selected, find the probability that their mean score is at least 559.5.
P(M > 559.5) =
To find the probability of a man's score being at least 559.5 on the standardized college aptitude test, we can calculate the z-score and find the area under the normal distribution curve. The same process applies to finding the probability of the mean score of a sample of 18 men being at least 559.5.
Explanation:To find the probability that a randomly selected man's score is at least 559.5, we need to calculate the z-score for this value and then find the area under the normal distribution curve to the right of that z-score.
To find the probability that the mean score of 18 randomly selected men is at least 559.5, we first need to find the mean and standard deviation of the sample mean. Then, we can calculate the z-score for the given mean score and find the area under the normal distribution curve to the right of that z-score.
P(X > 559.5) = 1 - P(X ≤ 559.5)
P(M > 559.5) = 1 - P(M ≤ 559.5)
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The probabilities of a score being above 559.5 are as follows: for a single randomly selected individual, the probability is approximately 0.3271; for a group of 18 randomly selected individuals, the probability that their mean score is above 559.5 is approximately 0.0287.
Explanation:This is a problem of statistics, more specifically Normal Distribution and Standard Deviation. In a Normal Distribution, the mean (average) is the center of the distribution and standard deviation measures how spread out the scores are from the mean. The Z-Score gives us a measure of how many standard deviations an element is from the mean.
Firstly, to find the probability that a randomly selected man scores at least 559.5, we find the Z-Score using the formula Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. Thus the Z-Score is Z = (559.5 - 512) / 106 = 0.448. From the Z-table or calculator, we find that P(Z > 0.448) ≈ 0.3271. Therefore, P(X > 559.5) = 0.3271.
Secondly, for a sample of 18 men, we use the formula for the standard deviation of a sample mean, σM = σ / sqrt(n), where σ is the standard deviation, and n is the size of the sample. The new standard deviation becomes σM = 106 / sqrt(18) = 25. This gives Z = (559.5 - 512) / 25 =1.90. From the Z-table or calculator, we find that P(Z > 1.90) ≈ 0.0287. Therefore, P(M > 559.5) = 0.0287.
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This year, a small business had a total revenue of $ 62,100 . If this is 15 % more than their total revenue the previous year, what was their total revenue the previous year?
Answer:
Their total revenue the previous year was $54,000.
Step-by-step explanation:
This question can be solved by a simple rule of three.
This year revenue was $62,100. It was 15% more than last year, so 115% = 1.15 of last year. How much was the revenue last year, that is, 100% = 1?
62,100 - 1.15
x - 1
[tex]1.15x = 62100[/tex]
[tex]x = \frac{62100}{1.15}[/tex]
[tex]x = 54000[/tex]
Their total revenue the previous year was $54,000.
Consider a Poisson probability distribution in a process with an average of 3 flaws every 100 feet. Find the probability of 4 flaws in 100 feet.
Answer:
16.80% probability of 4 flaws in 100 feet.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
[tex]e = 2.71828[/tex] is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
An average of 3 flaws every 100 feet.
So [tex]\mu = 3[/tex]
Find the probability of 4 flaws in 100 feet.
This is [tex]P(X = 4)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 4) = \frac{e^{-3}*(3)^{4}}{(4)!} = 0.1680[/tex]
16.80% probability of 4 flaws in 100 feet.
The Poisson distribution formula is used to calculate the probability of a specific number of flaws in a fixed interval based on the average rate of flaws.
Explanation:Poisson Probability Distribution:
Calculate the average rate of flaws: μ = np = 100(.03) = 3.Use the Poisson distribution formula: P(x ≤ 4) ≈ poissoncdf(3, 4) ≈ .8153.The probability of 4 flaws in 100 feet is approximately 0.8153.
(6, -12). (15. -3)
Find the slope
Answer:
The slope is 1.
Step-by-step explanation:
A first order function has the following format
[tex]y = ax + b[/tex]
In which a is the slope
(6, -12).
This means that when [tex]x = 6, y = -12[/tex]
So
[tex]y = ax + b[/tex]
[tex]-12 = 6a + b[/tex]
(15. -3)
This means that when [tex]x = 15, y = -3[/tex]
So
[tex]y = ax + b[/tex]
[tex]-3 = 15a + b[/tex]
We have to solve the following system of equations:
[tex]-12 = 6a + b[/tex]
[tex]-3 = 15a + b[/tex]
We have to find a
In the second equation i will write as:
[tex]b = -3 - 15a[/tex]
Replacing in the first
[tex]-12 = 6a + b[/tex]
[tex]-12 = 6a - 3 - 15a[/tex]
[tex]-9 = -9a[/tex]
[tex]9a = 9[/tex]
[tex]a = \frac{9}{9}[/tex]
[tex]a = 1[/tex]
The slope is 1.
