The length of the end table is 45 inches. The width is 15 inches. What is the area?

Pls help

The probabilities are:

(a) P(X = 0) ≈ 0.3139

(b) P(X ≥ 1) ≈ 0.6861

(c) P(X ≤ 1) ≈ 0.9862

We can model this situation using the hypergeometric distribution.

Let's define:

N = total number of manufacturing plants = 30

D = number of plants outside the country = 7

n = number of plants in the performance evaluation = 4

(a) Probability of including no plants outside the country:

We want to find P(X = 0), where X is the number of plants outside the country in the performance evaluation. This can be calculated using the hypergeometric distribution formula:

P(X = 0) = (C(23, 4) * C(7, 0)) / C(30, 4) = (23 choose 4) / (30 choose 4) ≈ 0.3139

(b) Probability of including at least 1 plant outside the country:

We want to find P(X ≥ 1). We can use the complement rule and find the probability of including no plants outside the country and subtract it from 1:

P(X ≥ 1) = 1 - P(X = 0) = 1 - (C(23, 4) * C(7, 0)) / C(30, 4) ≈ 0.6861

(c) Probability of including no more than 1 plant outside the country:

We want to find P(X ≤ 1). This can be calculated as the sum of P(X = 0) and P(X = 1):

P(X ≤ 1) = P(X = 0) + P(X = 1) = (C(23, 4) * C(7, 0)) / C(30, 4) + (C(23, 3) * C(7, 1)) / C(30, 4) ≈ 0.9862

Therefore, the probabilities are:

(a) P(X = 0) ≈ 0.3139

(b) P(X ≥ 1) ≈ 0.6861

(c) P(X ≤ 1) ≈ 0.9862

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The number of possible combinations of three coins from the bag of 16 coins with different dates can be calculated using the formula for combinations, which is nCr = n! / r!(n-r)!, where n is the total number of objects, r is the number of objects to be chosen, and ! denotes factorial (the product of all positive integers up to the given number).

In this case, we have n = 16 (the total number of coins in the bag) and r = 3 (the number of coins to be chosen for each combination). Using the formula for combinations, we can calculate the number of possible combinations as follows:

nCr = 16! / 3!(16-3)!
nCr = (16 x 15 x 14) / (3 x 2 x 1)
nCr = 560

Therefore, 560 possible combinations of three coins can be chosen from the bag of 16 coins with different dates. These combinations could represent different historical events, significant dates, or other symbolic meanings depending on the dates inscribed on the coins. The calculation of combinations is an important concept in combinatorics and probability theory, and it has many real-world applications in fields such as statistics, economics, and computer science.

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A teacher asked Dwayne to find the values of x and y in the triangles shown. The teacher provided the following information about the triangles. Triangle ABC is similar to triangle PQR. In triangle ABC, cos(C) = 0.92. Dwayne claims that the value of x can be determined.

Hence, the correct option is A.

Since triangles ABC and PQR are similar, their corresponding angles are congruent and their corresponding sides are proportional. Therefore, we can set up the following proportion we get

AB/BC = PQ/QR

We can also use the cosine law to relate the angle C in triangle ABC to the length of side AB and BC.

cos(C) = ([tex]AB^2 + BC^2 - AC^2[/tex])/(2AB*BC)

We are given that cos(C) = 0.92, and we know that AC = 20, AB = x, and BC = y, so we can substitute these values into the cosine law we get

0.92 = ([tex]x^2 + y^2[/tex] - 400)/(2xy)

Simplifying this equation, we get

([tex]x^2 + y^2[/tex] - 400) = 1.84xy

We can also use the given information to relate x and y we get

cos(R) = y/45

However, we cannot use this equation to solve for y because we do not know the value of cos(R).

Therefore, Dwayne is correct in claiming that we can determine the value of x using the cosine law, but we cannot determine the value of y with the information provided.  His claim is correct because cos(C) = x/20 and 0.92 can be substituted for cos(C), but the cosine of angle R is not given for triangle PQR.

Hence, the correct option is A.

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Answer:

(a) The total revenue from the sale of x thousand candy bars is equal to the product of the price charged for a candy bar and the number of candy bars sold. If p(x) is the price charged for a candy bar in cents, then the revenue R(x) in dollars is given by:

R(x) = (p(x) * 1000x) / 100

R(x) = (151 - x/10) * 100x

R(x) = 15100x - 1000x^2

Therefore, the expression for the total revenue from the sale of x thousand candy bars is R(x) = 15100x - 1000x^2 dollars.

