find the horizontal asymptote of the graph of y= -2x^6+5x+8/8x^6+6x+5

a: y= -1/4

b: y=0

c= y=1

D= no horizontal asymptote

NEED HELP ASAP!!!!

Answer:

a) The minimum head breadth that will fit the clientele = 4.105 inches to 3d.p = 4.1 inches to 1 d.p

b) The maximum head breadth that will fit the clientele = 8.905 inches to 3 d.p = 8.9 inches to 1 d.p

Step-by-step explanation:

This is normal distribution problem.

A normal distribution has all the data points symmetrically distributed around the mean in a bell shape.

For this question, mean = xbar = 6.1 inches

Standard deviation = σ = 1 inch

And we want to find the lowermost 2.3% and uppermost 2.3% of the data distribution.

The minimum head breadth that will fit the clientele has a z-score with probability of 2.3% = 0.023

Let that z-score be z'

That is, P(z ≤ z') = 0.023

Using the table to obtain the value of z'

z' = - 1.995

P(z ≤ - 1.995) = 0.023

But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.

z' = (x - xbar)/σ

- 1.995 = (x - 6.1)/1

x = -1.995 + 6.1 = 4.105 inches

The maximum head breadth that will fit the clientele has a z-score with probability of 2.3% also = 0.023

Let that z-score be z''

That is, P(z ≥ z'') = 0.023

Using the table to obtain the value of z''

P(z ≥ z") = P(z ≤ -z")

- z'' = - 1.995

z" = 1.995

P(z ≥ 1.995) = 0.023

But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.

z'' = (x - xbar)/σ

1.995 = (x - 6.1)/1

x = 1.995 + 6.1 = 8.905 inches

Using z-scores for the smallest and largest 2.3% of the normal distribution, and a formula that accounts for the mean and standard deviation of head breadths, we can calculate that the helmets need to accommodate head breadths between 4.1 inches (minimum) and 8.1 inches (maximum).

To determine the minimum and maximum head breadths that will fit the clientele, we need to find the z-scores that correspond to the smallest 2.3% and the largest 2.3% of the normal distribution. Then we can use these z-scores to calculate the specific head breadths.

First, we find the z-scores using a standard z-table or a calculator since the normal distribution table typically provides the area to the left of a z-score. For the smallest 2.3%, we need the z-score that has 0.023 to its left. This value is approximately z = -2. Conversely, for the largest 2.3%, since the normal distribution is symmetric, the z-score will have the same absolute value but be positive, which would be z = 2. However, to be precise, one should use statistical tables or software to find the exact z-scores close to these values.

To convert the z-scores to specific head breadths, we use the formula:

X = μ + (z * σ)

where X is the head breadth, μ is the mean, and σ is the standard deviation.

The minimum head breadth (Xmin) is calculated as follows:

The maximum head breadth (Xmax) is calculated as follows:

Therefore, the minimum head breadth that will fit the clientele is 4.1 inches, and the maximum head breadth is 8.1 inches.

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.

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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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]

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!

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.

From 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 function that determines the rock's height above the road in terms of the number of seconds since it was thrown is f(t) = -16t² + 24.32t + 57. This quadratic function accounts for the initial upward velocity and the constant downward acceleration due to gravity.

To determine the rock's height above the road as a function of time after it was thrown, we can use the standard equation of motion under constant acceleration due to gravity, which in this case is downward. The formula for the height f(t) of the rock at time t seconds after it is thrown is:

f(t) = -16t² + vt + s

where:

Since the rock reaches its maximum height 0.76 seconds after it is thrown, the initial velocity v can be calculated using the fact that the velocity at the peak height is 0 (the rock stops moving upward for an instant before descending).

At maximum height, the velocity v had decreased by 32 feet/second for 0.76 seconds, which is v - 32(0.76) = 0. Solving for v, we find that v = 32(0.76).

The rock contacts the road 3.15 seconds after it was thrown, meaning it falls 57 feet in that time. Plugging these values into the formula:

f(t) = -16t² + 32(0.76)t + 57

which simplifies to:

f(t) = -16t² + 24.32t + 57

This function f(t) gives the rock's height in feet above the road at any time t measured in seconds since it was thrown, assuming negligible air resistance.

