Given A = and -3 3 -5 B= 4 use the Frobenius inner product and the corresponding induced norm to determine the value of each of the following: La Fun - (A, B) = ||A||F 2 || B||F = A,B radians.

The slope for the fitted line for the Male group is 9 + 4*X1, where X1 is the value of the continuous variable for the Male group. The coefficient of Male*X1 in the equation indicates the change in slope for the Male group compared to the slope for the non-Male group.

To find the slope of the fitted line for the Male group, you need to consider the given equation:

Yhat = 100 + 9X1 + 9Male + 4Male*X1

Since Male is a 0/1 variable, we assign a value of 1 for the Male group:

Yhat = 100 + 9X1 + 9(1) + 4(1)*X1

Now, simplify the equation:

Yhat = 100 + 9X1 + 9 + 4X1

Combine the terms with the continuous variable X1:

Yhat = 100 + 9X1 + 4X1 + 9

Yhat = 100 + 13X1 + 9

The slope of the fitted line for the Male group is the coefficient of the continuous variable X1, which is 13.

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approximately $29.54 The average income (in dollars) over this 4-hour period. To find the average income over the 4-hour period, we need to integrate the income function I(t) = 20e^(0.5t) with respect to time (t) over the interval [0, 4] and then divide by the length of the interval (4 hours).

The average income A(t) can be calculated using the formula:

A(t) = (1/4) ∫(20e^(0.5t) dt) from 0 to 4

First, let's find the integral of 20e^(0.5t) with respect to t:

∫(20e^(0.5t) dt) = 40e^(0.5t)

Now, let's apply the definite integral from 0 to 4:

[40e^(0.5(4)) - 40e^(0.5(0))] = 40e^2 - 40

Next, divide by 4 to find the average income:

A(t) = (1/4)(40e^2 - 40) ≈ 29.54 (rounded to the nearest cent)

The average income (in dollars) over this 4-hour period is approximately $29.54.

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A true statement about contextual interference is: it is high if a practice schedule involves a random arrangement of trials.

Contextual interference is a phenomenon in motor learning where practicing multiple variations of a task in a random or mixed order results in interference or disruption during learning. The interference caused by practicing in a mixed order appears to be counterintuitive, as it creates a challenge for the learner to constantly switch between different task variations. However, this interference has been shown to lead to better long-term retention of the learned skill.

The reason why contextual interference works is related to the way our brains process and consolidate information. When we practice a motor skill in a blocked or consistent order, our brains can quickly and easily memorize the movements required for that skill. However, this type of learning tends to be shallow, and the skill is not retained as well over time. In contrast, when we practice a skill in a random or mixed order, our brains have to work harder to distinguish between the different variations of the task. This additional processing load strengthens the neural connections underlying the skill, leading to more durable and flexible memory representations.

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For having to pay the same amount for both memberships and walk-in classes, if memberships cost $51 per month for unlimited classes and walk-ins charge $3 per class, one has to attend classes and pay $51 as the total amount.

Let the number of classes that are attended be x

If the same amount is paid for unlimited membership classes and walk=in classes, then we get the following equation:

Membership cost = cost of walk-in classes = $51

Cost of walk-in classes = 3x

Thus, we get the following equation.

51 = 3x

x = 17

Thus, the number of classes attended is 17 and the final amount paid is $51.

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Answer:

4,000 pounds

Step-by-step explanation:

One ton is equal to 2,000 pounds.
Therefore, a boulder that weighs 2 tons would weigh:

2 tons x 2,000 pounds/ton = 4,000 pounds

So the boulder weighed 4,000 pounds.

The equation that represent the average price of a house as a function of the year can be presented as follows;

The average of a value, such as the average price of a houses is the ratio between the sum of the prices of the houses to the number of houses.

