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professor190 [17]
3 years ago
11

Otis tutors math students on sunday afternoons. he charges $35 for each of the first 10 students. he changed $40 for each additi

onal student the maximum number of students he will accept is 20.
answer each question below:

a. what did otis make if 8 students came on the first sunday?

b. what did otis made if 17 students came of the second sunday

c. if a is the amount that otis makes from S students that he tutors, write A as a piece wise function of S
Mathematics
1 answer:
lions [1.4K]3 years ago
7 0
For A it’ll be $280
B-$630
C- I don’t know
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Carol is ordering t-shirts for her club and there are two different patterns to choose from. The first shirt is green and blue s
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Answer:

14 Striped and 10 Flowered

Step-by-step explanation:

This can best be determined using a set of linear equations that are solved simultaneously.

This pair of linear equations may be solved simultaneously by using the elimination method. This will involve ensuring that the coefficient of one of the unknown variables is the same in both equations. It may be solved by substitution in that one of the variable is made the subject of the equation and the result is substituted into the second equation

Given that the green and blue striped shirt is $15 and the white with purple flowers is $13. She needs to order 24 shirts and has a total of $340 to spend, let the number of striped shirts be g and that of flowered be h then,

g + h = 24 and

15g + 13h = 340

g = 24 - h

15(24 - h) + 13h = 340

360 - 15h + 13h = 340

2h = 20

h = 10

g = 24 - h

g = 24 - 10

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What is 12- (-8) +7 in integers
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A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
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Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean \mu and standard deviation \sigma, we have these following probabilities

Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827

Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545

Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545

Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973

-----------

Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0
3 years ago
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