(b) To find the value of x that leads to maximum revenue, we need to find the value of x for which R(x) is maximum. We can do this by finding the derivative of R(x) with respect to x, setting it equal to zero, and solving for x. So:

R'(x) = 15100 - 2000x

Setting R'(x) equal to zero, we get:

15100 - 2000x = 0

Solving for x, we get:

x = 7.55

Therefore, the value of x that leads to maximum revenue is 7.55 thousand candy bars.

(c) To find the maximum revenue, we substitute x = 7.55 into the expression for R(x):

R(7.55) = 15100(7.55) - 1000(7.55)^2

R(7.55) = $57042.50

Therefore, the maximum revenue is $57,042.50 when 7.55 thousand candy bars are sold.

Step-by-step explanation:

The maximum revenue from the sale of candy bars in the city is $84,375. a. The expression for total revenue from the sale of x thousand candy bars can be found by multiplying the price per candy bar by the number of candy bars sold:

Total revenue = p(x) * x

Substituting the given equation for p(x), we get:

Total revenue = (151 - x/10) * x

b. To find the value of x that leads to maximum revenue, we need to take the derivative of the revenue function and set it equal to zero:

d/dx (151x - x^2/10) = 0

Simplifying and solving for x, we get:

x = 750

c. To find the maximum revenue, we substitute the value of x obtained in part (b) into the revenue function:

Total revenue = (151 - 750/10) * 750 = $84,375

Therefore, the maximum revenue from the sale of candy bars in the city is $84,375.

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The bracket makes an angle of approximately 51.48° degrees with the wall.

First, we can use the Pythagorean theorem to find the distance between the end of the bracket and the wall:

[tex]d = \sqrt{((3.2 ft)^2 - (2.5 ft)^2) }[/tex]≈ 1.99 ft

Now we can use the definition of the tangent function to find the angle x:

tan(x) = opposite / adjacent = 2.5 ft / 1.99 ft

Taking the arctangent of both sides, we get:

x = tan⁻¹(2.5 ft / 1.99 ft) ≈ 51.48°

Therefore, the bracket makes an angle of approximately 51.48° degrees with the wall.

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To test the claim that having lights on at night increases weight gain, we can conduct a two-sample t-test with unequal variances.

Let μ1 be the population mean weight change for mice living in bright light and μ2 be the population mean weight change for mice living in a normal light/dark cycle. The null hypothesis is H0: μ1 - μ2 = 0 (there is no difference in weight gain between the two groups) and the alternative hypothesis is Ha: μ1 - μ2 > 0 (mice in bright light gain more weight).

Using the given data, we can calculate the sample means and standard deviations:

x1 = 2.312 kg, s1 = 1.052 kg (for the sample of 10 mice in bright light)
x2 = 1.062 kg, s2 = 0.598 kg (for the sample of 8 mice in normal light/dark cycle)

We can then calculate the test statistic t:

t = (x1 - x2) / √(s1^2/n1 + s2^2/n2) = (2.312 - 1.062) / √(1.052^2/10 + 0.598^2/8) = 2.840

The degrees of freedom for the t-test is approximately given by the Welch-Satterthwaite equation:

df = (s1^2/n1 + s2^2/n2)^2 / (s1^4/(n1^2*(n1-1)) + s2^4/(n2^2*(n2-1))) = (1.052^2/10 + 0.598^2/8)^2 / (1.052^4/(10^2*9) + 0.598^4/(8^2*7)) = 14.867

Using a t-distribution table or calculator with df = 14.867 and a one-tailed test at α = 0.06 (equivalent to a critical t-value of 1.796), we find the p-value to be p = 0.006. Since this p-value is less than the significance level of 0.06, we reject the null hypothesis and conclude that there is evidence to support the claim that mice in bright light gain more weight than those in a normal light/dark cycle.

Note that the 6% level of significance is not a commonly used level and may be too liberal or too conservative depending on the context. It is important to consider the practical significance of the result and the potential for type I and type II errors.

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The notation of number 25.67 x [tex]10^{-2[/tex] is not written in the scientific notation.

Scientific notation is a way of writing very large or very small numbers using powers of 10. It has the form [tex]a X 10^n[/tex], where a is a number between 1 and 10 (or sometimes between -1 and -10), and n is an integer.