The average cost of a family house in 1997 = $156,100

The average cost of the family house in 2010 = $254,400

The equation for the price (p) of a house as a function of the year, t where t = 0 corresponds to 1997, can be found as follows;

The slope of the equation is; (254,400 - 156,100)/(2010 - 1997) = 98300/13

98300/13 = 7561 7/13

The linear equation is therefore;
p - 156100 = (98300/13) × (t - 0)

p = (98300/13) × t + 156100

p = (7561 7/13) × t + 156100

The average price of a house today, 2023 can be found by plugging in the value, t = 2023 - 1997 = 26 in the above equation as follows;

p = (98300/13) × 26 + 156100 = 352,700

The price of a house today is $352,700

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Answer: The general form of a sine function is:

y = A sin (Bx + C) + D

Where:

A = amplitude

B = 2π/period

C = phase shift

D = vertical shift or midline

Given the values in the problem, we can substitute and simplify:

A = 3

midline = 2, so D = 2

period = 8π/7, so B = 2π/(8π/7) = 7/4

y = 3 sin (7/4 x + C) + 2

To find the phase shift, we need to use the fact that the sine function is at its maximum when the argument of the sine function is equal to π/2. That is:

Bx + C = π/2

We can solve for C:

C = π/2 - Bx

C = π/2 - (7/4) x

Substituting back the value of C in the equation, we get:

y = 3 sin (7/4 x + π/2 - 7/4 x) + 2

y = 3 sin (7/4 x - 7π/8) + 2

Therefore, the sine function with an amplitude of 3, a midline of y=2, and a period of 8π/7 is:

y = 3 sin (7/4 x - 7π/8) + 2

Answer:

Step-by-step explanation:

B. The statement is false. The transpose of the product of two matrices is the product of the transposes of the individual matrices in reverse order, or (AB)^T=B^TA^T.

Justification:
1. Let A and B be arbitrary matrices for which the product AB is defined.
2. To find the transpose of the product (AB)^T, we need to first understand the properties of transposes.
3. According to the property of transposes, (AB)^T is equal to the product of the transposes of the individual matrices in reverse order.
4. So, (AB)^T = B^TA^T, which contradicts the statement (AB)^T = A^TB^T.

Hence, the correct answer is option B.

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The number of mezzanine tickets that were sold is: 540

Let the regular seats be x

Let the mezzanine seats be y

Thus:

x + y = 910   -----(1)

18.50x + 16.25y = 15620   -----(2)

From eq 1,

y = 910 - x

Thus:

18.50x + 16.25(910 - x) = 15620

18.50x + 14,787.5 - 16.25x = 15620

2.25x = 15620 - 14,787.5

2.25x = 832.5

x = 832.5/2.25

x = 370 tickets

Thus:

y = 910 - 370

y = 540 tickets

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An example of a triangle with the same area as the shaded rectangle has been attached in the folder below.

Looking at the shaded diagram, we can tell that the rectangle has a length of 4cm and a width of 2 cm. The area of a rectangle can be calculated by the formula A = L x W, This amounts to 8 cm.

The area of a triangle can be calculated by multiplying the base by the height and then dividing by two. The formula is: A = 1/2 x base x height.

However, since we know the area of the rectangle is 8cm, we can decide that the height is 4cm and base 4 cm. it becomes 4 x 4 x 1/2 = 8

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Answer:

392.69908169872 feet3

Step-by-step explanation:

1/3πr^2h

1/3π*5^2*15

125π

392.69908169872 feet3

The standard error is found to be 0.024, when constructing a 90% confidence interval for the true proportion of students.

The point estimate of the proportion of students (sample) who dropped an online course is:

p' = 104/520 = 0.2

The standard error of the sample proportion is:

SE = √[p'(1 - p') / n]

where n is the sample size.

SE = √[0.2(1 - 0.2) / 520] = 0.024

To construct a 90% confidence interval, we can use the formula:

CI = p' ± z × SE

where z is the desired confidence level's associated z-score. The z-score with a 90% degree of confidence is 1.645.

CI = 0.2 ± 1.645(0.024) = (0.155, 0.245)

Therefore, we can be 90% confident that the true proportion of students who will drop an online course is between 0.155 and 0.245.

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These values are based on the t-distribution table and represent critical values for the given probabilities and degrees of freedom.

