Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.) {1/5, 1/2, 1/6, 1/3, 1/7, 1/4, 1/8, 1/5,...}

The equation for translation of the graph given is y = |x/2-2| + 4

Given is a function y = |x/2-2| + 3, it is translated one unit up, we need to find the translated function.

We know that, f(x) + k moves f(x) by k units up.

So, y = |x/2-2| + 3 → y = |x/2-2| + 3 + 1

y = |x/2-2| + 4

Hence, the equation for translation of the graph given is y = |x/2-2| + 4

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To formulate this problem as a linear programming problem, we need to identify the decision variables, objective functions, and constraints.

Decision Variables:
Let x be the number of model A printers manufactured per month, and y be the number of model B printers manufactured per month.

Objective Function:
The objective is to maximize monthly profits, which can be expressed as P = 25x + 40y.

Constraints:
1. The total number of printers demanded per month cannot exceed 3000, so we have the constraint x + y ≤ 3000.
2. The company has earmarked not more than $600,000/month for manufacturing costs, so the cost constraint is 120x + 140y ≤ 600,000.
3. The number of printers manufactured must be non-negative, so x ≥ 0 and y ≥ 0.

Therefore, the linear programming problem is:
Maximize P = 25x + 40y
Subject to:
x + y ≤ 3000
120x + 140y ≤ 600,000
x ≥ 0, y ≥ 0

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The 90% CI for the change in blood pressure is (-5.843, 7.977)., the 99.9% CI for the change in blood pressure is (-14.077, 16.211).

a) To calculate a 90% confidence interval (CI) for the change in blood pressure, we use the formula:

CI = x ± t(α/2, df) * (s/√n)

where x is the sample mean of the change in blood pressure, t(α/2, df) is the t-value for the desired confidence level and degrees of freedom, s is the sample standard deviation of the change in blood pressure, and n is the sample size.

Using the given data, we have:

x = 1.067
s = 26.691
n = 15
df = n - 1 = 14

From the t-distribution table, the t-value for a 90% confidence level and 14 degrees of freedom is 1.761.

Plugging in the values, we get:

CI = 1.067 ± 1.761 * (26.691/√15) = (-5.843, 7.977)

Therefore, the 90% CI for the change in blood pressure is (-5.843, 7.977).

b) To calculate a 99.9% CI for the change in blood pressure, we use the same formula but with a different t-value:

t(α/2, df) = 3.922 for a 99.9% confidence level and 14 degrees of freedom.

Plugging in the values, we get:

CI = 1.067 ± 3.922 * (26.691/√15) = (-14.077, 16.211)

Therefore, the 99.9% CI for the change in blood pressure is (-14.077, 16.211).

c) The interval calculated in part (a) does include 0, while the interval calculated in part (b) does not include 0. This is important because if the interval includes 0, it means we cannot conclude that there is a significant difference in blood pressure before and after the medication. On the other hand, if the interval does not include 0, it means we can be confident that there is a significant difference.

d) To conduct a one sample t-test with p = 0 and a = 10, we first calculate the t-statistic using the formula:

t = (x - p) / (s/√n)

where x is the sample mean of the change in blood pressure, s is the sample standard deviation of the change in blood pressure, and n is the sample size.

Plugging in the values, we get:

t = (1.067 - 0) / (26.691/√15) = 0.444

From the t-distribution table, the t-value for a one-tailed test with 14 degrees of freedom and a significance level of 10% is 1.345.

Since our calculated t-value (0.444) is less than the critical t-value (1.345), we fail to reject the null hypothesis that there is no significant difference in blood pressure before and after the medication. Therefore, the results are not consistent with the 90% CI calculated in part (a).

e) To conduct a one sample t-test with y = 0 and a = 0.001, we use the same formula as in part (d):

t = (x - y) / (s/√n)

Plugging in the values, we get:

t = (1.067 - 0) / (26.691/√15) = 0.444

From the t-distribution table, the t-value for a one-tailed test with 14 degrees of freedom and a significance level of 0.001 is 3.746.

Since our calculated t-value (0.444) is less than the critical t-value (3.746), we fail to reject the null hypothesis that there is no significant difference in blood pressure before and after the medication. Therefore, the results are not consistent with the 99.9% CI calculated in part (b).

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A function showing the value of the account after t years is [tex]f(t) = 690(1.021)^{t}[/tex]

The percentage of growth per year (APY) is 21.22%.

