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goldenfox [79]
3 years ago
5

(b) What is the probability that distance exceeds the mean distance by more than 2 standard deviations?

Mathematics
1 answer:
Flura [38]3 years ago
7 0

Answer:

>0.95

Step-by-step explanation:

This probably refers to a normal distribution. In a normal distribution, according to the empirical rule,

68% of observed data falls within 1 standard deviation of the mean

95% falls within 2 standard deviations of the mean

99.7% falls within 3 standard deviations of the mean

In this case, the observed data is distance and the average figure for all observed distances is the mean distance.

So, the probability that a/any distance observed exceeds the mean distance by more than 2 standard deviations is >95%

In other words, the probability is >0.95

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A company estimates that its cost and revenue can be modeled by the functions C)-0.75x+20,000
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P(x)  = -0.75x + 20000

Therefore, the profit function exists -0.75x + 20000.

<h3>How to find profit function?</h3>

The profit function can be estimated by subtracting the cost function from the revenue function. Let profit be expressed as P(x), the revenue as R(x), the cost as C(x), and x as the number of items traded. Then the profit function exists noted as P(x) = R(x) - C(x).

Given:

C(x) = -0.75x+20,000 and R(x)= -1.50x

P(x) = R(x) - C(x)

= -1.50x - (-0.75)x + 20,000

= -1.50x + 0.75x + 20,000

Apply rule -(-a) = a

= -1.5x + 0.75x + 20000

Add similar elements:

-1.5 x + 0.75x = -0.75x

P(x)  = -0.75x + 20000

Therefore, the profit function exists -0.75x + 20000.

To learn more about profit function refer to:

brainly.com/question/16866047

#SPJ9

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