To answer this question, we need to use a t-table. The values we are looking for are the t-values associated with specific probabilities and degrees of freedom.

(a) t0.10, 10 = 1.372 (from the t-table, using 10 degrees of freedom and the probability of 0.10)

(b) t0.05, 10 = 1.812 (from the t-table, using 10 degrees of freedom and the probability of 0.05)

(c) t0.05, 21 = 1.721 (from the t-table, using 21 degrees of freedom and the probability of 0.05)

(d) t0.05, 60 = 1.671 (from the t-table, using 60 degrees of freedom and the probability of 0.05)

(e) t0.005, 60 = 2.660 (from the t-table, using 60 degrees of freedom and the probability of 0.005)

Note that we round all answers to three decimal places, as specified in the question. The values we have found are the t-values for the specified probabilities and degrees of freedom. These values can be used in calculations involving t-distributions.

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There is sufficient evidence to suggest that drivers over the age of 60 drive less on average than 24 miles per day. To answer this question, we can use a one-sample t-test. The null hypothesis is that the mean number of miles driven by drivers over 60 is equal to 24, while the alternative hypothesis is that it is less than 24.

Using the given information, we can calculate the t-statistic as follows:

t = (23.4 - 24) / (4.1 / sqrt(25)) = -1.95

The degrees of freedom for this test are 24 (n-1).

Using a t-table or calculator, we can find the critical value for a one-tailed test at a significance level of 0.05 with 24 degrees of freedom to be -1.711.

Since our calculated t-value of -1.95 is less than the critical value of -1.711, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that drivers over the age of 60 drive less on average than 24 miles per day.

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The simplification of the given expression, [tex]\sqrt{\frac{2^{x+2} + 2^{x}}{2^{x-3}} + 9}[/tex], is 7

From the question, we are to simplify the given expression

The given expression is

[tex]\sqrt{\frac{2^{x+2} + 2^{x}}{2^{x-3}} + 9}[/tex]

The expression can be simplified as follows:

[tex]\sqrt{\frac{2^{x+2} + 2^{x}}{2^{x-3}} + 9}[/tex]

[tex]\sqrt{\frac{2^{x} \times 2^{2} + 2^{x}}{2^{x} \div 2^{3}} + 9}[/tex]

[tex]\sqrt{\frac{2^{x} (2^{2} + 1)}{2^{x} \times \frac{1}{2^{3}}} + 9}[/tex]

[tex]\sqrt{\frac{(2^{2} + 1)}{ \frac{1}{2^{3}}} + 9}[/tex]

[tex]\sqrt{\frac{(4 + 1)}{ \frac{1}{8}} + 9}[/tex]

[tex]\sqrt{\frac{(5)}{ \frac{1}{8}} + 9}[/tex]

[tex]\sqrt{(5) \div \frac{1}{8} + 9}[/tex]

[tex]\sqrt{(5) \times \frac{8}{1} + 9}[/tex]

[tex]\sqrt{5 \times 8+ 9}[/tex]

[tex]\sqrt{40+ 9}[/tex]

[tex]\sqrt{49}[/tex]

= 7

Hence, the simplified expression is 7

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(i) The number of infected plants among 5 plants follows a binomial distribution with parameters n = 5 and p.

We observed 4 infected plants. To determine the probability of a result at least as extreme as the observed data, we need to calculate the probabilities of 4 and 5 infected plants for a range of values of p and add them up.

The probability of 4 infected plants is:

P(X = 4) = 5C4 * p^4 * (1-p)^1 = 5p^4 * (1-p)

The probability of 5 infected plants is:

P(X = 5) = 5C5 * p^5 * (1-p)^0 = p^5

Therefore, the probability of a result at least as extreme as the observed data is:

P(X >= 4) = P(X = 4) + P(X = 5) = 5p^4 * (1-p) + p^5

(ii) To determine if p = 0.3 or p = 0.2 lie within the 95% confidence interval, we need to find the range of values of p such that the probability of a result at least as extreme as the observed data is less than or equal to 0.025. We can solve this numerically using the expression derived in part (i):

For p = 0.3:

P(X >= 4) = 5(0.3)^4 * (1-0.3) + (0.3)^5 = 0.0765

Since 0.0765 > 0.025, we cannot conclude that p = 0.3 lies within the 95% confidence interval.