Answer:
1
Step-by-step explanation:
Slope = (y2-y1)/(x2-x1)
= (-3-(-12))/(15-6)
= (-3+12)/(9)
= 9/9
= 1
Solve the equation. StartFraction dy Over dx EndFraction equals5 x Superscript 4 Baseline (1 plus y squared )Superscript three halves An implicit solution in the form F(x,y)equalsC is nothingequalsC, where C is an arbitrary constant.
Answer:
Step-by-step explanation:
To solve the differential equation
dy/dx = 5x^4(1 + y²)^(3/2)
First, separate the variables
dy/(1 + y²)^(3/2) = 5x^4 dx
Now, integrate both sides
To integrate dy/(1 + y²)^(3/2), use the substitution y = tan(u)
dy = (1/cos²u)du
So,
dy/(1 + y²)^(3/2) = [(1/cos²u)/(1 + tan²u)^(3/2)]du
= (1/cos²u)/(1 + (sin²u/cos²u))^(3/2)
Because cos²u + sin²u = 1 (Trigonometric identity),
The equation becomes
[1/(1/cos²u)^(3/2) × 1/cos²u] du
= cos³u/cos²u
= cosu
Integral of cosu = sinu
But y = tanu
Therefore u = arctany
We then have
cos(arctany) = y/√(1 + y²)
Now, the integral of the equation
dy/(1 + y²)^(3/2) = 5x^4 dx
Is
y/√(1 + y²) = x^5 + C
So
y - (x^5 + C)√(1 + y²) = 0
is the required implicit solution
Let X1 and X2 be two random variables following Binomial distribution Bin(n1,p) and Bin(n2,p), respectively. Assume that X1 and X2 are independent.
(a) The mgf of binomial distribution Bin(n, p) is (1 − p + pet)n. Use this fact to obtain the distribution of X1 + X2.
(b) Find the probability P(X1 + X2 = 1|X2 = k) for k = 0 and 1. Then use the law of total probability to find P (X1 + X2 = 1)
Answer:
a) X1+X2 have distribution Bi(n1+n2, p)
b)
P(X1+X2 = 1 | X2 = 0) = np(1-p)ⁿ¹⁻¹
P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹
P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
Step-by-step explanation:
Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).
b)
Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).
If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹
If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹
Thus,
P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
A disease is infecting a colony of 1000 penguins living on a remote island. Let P(t) be the number of sick penguins t days after the outbreak. Suppose that 50 penguins had the disease initially, and suppose that the disease is spreading at a rate proportional to the product of the time elapsed and the number of penguins who do not have the disease.
(a) Give the mathematical model(differential equation and initial condition) for P.
(b) Find the generalsolution of the differential equation in (a).
(c) Find the particular solution that satisfies the initial condition.
Answer:
a. [tex]P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]
b. [tex]C = 950[/tex]
c. [tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]
Step-by-step explanation:
a. Let the number of penguins who have the disease t days after the outbreak be P
Initial number of penguins = 1000
Therefore, current number of penguins = 1000 - P
And the rate of spread of disease according to the statement is
[tex]\frac{dP}{dt}\alpha t(1000-P)\\\frac{dP}{dt}=kt(1000-P)[/tex]
where k is the constant of proportionality
[tex]\frac{dP}{1000-P}=kt.dt[/tex]
Integrating both sides
[tex]-ln(1000-P) = \frac{kt^2}{2}+c\\\frac{1}{(1000-P)} = Ce^{\frac{kt^2}{2} }\\ (1000-P) = Ce^{-\frac{kt^2}{2} }\\P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]
b. Seeing as 50 penguins had the disease initially,
t = 0
P = 50
The general solution of the differential solution becomes
50 = 1000 - C (anything raised to the power of 0 is 1, hence e is equal to 1)
[tex]C = 1000 - 50 = 950[/tex]
c. Therefore, the solution that satisfies the initial condition is
[tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]
suppose that you made four measurement of a speed of a rocket: 12.7 km/s, 13.4 km/s, 12.6 km, and 13.3 km/s. compute: the mean, the standard deviations, and the standard deviation of the mean
the mean speed is [tex]\( 12.75 \)[/tex] km/s, the standard deviation is approximately [tex]\( 0.433 \)[/tex] km/s, and the standard deviation of the mean is approximately [tex]\( 0.217 \)[/tex] km/s.