As mentioned earlier, scientific notation has the form [tex]a X 10^n[/tex], where a is a number between 1 and 10 (or sometimes between -1 and -10), and n is an integer. But is should be reduced to one decimal number.

Thus, 25.67 x [tex]10^{-2[/tex] is not written in scientific notation.

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Using permutation, the soccer coach can choose one starter and one reserve player for the certain position in 56 different ways.

The coach wants to choose and order two players from a group of 8. This is a permutation problem since order matters.

To find the number of ways to choose one starter and one reserve player from 8 candidate players, you can use the following steps:

The formula for the number of permutations of n objects taken r at a time is nPr = n!/(n-r)!.

Using this formula, we can calculate the number of ways the coach can choose and order two players:

8P2 = 8!/(8-2)! = 8!/6! = 8x7 = 56

Therefore, there are 56 ways the coach can choose and order a starter and reserve player for the position from a group of 8 candidates.

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Answer:

44/35

Step-by-step explanation:

this answer cannot be further simplified*

Answer:

24 is the answer

Step-by-step explanation:

add the numbers and divide them by 7

The product of two term 88 x 45  would be equal to 3960.

Since Multiplication is the mathematical operation that is used to determine the product of two or more numbers.

When an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.

Given that 88 x 45

We need to simply multiply the term;

88 x 45

= 3960

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Answer: When we use the change of variables u=x2,v=y3,w=z�=�2,�=�3,�=� to find the volume of the solid enclosed by the ellipsoid x24+y29+z2=1�24+�29+�2=1 above the xy−��− plane, we are essentially transforming the original equation of the ellipsoid into a new equation that is easier to work with.

Step-by-step explanation:

The new equation becomes u/4+v/9+w/1=1. We can now use this equation to find the volume of the solid by integrating over the region in uvw-space that corresponds to the region in xyz-space above the xy−��− plane. This region is a solid bounded by a plane, two planes perpendicular to the uvw-axes, and the surface of the ellipsoid. To integrate over this region, we can use triple integrals in uvw-space.
The triple integral would have limits of integration of 0 to 1 for u, 0 to (1-4u/9) for v, and 0 to sqrt(1-4u/9-v) for w. Integrating this triple integral would give us the volume of the solid enclosed by the ellipsoid above the xy−��− plane. In summary, the change of variables transforms the original equation into a simpler equation that can be used to set up a triple integral to find the volume of the solid. The region of integration in uvw-space corresponds to the region in xyz-space above the xy−��− plane, and we can use triple integrals to integrate over this region and find the volume.
Using the change of variables u = x^2, v = y^3, w = z, we can rewrite the equation for the ellipsoid as u/24 + v/29 + w^2 = 1. We want to find the volume of the solid enclosed by this ellipsoid above the xy-plane, which means we're looking for the region where w ≥ 0.

To do this, we will set up a triple integral over the given region using the Jacobian determinant to transform from (x, y, z) coordinates to (u, v, w) coordinates. The Jacobian determinant is given by:

J = |(∂(x,y,z)/∂(u,v,w))| = |(∂x/∂u, ∂x/∂v, ∂x/∂w; ∂y/∂u, ∂y/∂v, ∂y/∂w; ∂z/∂u, ∂z/∂v, ∂z/∂w)|

Computing the partial derivatives, we get J = |(1/2, 0, 0; 0, 1/3, 0; 0, 0, 1)| = 1/6.

Now, we can set up the triple integral:

Volume = ∫∫∫(u, v, w) dudvdw

The limits of integration for u will be 0 to 24, for v will be 0 to 29, and for w will be 0 to 1.

Volume = (1/6) ∫(0 to 24) ∫(0 to 29) ∫(0 to 1) dudvdw

Calculating this triple integral, we find the volume of the solid enclosed by the ellipsoid above the xy-plane:

Volume ≈ 58.8 cubic units.

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Answer:

To solve this problem, we need to use the concept of hypergeometric distribution, which gives the probability of selecting a certain number of objects with a specific characteristic from a population of known size without replacement. We will use the formula:

P(X = k) = [ C(M, k) * C(N - M, n - k) ] / C(N, n)

where:

P(X = k) is the probability of selecting k objects with the desired characteristic;

C(M, k) is the number of ways to select k objects with the desired characteristic from a population of M objects;

C(N - M, n - k) is the number of ways to select n - k objects without the desired characteristic from a population of N - M objects;

C(N, n) is the total number of ways to select n objects from a population of N objects.