In Mathematics and Financial accounting, continuous compounding interest can be determined or calculated by using this mathematical equation (formula):

[tex]f(t) = P_{0}e^{rt}[/tex]

Where:

Based on the information provided above, we can reasonably infer and logically deduce that the function for the future value after t years is given by;

[tex]f(t) = 690(1 + 0.021)^{t}\\\\f(t) = 690(1.021)^{t}[/tex]

Growth per year (APY) =  [tex]e^{r} =e^{0.021}[/tex]

Growth per year (APY) = 1.02122 - 1 = 0.2122 = 21.22%

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Answer: Let X be the random variable denoting the loss due to burglary. Then, we know that X is exponentially distributed with mean 100, which implies that the probability density function of X is given by:

f(x) = (1/100) * exp(-x/100) for x ≥ 0

The probability that a loss is between 40 and 50 can be expressed as:

P(40 ≤ X ≤ 50) = ∫40^50 f(x) dx

= ∫40^50 (1/100) * exp(-x/100) dx

= [-exp(-x/100)]40^50

= exp(-2/5) - exp(-1/2)

Similarly, the probability that a loss is between 60 and r can be expressed as:

P(60 ≤ X ≤ r) = ∫60^r f(x) dx

= ∫60^r (1/100) * exp(-x/100) dx

= [-exp(-x/100)]60^r

= exp(-3/5) - exp(-r/100)

Given that the above two probabilities are equal, we have:

exp(-2/5) - exp(-1/2) = exp(-3/5) - exp(-r/100)

Solving for r, we get:

r = -100 * ln [exp(-2/5) - exp(-1/2) + exp(-3/5)]

r ≈ 85.863

Therefore, the value of r is approximately 85.863.

Using a numerical solver, we can find the value of r: r ≈104.59. So, the value of r is approximately 67.35. To solve this problem, we can use the fact that the exponential distribution is memoryless, which means that the probability of a loss being between two values is the same no matter how much time has elapsed since the last loss.

Let X be the amount of the loss due to burglary. We know that X is exponentially distributed with mean 100, which implies that its probability density function (pdf) is given by f(x) = (1/100) * e^(-x/100) for x > 0.

The probability that a loss is between 40 and 50 is given by the integral of f(x) over that interval:

P(40 < X < 50) = integral from 40 to 50 of f(x) dx
               = integral from 40 to 50 of (1/100) * e^(-x/100) dx
               = e^(-40/100) - e^(-50/100)
               = 0.3679 - 0.3679 * e^(-1/10)
               = 0.0511 (rounded to four decimal places)

Now, let's find r such that the probability that a loss is between 60 and r is also 0.0511. We can set up the following equation:

P(60 < X < r) = integral from 60 to r of f(x) dx
              = e^(-60/100) - e^(-r/100)
              = 0.0511

Solving for r, we get:

e^(-r/100) = e^(-60/100) - 0.0511
r = -100 * ln(e^(-60/100) - 0.0511)
 = 104.59 (rounded to two decimal places)

Therefore, r is approximately 104.59.

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$138.98 was the list price of the sneakers.

Let's use "x" to represent the trainers' list price. Given that the sales tax rate was 7%,

the tax amount corresponded to 0.07x, or 7% of the list price.

The total cost including tax was $148.61, so we can set up the following equation:

x + 0.07x = 148.61

Simplifying and solving for x, we get:

1.07x = 148.61

x = 138.98

Therefore, the list price of the sneakers was $138.98.

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Based on the information given, we know that 14% of the sample of 1000 adults reported meeting the daily serving recommendations for fruits and vegetables set by the CDC. To construct a 90% confidence interval, we need to calculate the margin of error.

We can use the formula:

[tex]Margin of error = z* (sqrt(p*(1-p)/n))[/tex]

Where:

z* is the critical value for a 90% confidence interval, which is 1.645
p is the proportion of the sample that met the CDC recommendations, which is 0.14
n is the sample size, which is 1000

Plugging in the values, we get:

Margin of error = 1.645 * (sqrt(0.14*(1-0.14)/1000))
Margin of error ≈ 0.022

Therefore, the answer is e. 0.022.

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It is concluded that TU ≅ UV is by the definition of congruence as shown in the solution part.

As per the given figure, the required proof would be as:

Statements:

1. U is the midpoint of SW.

2. T is the midpoint of SU.

3. V is the midpoint of UW.

4. TU || WV (by midpoint theorem)

5. TV || UW (by midpoint theorem)

6. UT = TV (by midpoint theorem)

7. UV = 2VT (by midpoint theorem)

Reasons:

1. Given.

2. Given.

3. Given.

4. Midpoint theorem.

5. Midpoint theorem.

6. Midpoint theorem.

7. Midpoint theorem.

Therefore, TU ≅ UV by the definition of congruence.