For p = 0.2:

P(X >= 4) = 5(0.2)^4 * (1-0.2) + (0.2)^5 = 0.01024

Since 0.01024 < 0.025, we can conclude that p = 0.2 lies within the 95% confidence interval.

(iii) If we observed five infected plants out of five, then the probability of a result at least as extreme as the observed data is simply P(X = 5) = p^5. To determine if p = 0.3 lies within the 95% confidence interval, we need to find the range of values of p such that the probability of observing five infected plants is less than or equal to 0.025:

p^5 <= 0.025

p <= (0.025)^(1/5)

p <= 0.551

Since 0.3 < 0.551, we can conclude that p = 0.3 does not lie within the 95% confidence interval.

(i) The number of infected plants among 5 plants follows a binomial distribution with parameters n = 5 and p. We observed one infected plant. To determine if p = 0.6 or p = 0.7 lie within the 95% confidence interval, we need to find the range of values of p such that the probability of a result at least as extreme as the observed data (i.e., zero or one infected plants) is less than or equal to 0.025:

For p = 0.6:

P(X <= 1) = P(X = 0) + P(X = 1) = (0.4)^5 + 5(0.6)(0.4)^4 = 0.07808

Since 0.07808 > 0.025, we cannot conclude that p = 0.6

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Total shaded area is the sum of the areas of the two triangles is 64.5 in²

The area of each of the two triangles can be found from the formula ...

A = (1/2)bh

left triangle

A = (1/2)(9.8 in)(8.6 in) = 42.14 in²

right triangle

A = (1/2)(5.2 in)(8.6 in) = 22.36 in²

Total shaded area is the sum of the areas of the two triangles:

(42.14 +22.36) in² = 64.5 in²

Hence, total shaded area is the sum of the areas of the two triangles is 64.5 in²

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The volume of the cuboids and triangular prisms are:

d). 2139 cm³, e). 1.6 mm³, f). 1428.84 m³, g). 264 yd³, and h). 96 m³

Volume of cuboid = length × width × height

d). 15.5 cm × 9.2 cm × 15 cm = 2139 cm³

e). 2 mm × 0.8 mm × 1 mm = 1.6 mm³

f). 6.3 m × 6.3 m × 3.6 m = 1428.84 m³

Volume of triangular prism = base area × height

g). 11 yd × 6 yd × 4 yd = 264 yd³

h). 4 m × 4 m × 6 m = 96 m³

Therefore, the volume of the cuboids and triangular prisms are:

d). 2139 cm³, e). 1.6 mm³, f). 1428.84 m³, g). 264 yd³, and h). 96 m³

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Best Buy's expected value on the warranty is $1.25.

To calculate Best Buy's expected value on the warranty, we need to consider the potential outcomes and their probabilities.

If the customer doesn't return the item for a claim on the warranty, Best Buy receives $25 for the warranty but doesn't have to pay anything out. The probability of this happening is 95% (100% - 5%).

If the customer does return the item for a claim on the warranty, Best Buy has to refund the $250 purchase price but received $25 for the warranty. The probability of this happening is 5%.

So, to calculate the expected value, we can multiply the probability of each outcome by its value and add them together:

(0.95 x $25) + (0.05 x -$250) = $1.25

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The missing side of triangle is 11.66 cm, under the condition that the given triangle is a right angled triangle and the other sides of the triangle are 10cm, 6cm respectively.

In order to evaluate the other side of the triangle, we have to rely on the principles of Pythagorean Theorem which states that in a right triangle, the sum of the square sides are equal to the square of the hypotenuse side.

Then, let us consider that x be the length of the missing side then

x² = 10² + 6²

x² = 100 + 36

x² = 136

x = √(136)

x ≈ 11.66 cm

Then, the missing side of this triangle is approximately 11.66 cm.

