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s2008m [1.1K]
2 years ago
11

In Drew's town the average price for a tutor is $15 per hour, and the standard deviation is $5.25 per hour. In Drew's town, what

is the probability that a tutor charges between $10 and $20 per hour? Assume that tutoring prices are normally distributed.
Mathematics
1 answer:
alukav5142 [94]2 years ago
5 0

Answer:

0.659 is the  probability that a tutor charges between $10 and $20 per hour.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $15 per hour

Standard Deviation, σ = $5.25 per hour

We are given that the distribution of tutoring prices is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P( tutor charges between $10 and $20 per hour)

P(10 \leq x \leq 20)\\\\ = P(\displaystyle\frac{10 - 15}{5.25} \leq z \leq \displaystyle\frac{20-15}{5.25})\\\\ = P(-0.9523 \leq z \leq 0.9523)\\\\= P(z \leq 0.9523) - P(z < -−0.9523)\\= 0.8295 - 0.1705= 0.659

0.659 is the  probability that a tutor charges between $10 and $20 per hour.

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The equations are:

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Step-by-step explanation:

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T = 2Y and T + 2Y = 15

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By definition, the average rate of change is given by:

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We evaluate each of the functions in the given interval.

We have then:

For f (x) = x ^ 2 + 3x:

Evaluating for x = -2:

f (-2) = (-2) ^ 2 + 3 (-2)\\f (-2) = 4 - 6\\f (-2) = - 2

Evaluating for x = 3:

f (3) = (3) ^ 2 + 3 (3)\\f (3) = 9 + 9\\f (3) = 18

Then, the AVR is:

AVR = \frac{18-(-2)}{3-(-2)}

AVR = \frac{18+2}{3+2}

AVR = \frac{20}{5}

AVR = 4


For f (x) = 3x - 8:

Evaluating for x =4:

f (4) = 3 (4) - 8\\f (4) = 12 - 8\\f (4) = 4

Evaluating for x = 5:

f (5) = 3 (5) - 8\\f (5) = 15 - 8\\f (5) = 7

Then, the AVR is:

AVR = \frac{7-4}{5-4}

AVR = \frac{3}{1}

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For f (x) = x ^ 2 - 2x:

Evaluating for x = -3:

f (-3) = (-3) ^ 2 - 2 (-3)\\f (-3) = 9 + 6\\f (-3) = 15

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f (4) = (4) ^ 2 - 2 (4)\\f (4) = 16 - 8\\f (4) = 8

Then, the AVR is:

AVR = \frac{8-15}{4-(-3)}

AVR = \frac{-7}{4+3}

AVR = \frac{-7}{7}

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For f (x) = x ^ 2 - 5:

Evaluating for x = -1:

f (-1) = (-1) ^ 2 - 5\\f (-1) = 1 - 5\\f (-1) = - 4

Evaluating for x = 1:

f (1) = (1) ^ 2 - 5\\f (1) = 1 - 5\\f (1) = - 4

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AVR = \frac{-4-(-4)}{1-(-1)}

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AVR = \frac{0}{2}

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from the greatest to the least value based on the average rate of change in the specified interval:


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f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]


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