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t = (sample mean - hypothesized mean) / (standard deviation / square root of sample size)
Substituting the values given in the question, we get:
t = (6.2 - 6) / (0.9 / sqrt(160))
t = 5.46
Since the p-value is less than the significance level of 0.05, we can reject the null hypothesis and conclude that there is sufficient evidence at the 0.05 level that the valve performs above the specifications.

To determine if there is sufficient evidence at the 0.05 level that the valve performs above the specifications, we will perform a hypothesis test using the given information.

Step 1: State the null and alternative hypotheses.
Null hypothesis (H0): The mean water pressure is equal to 6 lbs/square inch (µ = 6).
Alternative hypothesis (H1): The mean water pressure is greater than 6 lbs/square inch (µ > 6).
The null hypothesis (H0) is that the mean pressure produced by the valve is 6 lbs/square inch, and the alternative hypothesis (Ha) is that the mean pressure produced by the valve is greater than 6 lbs/square inch.
We can use a one-sample t-test to test this hypothesis.
The test statistic is calculated as:
t = (sample mean - hypothesized mean) / (standard deviation / square root of sample size)

Step 2: Calculate the test statistic.
The test statistic (z) = (sample mean - population mean) / (standard deviation/sqrt (sample size))
z = (6.2 - 6) / (0.9 / sqrt(160))
z ≈ 3.11

Step 3: Determine the critical value and make a decision.
Using a t-distribution table with 159 degrees of freedom (160-1), we find that the probability of getting a t-value of 5.46 or greater is very small, less than 0.0001.

Since this is a one-tailed test at the 0.05 significance level, we will compare the test statistic (z) to the critical value from the z-table. The critical value for a one-tailed test at a 0.05 significance level is 1.645.

Since the test statistic (3.11) is greater than the critical value (1.645), we reject the null hypothesis in favor of the alternative hypothesis.

Conclusion: There is sufficient evidence at the 0.05 level that the valve performs above the specifications, as the mean water pressure is greater than 6 lbs/square inch.

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The Chain Rule to find dz/ds and Oz/ot is

[tex]Oz/ot = dz/dt = -2s^2 cos(t) sin(t) = -2s^2 cos(t) * (y/x) = -2s^2 y sin(t) / x[/tex]

Thus, [tex]Oz/ot = -2s^2 y sin(t) / x.[/tex]

The chain rule is a rule in calculus that describes how to find the derivative of a composite function. If we have a function that is made up of two or more functions, the chain rule tells us how to find the derivative of the composite function.

To find dz/ds, we need to apply the Chain Rule, which states that if z is a function of u and u is a function of s, then:

dz/ds = dz/du * du/ds

In this case, we have:

[tex]z = x^2 * y^0 = x^2 = (s cos(t))^2 = s^2 cos^2(t)[/tex]

x = s cos(t)

y = s sin(t)

So, we can rewrite z as:

[tex]z = s^2 cos^2(t)[/tex]

Now, let's find du/ds and dz/du:

du/ds = d/ds (s cos(t)) = cos(t)

[tex]dz/du = d/ds (s^2 cos^2(t)) = 2s cos^2(t)[/tex]

Using the Chain Rule formula, we can now find dz/ds:

[tex]dz/ds = dz/du * du/ds = 2s cos^2(t) * cos(t) = 2s cos^3(t)[/tex]

Therefore,[tex]dz/ds = 2s cos^3(t).[/tex]

To find Oz/ot, we need to find dz/dt using the Chain Rule, and then substitute for z in terms of x and y:

dz/dt = dz/dx * dx/dt + dz/dy * dy/dt

[tex]dz/dx = 2xy^0 = 2x = 2s cos(t)[/tex]

[tex]dz/dy = x^2 * 0y^-1 = 0[/tex]

dx/dt = -s sin(t)

dy/dt = s cos(t)

Substituting these values in the Chain Rule formula, we get:

dz/dt = 2s cos(t) * (-s sin(t)) + 0 * (s cos(t)) = -2s^2 cos(t) sin(t)

Now, we can substitute for z in terms of x and y:

[tex]z = x^2 * y^0 = x^2 = (s cos(t))^2 = s^2 cos^2(t)[/tex]

Therefore, we have:

[tex]Oz/ot = dz/dt = -2s^2 cos(t) sin(t) = -2s^2 cos(t) * (y/x) = -2s^2 y sin(t) / x[/tex]

Thus, Oz/ot = -2s^2 y sin(t) / x.

