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BlackZzzverrR [31]
1 year ago
13

Can someone help me with this problem from Khan Academy?

Mathematics
1 answer:
ANTONII [103]1 year ago
5 0

Given that the graph shows P as a function of n, we have:

a. The more hot dogs that is sold, his profit increases.

b. Jimmy sells the hot dog for $1 each.

c. Jimmy needs to sell 16 hot dogs to recover the $8 he invested.

d. He would make a profit of $7.

e. The number of hot dogs he needs to sell is: 20.

<h3>What is a Linear Function?</h3>

A linear  function is expressed as, y = mx + b, where y is a function of x, m is the unit rate, and b is the starting value or initial value of the function.

We are given that profit, P, is a function of the number of hot dogs sold, n.

a. The graph of the function slopes upwards, so this means that the more hot dogs that is sold, his profit increases.

b. Cost of 1 hot dog = Unit rate = change in y / change in x = 2 units/2 units

Cost of 1 hot dog = Unit rate = 1

Thus, Jimmy sells the hot dog for $1 each.

c. If you trace profit (P) on the graph when it is $8, the corresponding number of hot dogs, n, would be 16.

So, Jimmy needs to sell 16 hot dogs to recover the $8 he invested.

d. When n = 15, P, from the graph would be 7.

Thus, if Jimmy sells 15 hot dogs, he would make a profit of $7.

e. When P = 12, from the graph, the corresponding n value is 20.

Therefore for Jimmy to make $12, the number of hot dogs he needs to sell is: 20.

Learn more about linear function on:

brainly.com/question/15602982

#SPJ1

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Answer:

The probability that there are 3 or less errors in 100 pages is 0.648.        

Step-by-step explanation:

In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.

For the given Poisson distribution the mean is p = 0.03 errors per page.

We have to find the probability that there are three or less errors in n = 100 pages.

Let us denote the number of errors in the book by the variable x.

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     P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)

                 

   Using the formula for Poisson distribution for P(x = X ) = \frac{e^{-\lambda}\lambda^X}{X!}

Therefore P( X ≤ 3) = \frac{e^{-3} 3^0}{0!} + \frac{e^{-3} 3^1}{1!} + \frac{e^{-3} 3^2}{2!} + \frac{e^{-3} 3^3}{3!}

                                 = 0.05 + 0.15 + 0.224 + 0.224

                                 = 0.648

 The probability that there are 3 or less errors in 100 pages is 0.648.                                

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