In our case, we want to select 5 rats out of a population of 80, and we want exactly 3 of them to have 2 genetic mutations. We can calculate this probability as follows:

P(3 rats have exactly 2 mutations) = [ C(12, 3) * C(68, 2) ] / C(80, 5)

where:

M is the number of rats that have exactly 2 mutations, which is the sum of the rats that have mutations AB, AC, AD, BC, BD, and CD, or M = 5 + 6 + 4 + 3 + 3 + 1 = 22;

N - M is the number of rats that do not have exactly 2 mutations, which is the remaining population of 80 - 22 = 58 rats;

n is the number of rats we want to select, which is 5.

We can simplify this expression as follows:

P(3 rats have exactly 2 mutations) = [ C(12, 3) * C(68, 2) ] / C(80, 5)

= [ (12! / (3! * 9!)) * (68! / (2! * 66!)) ] / (80! / (5! * 75!))

= 0.03617

Therefore, the probability that if 5 rats are selected at random, 3 will have exactly 2 genetic mutations is 0.03617 (rounded to five decimal places).

The sandbox's 3/4ths fraction is full, the volume of sand can be found by multiplying the volume of the entire sandbox (0.33 ft³) by 3/4, which gives us 0.2475 ft³ of sand.

The area given (64 cm²) is the base of the sandbox, we can find the height of the sandbox in cm using the formula for the area of a rectangle: A = l × w.

Since the area is 64 cm², and we know that the sandbox has a rectangular base, we can assume the length and width are equal and each measure 8 cm.

Next, we need to convert the height of the sandbox from cm to feet. One foot is equal to 30.48 cm, so the sandbox is 0.33 feet tall (approximately). Since the sandbox is 3/4ths full, the volume of sand can be found by multiplying the volume of the entire sandbox (0.33 ft³) by 3/4, which gives us 0.2475 ft³ of sand.

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The sandbox was a foot tall, but the sand only was only filled up 3/4ths of the way with an area of 64 cm². How many cubic feet of sand are there in the box?

the answer to your question is  130

The class limits for the class with the highest frequency is 65 up to 75. The correct answer is A.

The frequency distribution given in the question represents the number of textbooks and their corresponding costs. The distribution is divided into several classes, each representing a range of costs. The frequency for each class indicates how many textbooks fall within that range of costs.

The question asks us to find the class limits for the class with the highest frequency. We can see from the distribution that the class with the highest frequency is "65 up to 75", which has a frequency of 20.

The class limits for a given class are the lowest and highest values included in that class. In this case, the lower limit of the class "65 up to 75" is 65 (because it is the lowest value in that range), and the upper limit of the class is 75 (because it is the highest value in that range).

Therefore, the class limits for the class with the highest frequency are 65 (the lower limit) and 75 (the upper limit), and the correct answer is "65 up to 75".  The correct answer is A.

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Answer: 132 Ounces.

Step-by-step explanation: 1 pound = 16 ounces. 8 1/4 or 8.25 x 16 = 132

Seth weigh 132 ounce when he was born.

The unitary technique involves first determining the value of a single unit, followed by the value of the necessary number of units.

For example,Let's say Ram spends 36 Rs. for a dozen (12) bananas.

12 bananas will set you back 36 Rs. 1 banana costs 36 x 12 = 3 Rupees.

As a result, one banana costs three rupees. Let's say we need to calculate the price of 15 bananas.

This may be done as follows: 15 bananas cost 3 rupees each; 15 units cost 45 rupees.

We have,

Seth weighed 8 pounds when he was 8 1/4 pounds born.

So, the weight in ounce

= 8 1/4 x 16

= 33/4 x 16

= 33 x 4

= 132 ounce

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If we choose ε > 0, we can find a δ such that ||f – ft||1 < ε for all t with 0 < |t - 1| < δ, and we have shown that lim||f - ft||1 = 0 as t -> 1.

(a) To prove that lim||f – ft||1 = 0 as t -> 0, we need to show that for any ε > 0, there exists a δ > 0 such that ||f – ft||1 < ε for all t with 0 < |t| < δ.