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The diameter of the cone is 40 in. Option D

We can see from the diagram shown, that the shape inside the cone is a triangle.

Thus, using the Pythagorean theorem which states that the square of the longest side which is the hypotenuse is equal to the sum of the squares of the other two sides.

We then have that;

36² = 30² + r²

find the squares

1296 = 900+ r²

collect like terms

r² = 396

Find the square root of the sides

r = 19. 89 in

Then,

Diameter = 2(radius)

Substitute

Diameter = 39. 79in

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The inequality that represents the given condition is "p > 12".

Let p be the amount of cheese produced in kg.

Then, the amount of yogurt produced is (p + 24) kg (since it is 24 kg more than cheese). And the amount of ice cream produced is 3p kg (since it is 3 times as much as cheese).

Now, we need to find the possible values of p that satisfy the condition "the dairy farm produced more kilograms of ice cream than yogurt last week." In other words, we need to compare the amount of ice cream produced (3p) with the amount of yogurt produced (p + 24) and ensure that the amount of ice cream is greater.

So, we can write the following inequality:

3p > p + 24

Simplifying this inequality, we get:

2p > 24

Dividing both sides by 2, we get:

p > 12

Therefore, the possible values of p that satisfy the given condition are all values of p greater than 12.

In summary, the inequality that represents the given condition is "p > 12".

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In this scenario, we have a uniformly charged thin rod extending along the x-axis from the origin (x = 0) to positive infinity (x = +∞).

The term "uniformly charged" means that the charge is distributed evenly throughout the entire length of the rod.

To analyze this situation, we can consider the following steps: 1. Determine the linear charge density (λ) of the rod. Since the rod is uniformly charged, λ remains constant along its entire length. λ is usually given in units of charge per length (e.g., coulombs per meter).

2. To find the electric field at a particular point along or outside the rod, we can break the rod into infinitesimally small segments (dx) and consider the contribution of the electric field (dE) from each of these segments.

3. Calculate the electric field (dE) produced by each segment at the desired point using Coulomb's equations , considering the linear charge density (λ) and distance between the segment and the point.

4. Integrate the electric field contributions (dE) from all segments along the entire length of the rod (from x = 0 to x = +∞) to find the total electric field (E) at the point of interest.

By following these steps, you can analyze the electric field and related properties of a uniformly charged thin rod extending along the x-axis from x = 0 to x = +∞.

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The specified secants, [tex]\overline{TUV}[/tex] and [tex]\overline{XWV}[/tex], evaluated using the intersecting secant theorem indicates that the length of the segment TU is 21 units

The intersecting secant theorem states where two secants have the same endpoint external to or outside a circle, then the product of a secant and its external segment is equivalent to the product of the other secant and its external segment.

The specified segment lengths are;

UV = 25

WV = 22

TU = XW - 9

The intersecting secant theorem indicates that in the circle, we get;

[tex]\overline{TUV}[/tex]× UV = [tex]\overline{XWV}[/tex] × WV

[tex]\overline{TUV}[/tex] = TU + UV

[tex]\overline{XWV}[/tex] = XW + WV

UV = 25, WV = 22

TU = XW - 9

Therefore;

[tex]\overline{TUV}[/tex] = XW - 9 + 25 = XW + 16

[tex]\overline{XWV}[/tex] = XW + WV = XW + 22

Let x represent XW, we get;

[tex]\overline{TUV}[/tex]× UV = [tex]\overline{XWV}[/tex] + WV

(x + 16) × 25 = (x + 22) × 22

25·x + 400 = 22·x + 484

25·x - 22·x = 484 - 400 = 84

3·x = 84

x = 84/3 = 28

XW = 28, therefore;

TU = 28 - 9 = 21

TU = 21

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The answer is 64.  To find the number of homozygous dominant individuals in the population.

We need to use the Hardy-Weinberg equation:

p^2 + 2pq + q^2 = 1

Where:
p = frequency of dominant allele
q = frequency of recessive allele

Since the dominant allele frequency is 0.8, we can assume that the recessive allele frequency is 0.2 (since p + q = 1).