To compute the mean, standard deviation, and standard deviation of the mean, we'll follow these steps:
1. Calculate the mean [tex](\( \mu \))[/tex]:
[tex]\[ \mu = \frac{\text{sum of all measurements}}{\text{number of measurements}} \][/tex]
2. Calculate the standard deviation [tex](\( \sigma \))[/tex]:
[tex]\[ \sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2}{n}} \][/tex]
3. Calculate the standard deviation of the mean [tex](\( \sigma_\bar{x} \))[/tex]:
[tex]\[ \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}} \][/tex]
Let's plug in the given measurements:
[tex]\[ x_1 = 12.7 \, \text{km/s} \][/tex]
[tex]\[ x_2 = 13.4 \, \text{km/s} \][/tex]
[tex]\[ x_3 = 12.6 \, \text{km/s} \][/tex]
[tex]\[ x_4 = 13.3 \, \text{km/s} \][/tex]
1. Mean (\( \mu \)):
[tex]\[ \mu = \frac{12.7 + 13.4 + 12.6 + 13.3}{4} \][/tex]
[tex]\[ \mu = \frac{51}{4} \][/tex]
[tex]\[ \mu = 12.75 \, \text{km/s} \][/tex]
2. Standard deviation (\( \sigma \)):
[tex]\[ \sigma = \sqrt{\frac{(12.7 - 12.75)^2 + (13.4 - 12.75)^2 + (12.6 - 12.75)^2 + (13.3 - 12.75)^2}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.05^2 + 0.65^2 + (-0.15)^2 + 0.55^2}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.0025 + 0.4225 + 0.0225 + 0.3025}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.75}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{0.1875} \][/tex]
[tex]\[ \sigma \approx 0.433 \, \text{km/s} \][/tex]
3. Standard deviation of the mean (\( \sigma_\bar{x} \)):
[tex]\[ \sigma_\bar{x} = \frac{0.433}{\sqrt{4}} \][/tex]
[tex]\[ \sigma_\bar{x} = \frac{0.433}{2} \][/tex]
[tex]\[ \sigma_\bar{x} \approx 0.217 \, \text{km/s} \][/tex]
So, the mean speed is [tex]\( 12.75 \)[/tex] km/s, the standard deviation is approximately [tex]\( 0.433 \)[/tex] km/s, and the standard deviation of the mean is approximately [tex]\( 0.217 \)[/tex] km/s.
Suppose the demand for X is given by Qxd = 100 - 2PX + 4PY + 10M + 2A, where PX represents the price of good X, PY is the price of good Y, M is income and A is the amount of advertising on good X. Good X is
Answer:
Normal Good
Step-by-step explanation:
A normal good is a good in which a rise in income comes with bigger increases in its quantity demanded. In the demand function, M which is the income is positive and has the highest value.
Therefore Good X is a Normal Good.
The equation represents the demand function for good X. The coefficients of the variables indicate how demand for X is influenced by changes in the price of X itself (PX), the price of a related good (PY), income (M), and advertising (A).
Explanation:The function Qxd = 100 - 2PX + 4PY + 10M + 2A represents the demand function for a particular good, X. PX represents the price of good X, PY the price of a related good (Y), M is income, and A is the amount of advertising on good X. The coefficients of these variables determine how the demand for good X responds to changes in these variables. For example, the demand for good X decreases with an increase in its own price (as indicated by the negative coefficient -2) and increases with an increase in the price of good Y, income, and the amount of advertising (as indicated by positive coefficients).
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Steve has $25,000 to invest and wishes to earn an overall annual rate of return of 8%. His financial advisor recommends that he invest some of the money in a 5-year CD paying 5% per annum and the rest in a corporate bond paying 9% per annum. How much should be placed in each investment in order for Steve to achieve his goal?
Answer:
Steve should place $6,250 in the 5-year CD and $18,750 in the corporate bond
Step-by-step explanation:
System of Equations
We need to find how Steve will distribute his investments between two possible options: one of them will pay 5% per annum and the other will pay 9% per annum. We know Steve has $25,000 to invest and wants to have an overall annual rate of return of 8%.
Let's call x to the amount Steve will invest in the CD paying 5% per annum and y to the amount he will invest in a corporate bond paying 9% per annum.