We have:

||f – ft||1 = ∫|f(x) – f(x – t)| dx

By the continuity of f, we know that for any ε > 0, there exists a δ > 0 such that |f(x) – f(x – t)| < ε whenever |t| < δ. Therefore:

||f – ft||1 = ∫|f(x) – f(x – t)| dx < ε∫dx = ε

This holds for all t with 0 < |t| < δ, so we have shown that lim||f – ft||1 = 0 as t -> 0.

(b) To prove that lim||f - ft||1 = 0 as t -> 1, we need to show that for any ε > 0, there exists a δ > 0 such that ||f – ft||1 < ε for all t with 0 < |t - 1| < δ.

We have:

||f – ft||1 = ∫|f(x) – f(tx)| dx

Using the change of variables y = tx, we can write this as:

||f – ft||1 = (1/t)∫|f(y/t) – f(y)| dy

Since f is integrable, it is also bounded. Let M be a bound on |f|. Then we have:

||f – ft||1 ≤ (1/t)∫|f(y/t) – f(y)| dy ≤ (1/t)∫M|y/t – y| dy = M|1 – t|

This holds for all t with 0 < |t - 1| < δ, where δ = ε/2M. Therefore, if we choose ε > 0, we can find a δ such that ||f – ft||1 < ε for all t with 0 < |t - 1| < δ, and we have shown that lim||f - ft||1 = 0 as t -> 1.

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The probability of a randomly selected hair salon rating having at most 2 stars is 0.44,

The probability of having more than 3 stars is 0.47.

We have,
(a) To find the probability that the randomly selected hair salon rating has at most 2 stars, we need to add the probabilities for 1-star and 2-star ratings.

Based on the provided probability distribution,

P(X=1) = 0.25 and P(X=2) = 0.19.

The probability of a rating having at most 2 stars.
P(X ≤ 2) = P(X=1) + P(X=2)

= 0.25 + 0.19

= 0.44

(b)

To find the probability that the randomly selected hair salon rating has more than 3 stars, we need to add the probabilities for 4-star and 5-star ratings.

Based on the provided probability distribution, P(X=4) = 0.21 and P(X=5) = 0.26.

The probability of a rating having more than 3 stars.
P(X > 3) = P(X=4) + P(X=5)

= 0.21 + 0.26

= 0.47

Thus,
The probability of a randomly selected hair salon rating having at most 2 stars is 0.44, and the probability of having more than 3 stars is 0.47.

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The short-run supply curve shows the quantity of output a firm is willing to supply at different market prices in the short run. It is typically upward sloping, meaning that as the price of the product increases, the firm is willing to produce and supply more units. This is because higher prices will allow the firm to cover its variable costs and potentially earn a profit.

When used in conjunction with the average variable cost (AVC) curve, the supply curve can give a firm valuable information about its profits. The AVC curve represents the average variable cost per unit of output, which includes the costs that vary with the level of production (such as labor and materials).

If the market price is above the AVC curve, the firm is covering all of its variable costs and may earn a profit. If the market price is below the AVC curve but still above the average total cost (ATC) curve, the firm is not covering all of its costs but is still producing because it is covering its variable costs. If the market price falls below the ATC curve, the firm is not covering all of its costs and is likely to shut down production in the short run.

Therefore, the supply curve in conjunction with the AVC curve allows a firm to determine whether it should produce and supply output in the short run based on the prevailing market price. If the market price is high enough to cover variable costs and potentially earn a profit, the firm will continue to produce. However, if the market price falls below the AVC curve, the firm will likely reduce or cease production to minimize losses.

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The value of the average rate of change is,

⇒ f ' (x) = 14

We have to given that;

The function is,

⇒ f (x) = 3x² - 4

Now, We can formulate;

The value of the average rate of change as;

⇒ f ' (x) = f (4) - f (2) / (4 - 2)

⇒ f ' (x) = (3 × 4² - 4) - (3 × 2² - 4) / 2

⇒ f ' (x) = 44 - 16 / 2

⇒ f ' (x) = 28/2

⇒ f ' (x) = 14

Thus.,  The value of the average rate of change is,

⇒ f ' (x) = 14

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The correct mean adjusts in the records for all 230 children is  96 pounds

The given issue includes finding the right cruel weight of the 230 children after adjusting for the scale blunder. The first mean weight is given as 98 pounds, but we got to alter for the scale blunder of 2 pounds that the scale was perusing over the genuine weight.