To find the frequency of homozygous dominant individuals (p^2), we simply square the frequency of the dominant allele:

p^2 = (0.8)^2 = 0.64

To find the number of homozygous dominant individuals, we multiply the frequency by the total population size:

0.64 x 100 = 64

Therefore, there are 64 homozygous dominant individuals in the population.

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The slope and the y intercepts of the given graph which shows height of the candlesticks over time is :

Slope = -2/3

Y intercept = 9

The graph is that of the height of the candlesticks over time.

The line passes through two points (0, 9) and (3, 7).

Slope of the line can be calculated as,

Slope = (7 - 9) / (3 - 0) = -2/3

Hence the slope id negative two thirds.

y intercept of a graph is the y coordinate of the point where the line touches the Y axis.

The x coordinate will be 0 there.

The line passes through (0, 9).

So y intercept = 9

Hence the slope and the y intercept are -2/3 and 9 respectively.

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The statement that is NOT true in a Chi-Square test is "Small values of the chi squared test statistic would lead to a decision to reject the null hypothesis."

This is because if the test statistic is small, it means that the observed values are close to the expected values, and there is no significant difference between the populations. Therefore, a small test statistic would lead to a failure to reject the null hypothesis. In a Chi-Square test, we compare the proportions of specified characteristics in different populations, and we wish to determine whether they are the same or not. If the test statistic is large, it means that the observed values are significantly different from the expected values, and we have evidence to reject the null hypothesis. Finally, the P-value will be small if the test statistic is large, indicating strong evidence against the null hypothesis.

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Answer:

x² + z² + 2xz - y²

Step-by-step explanation:

(x - y + z) (x + y + z)

Rearranging to make it an identity,

=> (x + z + y) (x + z - y)

=> ((x + z) + y) ((x-z) - y)

We know the identity,

(a + b)(a-b) = a² - b²

=> (x + z)² - y²

We know the identity,

(a + b)² = a² + b² + 2ab

Expanding,

=> x² + z² + 2xz - y²

The grocer's selection and weighing of the apples is the experiment, while the specific result of the experiment - an average weight of 165 grams - is the outcome.

An event is a combination of one or more outcomes, such as an average weight between 150 and 165 grams. Finally, the average weight of five apples is the numerical representation of an outcome, which is known as the random variable.
Hi! I'd be happy to help explain these terms in the context of your question:

1. Experiment: Selecting and weighing the apples. This is the activity undertaken to gather data.
2. Outcome: The 165 grams average weight. This is a specific result of the experiment.
3. Event: An average weight between 150 and 165 grams. An event is a combination of one or more outcomes.
4. Random variable: The average weight of five apples. This is a numerical representation of an outcome.

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(a) The probability that at least one individual with a reservation cannot be accommodated on the trip is 0.4202.

(b) The expected number of available places when the limousine departs is 0.338.

Let the random variable Y represent the number of passenger reserving the trip shows up.

The probability of the random variable Y is, p = 0.70.

Success in this case an be defined as the number of passengers who show up for the trip.

The random variable Y follows a Binomial distribution with probability of success as 0.70.

(a)

It is provided that n = 6 reservations are made.

Compute the probability that at least one individual with a reservation cannot be accommodated on the trip as follows:

P (At least one individual cannot be accommodated) = P (X = 5) + P (X = 6)

= 0.4202

Thus, the probability that at least one individual with a reservation cannot be accommodated on the trip is 0.4202.

The expected number of available places when the limousine departs is 0.338.

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So the vector form of the quadratic function y = x² - 6x + 7 is: y = (x - 3)² - 2.

To change from standard form to vertex form, we need to complete the square.

First, we group the x-terms together and factor out any common coefficient of x², giving:

y = x² - 6x + 7

y = 1(x² - 6x) + 7

Next, we need to add and subtract a constant inside the parentheses to complete the square. To determine this constant, we take half of the coefficient of x (-6) and square it:

(-6/2)² = 9

So we add and subtract 9 inside the parentheses:

y = 1(x² - 6x + 9 - 9) + 7

Now we can factor the quadratic expression inside the parentheses as a perfect square:

y = 1[(x - 3)² - 9] + 7

Simplifying and rearranging terms, we get:

y = (x - 3)² - 2

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The degrees of freedom and the critical t value (t*) for the given sample size and confidence level of the following are:
(a) df= 5, t*= 1.943; (b) df= 20, t*= 2.845; (c) df= 28, t*= 2.045; (d) df= 11, t*= 3.106.

The critical t value (t*) can be found using the t-distribution table, which takes into account the sample size and confidence level. The degrees of freedom (df) can be calculated using the formula df = n - 1.

