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storchak [24]
4 years ago
15

Keisha is a software saleswoman. Let y represent her total pay (in dollars). Let x represent the number of copies of English is

Fun she sells. Suppose that x and y are related by equation 90x+2500=y.
Answer the questions below.
Note that a change can be an increase or a decrease.
For an increase, use a positive number. For a decrease, use a negative number.

What is the change in Keisha's total pay for each copy of English is fun she sells?

What is Kiesha's total pay if she doesn't sell any copies of English is fun?
Mathematics
1 answer:
ad-work [718]4 years ago
7 0

Answer:

Step-by-step explanation:

Given that suppose x and y are related by equation 90x+2500=y.

Let y represent her total pay (in dollars). Let x represent the number of copies of English is Fun she sells.

a) change in Keisha total pay for each copy of English is fun she sells=y'=90

b) If she doesn't sell any copies of English is fun, then x=0

y (0) = 2500

She will get 2500 even if she does not sell any copy.

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In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume
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Answer:

The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018

Step-by-step explanation:

The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula

W = \frac{X - \mu}{\sigma} = \frac{X - 182}{6.5465}

In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.

The cummulative distribution function of W will be denoted by \phi . The values of \phi can be found in the attached file.

P(X > 201.0476) = P(\frac{X-182}{6.5465} > \frac{201.0476 - 182}{6.5465}) = P(W > 2.91) = 1-\phi(2.91) = \\1-0.9982 = 0.0018

Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.  

Download pdf
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