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yKpoI14uk [10]
3 years ago
10

Justin started a lawn service business. His rate to maintain an average size lawn is $35. It costs Justin $750 to start the busi

ness. And he estimates that his cost per lawn is $1.50. Write the cost function in terms of c(x) the revenue terms of r(x) and the profit function in terms p(x)
Mathematics
1 answer:
Ronch [10]3 years ago
4 0

Answer:

Step-by-step explanation:

Cost function: Fixed costs+ Variable costs

In this case Justin will have to pay $750 to star the business, this is his fixed cost. And then, he will have to pay $1,50 per each lawn, this is his variable cost because it depends on the number of lawns he sells.

C(x)= $750+$1,50x

Revenue function (R(x)):

R(x)= Price * Number of units sold (x)

R(x)= $35*x

Profit function ((P(x))= Revenue function (R(x))-Cost function (C(x))

P(x)= $35x- [$750+$1,50x]

P(x)= $33,5-$750

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The first three terms of a sequence are given. Round to the nearest thousandth (if necessary).21,42,84,...Find the 7th term.
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Answer:

21,42,84,168,336,672, 1,344

Step-by-step explanation:

The pattern is to multiply the previous term by 2 to find the next term. The seventh term should be 1,344.

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Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a
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Answer:

Mean = 68.9

s^2 =18.1 --- Variance

Step-by-step explanation:

Given

\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5

For 59 - 66

x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5

For 67 - 74

x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5

For 75 - 82

x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5

For 83 - 90

x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5

So, the table becomes:

\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

The mean is then calculated as:

Mean = \frac{\sum fx}{\sum f}

Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}

Mean = \frac{2547.5}{37}

Mean = 68.9 -- approximated

Solving (b): The sample variance:

This is calculated as:

s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}

So, we have:

s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}

s^2 =\frac{652.8}{36}

s^2 =18.1 -- approximated

5 0
2 years ago
According to the National Association of Colleges and Employers, the average starting salary for new college graduates in health
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Answer:

a) The probability that a new college graduate in business will earn a starting salary of at least $65,000 is P=0.22965 or 23%.

b) The probability that a new college graduate in health sciences will earn a starting salary of at least $65,000 is P=0.11123 or 11%.

c) The probability that a new college graduate in health sciences will earn a starting salary of less than $40,000 is P=0.14686 or 15%.

d) A new college graduate in business have to earn at least $77,133 in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences.

Step-by-step explanation:

<em>a. What is the probability that a new college graduate in business will earn a starting salary of at least $65,000?</em>

For college graduates in business, the salary distributes normally with mean salary of $53,901 and standard deviation of $15,000.

To calculate the probability of earning at least $65,000, we can calculate the z-value:

z=\frac{x-\mu}{\sigma} =\frac{65000-53901}{15000} =0.74

The probability is then

P(X>65,000)=P(z>0.74)=0.22965

The probability that a new college graduate in business will earn a starting salary of at least $65,000 is P=0.22965 or 23%.

<em>b. What is the probability that a new college graduate in health sciences will earn a starting salary of at least $65,000?</em>

<em />

For college graduates in health sciences, the salary distributes normally with mean salary of $51,541 and standard deviation of $11,000.

To calculate the probability of earning at least $65,000, we can calculate the z-value:

z=\frac{x-\mu}{\sigma} =\frac{65000-51541}{11000} =1.22

The probability is then

P(X>65,000)=P(z>1.22)=0.11123

The probability that a new college graduate in health sciences will earn a starting salary of at least $65,000 is P=0.11123 or 11%.

<em>c. What is the probability that a new college graduate in health sciences will earn a starting salary less than $40,000?</em>

To calculate the probability of earning less than $40,000, we can calculate the z-value:

z=\frac{x-\mu}{\sigma} =\frac{40000-51541}{11000} =-1.05

The probability is then

P(X

The probability that a new college graduate in health sciences will earn a starting salary of less than $40,000 is P=0.14686 or 15%.

<em />

<em>d. How much would a new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences?</em>

The z-value for the 1% higher salaries (P>0.99) is z=2.3265.

The cut-off salary for this z-value can be calculated as:

X=\mu+z*\sigma=51,541+2.3265*11,000=51,541+25,592=77,133

A new college graduate in business have to earn at least $77,133 in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences.

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