To correct the scale mistake, we ought to subtract 2 pounds from each child's recorded weight. This will shift the complete conveyance of weights by 2 pounds to the cleared out, so the unused cruel weight will be lower than the first cruel weight.

The first cruel weight is given as 98 pounds, but we got to alter for the scale blunder:

Rectified mean weight = Original cruel weight - Scale blunder

Rectified mean weight = 98 - 2

Rectified mean weight = 96 pounds

Subsequently, the proper cruel weight of the children after altering the scale blunder is 96 pounds. This implies that on normal, the children weighed 96 pounds rather than 98 pounds as initially recorded. 

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The final value is  0.67

(a) For a perpendicular rectangle without a common edge, the shape factor can be calculated using the formula:

f12 = min(w1, w2) / (h1 + h2)

where w1 and w2 are the widths of the rectangles and h1 and h2 are the heights.

In this case, the minimum width is 2 units, and the sum of the heights is 5 + 3 = 8 units. Therefore,

f12 = 2 / 8 = 0.25

(b) For parallel rectangles of unequal areas, the shape factor can be calculated using the formula:

f12 = (A2 / A1) * (h1 / h2)

where A1 and A2 are the areas of the rectangles and h1 and h2 are the heights.

In this case, the area of rectangle 1 is 24 square units (4 x 6) and the area of rectangle 2 is 16 square units (4 x 4). The heights are the same at 4 units. Therefore,

f12 = (16 / 24) * (4 / 4) = 0.67

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Answer:

I assume you trying to find a surface area (tell me if I'm wrong. okay?

Step-by-step explanation:

V = (1/2)bhL

where b is the base of the triangle, h is the height of the triangle, and L is the length of the prism.

The formula for the surface area of a triangular prism is:

SA = bh + 2(L + b)s

where b and h are the same as above, L is the length of the prism, and s is the slant height of the triangle.

To use these formulas, we need to identify the values of b, h, L, and s from the given dimensions. The base of the triangle is 8 units, the height of the triangle is 6 units, and the length of the prism is 15 units. The slant height of the triangle can be found using the Pythagorean theorem:

s^2 = b^2 + h^2 s^2 = 8^2 + 6^2 s^2 = 64 + 36 s^2 = 100 s = sqrt(100) s = 10

Now we can plug these values into the formulas and simplify:

V = (1/2)bhL V = (1/2)(8)(6)(15) V = (1/2)(720) V = 360

SA = bh + 2(L + b)s SA = (8)(6) + 2(15 + 8)(10) SA = 48 + 2(23)(10) SA = 48 + 460 SA = 508

Therefore, the volume of the triangular prism is 360 cubic units and the surface area is 508 square units.

Answer:

A, C, E

Step-by-step explanation:

To determine which expressions have a value greater than 1, evaluate the expressions following the order of operations (PEMDAS) and remembering the following:

[tex]\;\;\;\:-\frac{1}{3} \div (-2)+4\\\\= -\frac{1}{3} \cdot \left(-\frac{1}{2}\right)+4\\\\=\frac{(-1) \cdot (-1)}{3 \cdot 2}+4\\\\= \frac{1}{6}+4\\\\= 4\frac{1}{6}[/tex]

[tex]\;\;\;-\frac{1}{3} \cdot (-2)-4\\\\= \frac{2}{3} -4\\\\= \frac{2}{3} -\frac{12}{3}\\\\= \frac{2-12}{3}\\\\= -\frac{10}{3}\\\\=-3\frac{1}{3}[/tex]

[tex]\;\;\:\:-\frac{1}{3} \cdot (-2-4)\\\\= -\frac{1}{3} \cdot (-6)\\\\=\frac{(-1)\cdot (-6)}3{}\\\\= \frac{6}{3} \\\\= 2[/tex]

[tex]\;\;\:\:-\frac{1}{3} \cdot (-2)(-4)\\\\= -\frac{1}{3} \cdot (8)\\\\=\frac{(-1) \cdot 8}{3}\\\\= -\frac{8}{3} \\\\= -2\frac{2}{3}[/tex]

[tex]\;\;\:\:-\frac{1}{3} + (-2)-(-4)\\\\= -\frac{1}{3} -2+ 4\\\\= -2\frac{1}{3} + 4\\\\=4 -2\frac{1}{3}\\\\= 1\frac{2}{3}[/tex]

Therefore, the expressions that have a value greater than 1 are:

The probability that a random point on AK will be on CH is 1/2

A probability is a number that reflects the chance or likelihood that a particular event will occur. The certainty of an event is 1 and it is equal to 100% in percentage.

probability = sample space /total outcome

The total outcome is the range of AK.

range = highest - lowest

= 10-(-10) = 10+10 = 20

sample space of CH = 4-(-6)

= 4+6 = 10

Therefore probability that a random point on AK will be on CH is 10/20

= 1/2

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The probability that the mean weight of 20 randomly selected men is between 170 and 185 lbs is approximately 0.7189 or approximately 72%.

