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ludmilkaskok [199]
3 years ago
8

Luisa earns money mowing her neighbors' lawns.

Mathematics
2 answers:
Irina-Kira [14]3 years ago
4 0

Answer:

p(x)=16x-25

C is the correct option.

Step-by-step explanation:

We have been given that

r(x)=20x\\c(x)=4x+25

The profit is the difference of revenue and cost function.

Mathematically, we can write profit function as

p(x)=(r-c)x

We know that (f-g)(x)=f(x)-g(x). Thus, we have

p(x)=r(x)-c(x)

Plugging, the r(x) and c(x) functions

p(x)=20x-(4x+25)

Distribute negative over the parenthesis

p(x)=20x-4x-25

Combine the like terms. 20x-4x = 16x

p(x)=16x-25

C is the correct option.

dedylja [7]3 years ago
4 0

Answer:

C. p(x) =  16x - 25

Step-by-step explanation:

1. First define the equation that describes the profit Luisa earns, that is the revenue for mowing minus the Luisa´s cost for gas and the mower rental. So the equation is:

p(x) = (r – c)(x) (Eq.1)

2. Write the equations for the revenue for mowing and the cost for gas and the mower rental:

- Equation for the revenue for mowing:

r(x) = 20x (Eq.2)

- Equation for the cost for gas and the mower rental:

c(x) = 4x + 25 (Eq.3)

3. Replace the equations Eq. 2 and Eq. 3 in Eq. 1:

p(x) = (r – c)(x)

p(x) = 20x - (4x + 25)

p(x) = 20x - 4x - 25

p(x) = 16x - 25

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p_v represent the p value for the test (variable of interest)  

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If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

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We can replace in formula (1) the info given like this:  

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

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The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(1699)}>1.971)=0.049  

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If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

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