The correct function for the rock's height above the road in terms of the number of seconds [tex]\( t \)[/tex] since the rock was thrown is: [tex]\[ f(t) = -16t^2 + v_0t + 57 \][/tex] where [tex]\( v_0 \)[/tex] is the initial velocity of the rock in feet per second.

To derive this function, we use the kinematic equation for the vertical motion of an object under constant acceleration due to gravity, which is:

[tex]\[ h(t) = h_0 + v_0t - \frac{1}{2}gt^2 \][/tex]

Given that the bridge is 57 feet above the road, [tex]\( h_0 = 57 \)[/tex] feet. The acceleration due to gravity is [tex]\( g = 32 \)[/tex] feet per second squared. Plugging these values into the kinematic equation, we get:

[tex]\[ h(t) = 57 + v_0t - \frac{1}{2}(32)t^2 \][/tex]

Simplifying the equation by multiplying [tex]\( \frac{1}{2} \)[/tex] with 32 gives us:

[tex]\[ h(t) = 57 + v_0t - 16t^2 \][/tex]

This simplifies to the function [tex]\( f(t) \)[/tex] as provided above. Note that [tex]\( v_0 \)[/tex] is still unknown and will need to be determined using additional information given in the problem, such as the time it takes for the rock to reach its maximum height.

To find [tex]\( v_0 \)[/tex], we can use the fact that at the maximum height, the velocity of the rock is zero. The kinematic equation relating velocity and time under constant acceleration is:

[tex]\[ v(t) = v_0 - gt \][/tex]

Setting [tex]\( v(t) = 0 \)[/tex] at the maximum height and solving for [tex]\( v_0 \)[/tex] gives us:

[tex]\[ 0 = v_0 - gt_{\text{max}} \][/tex]

[tex]\[ v_0 = gt_{\text{max}} \][/tex]

Given that the rock reaches its maximum height 0.76 seconds after it is thrown, [tex]\( t_{\text{max}} = 0.76 \)[/tex] seconds. Plugging this into the equation for [tex]\( v_0 \)[/tex], we get:

[tex]\[ v_0 = 32 \times 0.76 \][/tex]

[tex]\[ v_0 = 24.32 \][/tex]

Now we can write the complete function with the known values:

[tex]\[ f(t) = -16t^2 + 24.32t + 57 \][/tex]

This function represents the height of the rock above the road as a function of time since it was thrown.

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.

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.

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.

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

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)

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.

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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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

Answer:

Probability that the weight of a randomly selected steer is between 939 and 1417 lbs is 0.68389 .

Step-by-step explanation:

We are given that the weights of steers in a herd are distributed normally. The standard deviation is 200 lbs and the mean steer weight is 1300 lbs.

So, Let X = weights of steers in a herd ,i.e.; X ~ N([tex]\mu,\sigma^{2}[/tex])

Here, [tex]\mu[/tex] = population mean = 1300 lbs

         [tex]\sigma[/tex] = population standard deviation = 200 lbs

The z score area distribution is given by;

            Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

So, Probability that the weight of a randomly selected steer is between 939 and 1417 lbs = P(939 lbs < X < 1417 lbs)

P(939 lbs < X < 1417 lbs) = P(X < 1417) - P(X <= 939)

P(X < 1417) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{1417-1300}{200}[/tex] ) = P(Z < 0.58) = 0.71904

P(X <= 939) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{939-1300}{200}[/tex] ) = P(Z <= -1.81) = 1 - P(Z < 1.81)

                                                       = 1 - 0.96485 = 0.03515

Therefore,  P(939 lbs < X < 1417 lbs) = 0.71904 - 0.03515 = 0.68389 .

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

SIMILARLY,

The value of $1 invested in each share class if investment horizon is 20 years

Answer:

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.

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

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.

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]

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]

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]

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.

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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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]

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.

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.

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.

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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.

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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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.

Answer:

See the attached file.

Step-by-step explanation:

see the attached file for explanation.

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.

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

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.

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."

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.