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The complete question is

Find the missing siside of the following right triangle in the figure





The graph that shows the solution to the system of equations is given by the image presented at the end of the answer.

The system of equations for this problem is defined as follows:

Replacing y = 0.5x on the second equation, the x-coordinate of the solution is given as follows:

x + 2(0.5x) = -8

x + x = -8

2x = -8

x = -4.

The y-coordinate of the solution is given as follows:

y = 0.5(-4)

y = -2.

Hence the graph will show the two lines intersecting at the point (-2,-4).

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A. For z = 0.50 = 66; B. For z = -0.25 = 57 and the probability of randomly selecting a score less than z = -0.25, you can use a standard normal (z-score) table or calculator. For z = -0.25, the probability is approximately 0.4013, meaning there's a 40.13% chance of randomly selecting a score less than z = -0.25.

To find the X value for each z-score, we use the formula:

X = u + (z * a)

where u is the mean, a is the standard deviation, and z is the z-score.

A. For z = 0.50:

X = 60 + (0.50 * 12)
X = 66

Therefore, the X value for z = 0.50 is 66.

B. For z = -0.25:

X = 60 + (-0.25 * 12)
X = 57

Therefore, the X value for z = -0.25 is 57.

To find the probability of randomly selecting a score that is less than z = -0.25, we use a z-table or a calculator that can calculate normal distribution probabilities. The probability of a score being less than z = -0.25 is the same as the area under the curve to the left of z = -0.25.

Using a z-table, we find that the probability of a score being less than z = -0.25 is 0.4013.

Therefore, the probability of randomly selecting a score that is less than z = -0.25 is 0.4013 or approximately 40.13%.

A. For z = 0.50:
To find the X value, use the formula X = μ + (z * σ), where μ is the mean and σ is the standard deviation.
X = 60 + (0.50 * 12) = 60 + 6 = 66

B. For z = -0.25:
X = 60 + (-0.25 * 12) = 60 - 3 = 57

To find the probability of randomly selecting a score less than z = -0.25, you can use a standard normal (z-score) table or calculator. For z = -0.25, the probability is approximately 0.4013, meaning there's a 40.13% chance of randomly selecting a score less than z = -0.25.

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Probability is the study of the chances of occurrence of a result, which are obtained by the ratio between favorable cases and possible cases.

To find the probability of (a or b), we can use the formula:

P(a or b) = P(a) + P(b) - P(a and b)

Plugging in the given values:

P(a or b) = 0.21 + 0.35 - 0.12

= 0.44

Therefore, the probability of (a or b) is 0.44.

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The statistical measures for the given set of data are:

Population size: 8

Median: 10

Minimum: 7

Maximum: 14

First quartile: 7.25

Third quartile: 12.75

Interquartile Range: 5.5

Outliers: none

The statistical measures for the given set of data are:

Mean: (7+7+8+9+11+12+13+14)/8 = 10.125

Median: 10

Mode: 7 (since it appears twice)

Range: 14-7 = 7

Variance: 6.625

Standard deviation: √(6.625) = 2.6

To create a box and whiskers plot, we first need to order the data set from smallest to largest:

7, 7, 8, 9, 11, 12, 13, 14

To find the quartiles, we need to find the median (which we already know is 10), and then find the median of the lower half and upper half of the data set separately:

Lower half: 7, 7, 8, 9

Upper half: 11, 12, 13, 14

The median of the lower half is (7+8)/2 = 7.5, and the median of the upper half is (12+13)/2 = 12.5.

Therefore, the quartiles are:

Q1 = 7.25

Q2 (median) = 10

Q3 = 12.75

To find the range of values within 1.5 times the IQR, we first calculate the IQR:

IQR = Q3 - Q1 = 12.5 - 7.5 = 5

Then, we calculate the lower and upper bounds:

Lower bound = Q1 - 1.5IQR = 7.5 - 1.55 = 0

Upper bound = Q3 + 1.5IQR = 12.5 + 1.55 = 20

Since all of the observations fall within the bounds, there are no outliers in this data set.