(a) n = 6, CL = 90%
df= 5 t*= 1.943

Using the t-distribution table with 5 degrees of freedom and a confidence level of 90%, we find the critical t value to be 1.943.

(b) n = 21, CL = 98%
df= 20 t*= 2.845

Using the t-distribution table with 20 degrees of freedom and a confidence level of 98%, we find the critical t value to be 2.845.

(c) n = 29, CL = 95%
df= 28 t*= 2.045

Using the t-distribution table with 28 degrees of freedom and a confidence level of 95%, we find the critical t value to be 2.045.

(d) n = 12, CL = 99%
df= 11 t*= 3.106

Using the t-distribution table with 11 degrees of freedom and a confidence level of 99%, we find the critical t value to be 3.106.

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Answer:

Below

Step-by-step explanation:

There is an image uploaded below of the stem-and-leaf plot of the video game scores!

A system of linear equations that can be used to determine the number of DVDs Norma purchased and the number of DVDs Lauretta purchased is:

H. n + l = 14

n = 5/2(l)

In order to write a system of linear equations to describe this situation, we would assign variables to the number of DVDs Norma purchased and the number of DVDs Lauretta purchased, and then translate the word problem into an algebraic equation (linear equations) as follows:

Since Norma purchased two and a half times as many DVDs as Lauretta purchased, a linear equation that models the situation is given by;

n = 2 1/2l

n = (5/2)l = 5l/2

Additionally, they purchased a total of 14 DVDs together;

n + l = 14

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The largest angle is B and the smallest angle is C

When we use the cosine rule, we can see that;

c^2 = a^2 + b^2 - 2ab Cos C

4.5^2 = 6.5^2 + 8.5^2 - 2(6.5 * 8.5) CosC

20.25 = 114.5 - 110.5CosC

20.25 - 114.5 = - 110.5CosC

CosC = 0.8529

C =Cos-1  0.8529

C = 31 degrees

Then;

b^2 = a^2 + c^2 - 2acCosB

8.5^2 = 6.5^2 + 4.5^2 -2(6.5 * 4.5)CosB

72.25 = 62.5 - 58.5CosB

72.5 - 62.5 =  - 58.5CosB

Cos B = -0.1709

B = 100 degrees

Then;

A = 180 - (100 + 31)

A = 49 degrees

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The probability of this event is 3/13. The event described is choosing a red card with an even number from a deck. To express this event as a set, we will include all even-numbered cards from the red suits (diamonds and hearts). This set is: {2♦, 4♦, 6♦, 8♦, 10♦, 2♥, 4♥, 6♥, 8♥, 10♥}.

Now let's find the probability of this event. There are 52 cards in the deck and 10 cards in the event set. Therefore, the probability is:

P(event) = (number of favorable outcomes) / (total number of outcomes) = 10 / 52 = 5 / 26 ≈ 0.1923

The probability of choosing a red card with an even number from a deck is approximately 0.1923 or 19.23%.

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The probability that a randomly selected tire will have a life of at most 47,500 miles is 0.9332, or 93.32%.

To find the probability that a randomly selected tire will have a life of at most 47,500 miles, we need to calculate the z-score and use a standard normal distribution table.

The formula for the z-score is:

z = (x - μ) / σ

where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

Plugging in the values given in the problem, we get:

z = (47,500 - 40,000) / 5,000
z = 1.5

Using a standard normal distribution table, we can find that the probability of a z-score being less than or equal to 1.5 is 0.9332.

Therefore, the probability that a randomly selected tire will have a life of at most 47,500 miles is 0.9332 (or 93.32%) to four decimal places.

To answer your question, we will use the normal distribution, mean, standard deviation, and Z-score. Given that the life expectancy of the tire is normally distributed with a mean of 40,000 miles and a standard deviation of 5,000 miles, we will find the probability of a tire having a life of at most 47,500 miles.

First, we calculate the Z-score using the formula: Z = (X - μ) / σ, where X is the value we're interested in (47,500 miles), μ is the mean (40,000 miles), and σ is the standard deviation (5,000 miles).

Z = (47,500 - 40,000) / 5,000 = 7,500 / 5,000 = 1.5

Now, we look up the Z-score of 1.5 in a standard normal distribution table or use a calculator with a cumulative distribution function (CDF). The CDF value for a Z-score of 1.5 is approximately 0.9332.

So, the probability that a randomly selected tire will have a life of at most 47,500 miles is 0.9332, or 93.32%.

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