Ask question

Ask question

All categories

4 years ago

14

Suppose the total benefit of watching 1 baseball game is 100, the total benefit of watching 2 games is 120, and the total benefi

t of watching 3 games is 125. In this case, the marginal benefit of watching the 3rd game is:A 5B 125C 41.67D 375

1 answer:

77julia77 [94]4 years ago

4 0

Answer:

A. 5

Step-by-step explanation:

The marginal benefit of an article, $B_m$ is calculated as:

$B_m=\frac{DB_t}{DQ}$

Where $DB_t$ is the change in the total benefit of the article, and $DQ$ is the change in the units of the article.

If we want to know the marginal benefit of watching the 3rd game, we need to use the information of total benefit of watching 2 games and 3 games.

In this case, the total benefit of watching 2 games is 120 and the total benefit of watching 3 games is 125, so $DB_t$ and $DQ$ are equal to:

$DB_t=125-120=5$

$DQ = 3 - 2 = 1$

Then, the marginal benefit of watching the third game is:

$B_m=\frac{5}{1}=5$

You might be interested in

According to Padgett Business Services, 20% of all small-business owners say the most important advice for starting a business i

elena-s [515]

Answer:

a) There is a 6.88% probability that none of the owners would say preparing for long hours and hard work is the most important advice.

b) There is a 1.93% probability that six or more owners would say preparing for long hours and hard work is the most important advice.

c) There is a 10.32% probability that exactly five owners would say having good financing ready is the most important advice.

d) The expected number of owners who would say having a good plan is the most important advice is 2.28

Step-by-step explanation:

Questions a), b), c) are all solved as binomial distribution problems.

Question d) is a simple calculation.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

For these problems

12 business owners were contacted, so $n = 12$ .

a. What is the probability that none of the owners would say preparing for long hours and hard work is the most important advice?

20% of all small-business owners say the most important advice for starting a business is to prepare for long hours and hard work, so $\pi = 0.2$ .

That is P(X=0) when $\pi = 0.2$ .

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{12,0}.(0.2)^{0}.(0.8)^{12} = 0.0688$

There is a 6.88% probability that none of the owners would say preparing for long hours and hard work is the most important advice.

b. What is the probability that six or more owners would say preparing for long hours and hard work is the most important advice?

20% of all small-business owners say the most important advice for starting a business is to prepare for long hours and hard work, so $\pi = 0.2$ .

This is:

$P = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 6) = C_{12,6}.(0.2)^{6}.(0.8)^{6} = 0.0155$

$P(X = 7) = C_{12,7}.(0.2)^{7}.(0.8)^{5} = 0.0033$

$P(X = 8) = C_{12,8}.(0.2)^{8}.(0.8)^{4} = 0.0005$

$P(X = 9) = C_{12,9}.(0.2)^{9}.(0.8)^{3} = 0.00006$

$P(X = 10) = C_{12,10}.(0.2)^{10}.(0.8)^{2} = 0.000004$

$P(X = 11) = C_{12,11}.(0.2)^{11}.(0.8)^{1} = 0.0000002$

$P(X = 12) = C_{12,12}.(0.2)^{12}.(0.8)^{0} = 0.000000004$

So:

$P = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.0155 + 0.0033 + 0.0005 + 0.000006 + 0.000004 + 0.0000002 + 0.000000004 = 0.0193$

There is a 1.93% probability that six or more owners would say preparing for long hours and hard work is the most important advice.

c. What is the probability that exactly five owners would say having good financing ready is the most important advice?

Twenty-five percent say the most important advice is to have good financing ready, so $\pi = 0.25$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 5) = C_{12,5}.(0.25)^{5}.(0.75)^{7} = 0.1032$

There is a 10.32% probability that exactly five owners would say having good financing ready is the most important advice.

d. What is the expected number of owners who would say having a good plan is the most important advice?

Nineteen percent say having a good plan is the most important advice, so that is $0.19*12 = 2.28$

The expected number of owners who would say having a good plan is the most important advice is 2.28

4 0

3 years ago

10n + 7 > n - 3 (5 - 3n)<br><br> plz

Ad libitum [116K]

Are you trying to figure out the value of n?

4 0

3 years ago

Read 2 more answers

What is the common denomination for 3/10 and 9/8

4vir4ik [10]

Answer:

80

Step-by-step explanation:

because if you want to find a common denimeter we have mulitly the deometer to find the answer

sorry if it is wrong

8 0

3 years ago

Verify that cos squared A plus Sin squared A is equal to 1 if A is equal to 90 degrees

Scilla [17]

Answer:

see explanation

Step-by-step explanation:

To verify cos²A + sin²A = 1 with A = 90° , then

cos²90° + sin²90°

= (0)² + (1)²

= 0 + 1

= 1

3 0

3 years ago

My question is in the attachment.<br>

Nikolay [14]

Answer:

5 and 25

Step-by-step explanation:

let the son's age be x then the father's age is x + 20

In 5 years

son = x + 5 and father = x + 20 + 5 = x + 25

Then

x + 25 = 3(x + 5) ← father is three times as old as son

x + 25 = 3x + 15 ( subtract x from both sides )

25 = 2x + 15 ( subtract 15 from both sides )

10 = 2x ( divide both sides by 2 )

5 = x and x + 20 = 5 + 20 = 25

si=on is 5 and father is 25

8 0

3 years ago

Read 2 more answers