Using this information, we can create a box and whiskers plot as follows:

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By arranging the population in a systematic pattern where every kth element has the same value, we ensure that the variance of the 1-in-k systematic sample mean is 0.

To construct a population such that V(īsy) = 0 for 1-in-k systematic sampling, we need to ensure that the population variance is evenly distributed throughout the sample. This can be achieved by selecting a population where the variability within each kth interval is the same.

For example, if we have a population of 1000 individuals, we can divide it into intervals of 10. Within each interval, we can ensure that the variance is the same by selecting individuals with similar characteristics or attributes. This ensures that when we take a 1-in-k systematic sample, the variance within each interval remains the same, resulting in V(īsy) = 0.

It is important to note that systematic sampling is a type of probability sampling where every kth individual is selected from the population. This method is often used when the population is large and spread out, and random sampling is not feasible. However, it is important to ensure that the systematic sampling is truly random and not biased towards any particular group within the population.

To construct a population where the variance of the 1-in-k systematic sample mean (V(īsy)) equals 0, we need to make sure that the sample means are the same for all possible systematic samples.

Step 1: Choose a population size, N, and a sampling interval, k. For simplicity, let's select N=12 and k=3.

Step 2: Arrange the population elements in a systematic pattern. Since we want the variance of the systematic sample mean to be 0, we'll arrange the elements such that every kth element has the same value. For example:

Population: 5, 2, 7, 5, 2, 7, 5, 2, 7, 5, 2, 7

Step 3: Perform 1-in-k systematic sampling. With k=3, we'll take every 3rd element starting from the first element:

Sample 1: 5, 5, 5, 5 (Mean: 5)
Sample 2: 2, 2, 2, 2 (Mean: 2)
Sample 3: 7, 7, 7, 7 (Mean: 7)

Step 4: Verify that V(īsy) = 0. The variance of the sample means is 0, as they are all equal for each possible systematic sample. Therefore, V(īsy) = 0 for this population.

In summary, by arranging the population in a systematic pattern where every kth element has the same value, we ensure that the variance of the 1-in-k systematic sample mean is 0.

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The probability of the ball not landing on green and landing on an even, black slot is:
36/38 * 18/38 = 9/19

To find the probability that the metal ball does not land on green, we first need to determine the total number of non-green slots.

There are 38 slots in total, with 0 and 00 being green. This means there are 36 slots that are not green (18 red odd-numbered and 18 black even-numbered slots).

Now, we can calculate the probability of the metal ball not landing on the green by dividing the number of non-green slots (36) by the total number of slots (38).

Probability = (Number of non-green slots) / (Total number of slots)
Probability = 36/38

Now, we can simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2.

Probability = (36/2) / (38/2)
Probability = 18/19
However, we need to take into account that the odd-numbered slots are red and the even-numbered slots are black. Since the question asks for the probability that the ball does not land on the green, we can focus on the even-numbered slots that are black. There are 18 even-numbered slots on the wheel, which means that the probability of the ball landing on an even, black slot is 18/38.

So, the probability that the metal ball does not land on the green is 18/19.

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Answer: hortest side of the triangle is 24 units.

Step-by-step explanation: To solve for the shortest side of a triangle, we first need to simplify the given equation:

s + s + 8 - 18 = 38

Combining like terms, we get:

2s - 10 = 38

Adding 10 to both sides, we get:

2s = 48

Dividing by 2, we get:

s = 24

Therefore, the value of the shortest side of the triangle is 24 units.

Answer:(94-7a)

Step-by-step explanation:

if 69 ate ur mom

The null hypothesis is that the true average age of the sporting goods store's customers is equal to or less than 39. The alternative hypothesis is that the true average age is greater than 39.

To test this hypothesis, we can use a one-sample t-test. The t-statistic for this sample is (41.3-39)/(7/sqrt(43)) = 3.06. With 42 degrees of freedom (43-1), the critical value for a one-tailed test with alpha=0.10 is 1.684. Since 3.06 > 1.684, we reject the null hypothesis and conclude that there is evidence to suggest that the true average age of the store's customers is greater than 39.

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