To solve this problem, we need to use the formula for the sampling distribution of the mean, which states that the mean of a sample of size n drawn from a population with mean μ and standard deviation σ is normally distributed with a mean of μ and a standard deviation of σ/sqrt(n).

In this case, we have a population of men with a mean weight of 178 lbs and a standard deviation of 26 lbs. We want to know the probability that 20 randomly selected men will have a mean weight between 170 and 185 lbs.

First, we need to calculate the standard deviation of the sampling distribution of the mean. Since we are taking a sample of size 20, the standard deviation of the sampling distribution is:

σ/sqrt(n) = 26/sqrt(20) = 5.82

Next, we need to standardize the interval between 170 and 185 lbs using the formula:

z = (x - μ) / (σ/sqrt(n))

For x = 170 lbs:

z = (170 - 178) / 5.82 = -1.37

For x = 185 lbs:

z = (185 - 178) / 5.82 = 1.20

Now we can use a standard normal distribution table (or a calculator) to find the probability of the interval between -1.37 and 1.20:

P(-1.37 < z < 1.20) = 0.8042 - 0.0853 = 0.7189

Therefore, the probability that 20 randomly selected men will have a mean weight between 170 and 185 lbs is 0.7189 or approximately 72%.

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Answer:

Sure, I can help you with that! First, let's simplify the expression inside the parentheses:

5 3/2 * 2 - 1/2 = 5 * 3 - 1/2 = 14.5

Now we can substitute this value back into the original expression and simplify:

(14.5)^2 = 210.25

Therefore, the simplified expression is 210.25.

Step-by-step explanation:

a) Samantha will have $20,359.68 in the account after the last $175 deposit is made in seven years.

b) Samantha makes 175-dollar deposits at the beginning of each month for seven years, she will have $21,372.77 in the account after the last deposit is made.

We can use the formula for the future value of an annuity with monthly compounding to solve this problem. The formula is:

FV = [tex]P * (((1 + r/n)^(n*t) - 1) / (r/n))[/tex]

Where:

FV is the future value of the annuity

P is the regular payment or deposit

r is the annual interest rate

n is the number of compounding periods per year (12 for monthly compounding)

t is the total number of years

a. If Samantha makes 175-dollar deposits at the end of each month for seven years, the total number of deposits she will make is:

7 years x 12 months/year = 84 deposits

The regular payment or deposit is P = $175, the annual interest rate is r = 5.4%, and the number of compounding periods per year is n = 12. The total number of years is t = 7.

Using the formula above, we can calculate the future value of the annuity:

FV = [tex]$175 *[/tex] [tex](((1 + 0.054/12)^(12*7) - 1) / (0.054/12))[/tex]

FV = [tex]$175 *[/tex] (((1.0045)[tex]^84 - 1[/tex]) / (0.0045))

FV =[tex]$175 * (116.2269)[/tex]

FV = $20,359.68

Therefore, Samantha will have $20,359.68 in the account after the last $175 deposit is made in seven years.

b. If Samantha makes 175-dollar deposits at the beginning of each month for seven years, we need to adjust the formula above to account for the timing of the deposits. One way to do this is to use the formula:

[tex]FV = P * (((1 + r/n)^(n*t) - 1) / (r/n)) * (1 + r/n)[/tex]

Where the additional factor (1 + r/n) accounts for the fact that the deposits are made at the beginning of each month.

Using this formula, we get:

FV = [tex]$175 * (((1 + 0.054/12)^(12*7)[/tex] - 1) / (0.054/12)) * (1 + 0.054/12)

FV = [tex]$175 * (((1.0045)^84 - 1)[/tex] / (0.0045)) * 1.0045

FV = $175 * (122.2837)

FV = $21,372.77

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