The total investment is $25,000 which leads to the first equation
[tex]x+y=25,000[/tex]
If x dollars are invested at 5%, then the interest return is 0.05x. Similarly, y dollars at 9% return 0.09y. The overall return is 8% on the total investment, thus
[tex]0.05x+0.09y=0.08(x+y)[/tex]
Rearranging:
[tex]0.05x+0.09y=0.08x+0.08y[/tex]
Simplifying
[tex]0.01y=0.03x[/tex]
Multiplying by 100
[tex]y=3x[/tex]
Substituting in the first equation
[tex]x+3x=25,000\\4x=25,000\\x=6,250[/tex]
And therefore
[tex]y=25,000-6,250=18,750[/tex]
Steve should place $6,250 in the 5-year CD and $18,750 in the corporate bond
Suppose the tank is halfway full of water. The tank has a radius of 2 ft and is 4 ft long. Calculate the force (in lb) on one of the ends due to hydrostatic pressure.
(Assume a density of water rho = 62.4 lb/ft3.)
Answer:
The answer is 332.8 lb
Step-by-step explanation:
See attached picture for the solution
The force on one of the ends due to hydrostatic pressure is 332.8lb
Data;
Density = 62.4 lb/ft^3length = 4ftradius = 2ftForce Due to PressureThe force due to hydrostatic pressure can be calculated as
From the attached diagram;
[tex]F = pressure * area\\density = 62.4 lb/ft^3\\depth of water = 2 - y\\pressure = (2 - y)(62.4)\\pressure = 124.8 - 62.4y\\[/tex]
We can proceed as
[tex]x^2 + (y - 2)^2 = 2^2\\x^2 = 4 - (y - 2)^2\\x = +- \sqrt{4y - y^2}\\[/tex]
this implies that
[tex]2x = 2\sqrt{4y - y^2}[/tex]
The area is given as
[tex]\delta A = (2x)*\delta y\\\delta A = 2\sqrt{4y - y^2 \delta y}[/tex]
The force would be given by
[tex]\delta F = (2-y)(62.4)(2\sqrt{4y - y^2})\delta y[/tex]
The total force is given by
[tex]F = \int\limits^2_0 {(2-y)(62.4)(2\sqrt{4y - y^2}) } \, dy\\F = 124.8\int\limits^2_0 {(2-y)(\sqrt{4y - y^2}) } \, dy\\F = 124.8[-\frac{1}{3}y(y -4)(\sqrt{4y -y^2}]_0^2\\F = 124.8[-\frac{1}{3}(2)(2-4)\sqrt{4(2)-2^2}\\ F = 332.8lb[/tex]
The force on one of the ends due to hydrostatic pressure is 332.8lb
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Bespin Car Rental predicts that the annual probability of one of its cars being destroyed in a crash is 1 in 1,000,000. If destroyed, the value of the property damage to the car equals $45,000. Assume that there are no partial losses; the car is either destroyed in a crash or suffers no loss. A) Show the physical damage loss distribution for Bespin Car Rental’s automobiles and calculate the expected value of the physical damage loss. B) Show the calculations for the variance and the standard deviation.
Answer:
(A) The expected loss is $0.045.
(B) The variance and standard deviation of physical damage loss are $2,025 and $45 respectively.
Step-by-step explanation:
The annual probability of Bespin Car Rental's cars being destroyed is 1 in a million, i.e 0.000001.
It is assumed that the car is either destroyed or there was no loss suffered.
The loss amount in case the car is destroyed is, $45,000.
(A)
The distribution for physical damage loss is displayed in the table below.
The Expected value of physical damage loss is:
[tex]E(X)=\sum xP(X)=(45000\times0.000001)+(0\times0.999999)=0.045[/tex]
Thus, the expected loss is $0.045.
(B)
The variance of a random variable X is: Var (X) = E (X²) - [E (X)]².
The variance of physical damage loss is:
Compute the variance as follows:
[tex]Var(X)=E(X^{2})-[E(X)]^{2}\\=\sum x^{2}P(X)-[\sum xP(X)]^{2}\\=[(45000^{2}\times0.000001)+(0^{2}\times0.999999)]-(0.045)^{2}\\=2025-0.002025\\=2024.997975\\\approx2025[/tex]
The standard deviation of physical damage loss is:
[tex]SD=\sqrt{Var(X)}=\sqrt{2025}=45[/tex]
Thus, the variance and standard deviation of physical damage loss are $2,025 and $45 respectively.