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V125BC [204]
3 years ago
5

A car rental company charges a one-time application fee of 40 dollars, 55 dollars per day, and 13 cents per mile for its cars. A

) Write a formula for the cost, C, of renting a car as a function of the number of days, d, and the number of miles driven, m. B) If C = f(d, m), then f(5, 600) =
Mathematics
2 answers:
lara [203]3 years ago
8 0

Answer:

A) C = 40 + 55d + 0.13m

B) f(5,600) = $393

Step-by-step explanation:

First we need to convert all the units into dollars.

A car rental company charges:

an application fee of $40

$55 per day

13 cents = 13/100 = $0.13 per mile

A) The cost C of renting a car can be expressed as a function of number of days, d and the number of miles driven, m.

C = 40 + 55d + 0.13m

since $40 is a fixed cost which does not depend on time so we have added it as it is. The rest are added as a multiple of the parameter on which they depend i.e. days and miles.

B) C = f(d,m)

f(d,m) = 40 + 55d + 0.13m and we need to find f(5,600) i.e. here d = 5 and m = 600. So, we will substitute the values in the equation to find the total cost

f(5,600) = 40 + 55(5) + 0.13(600)

             = 40 + 275 + 78

f(5,600) = $393

liberstina [14]3 years ago
3 0

Answer:

A) C(d,m) = 40 + 55d + 0.13m

B) $448

Step-by-step explanation:

Let 'd' be the number of days and 'm' the number of miles driven.

A) The cost function that describes a fixed amount of $40, added to a variable amount of $55 per day (55d) and a variable amount of 13 cents per mile (0.13m) is:

C(d,m) = 40 +55d +0.13m

B) If  d = 5 and m =600, the total cost is:

C(5,600) = 40 +55*6 +0.13*600\\C(5,600)=\$448

The cost is $448.

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Starting in the 1970s, medical technology allowed babies with very low birth weight (VLBW, less than 1500 grams, about 3.3 pound
cluponka [151]

Answer:

The test statistic value using the VLBW babies as group 1 is z=-2.76±0.01 and the P-value for the test (±0.0001) is 0.0030.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.

Then, the null and alternative hypothesis are:

H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0

The significance level is 0.05.

The sample 1 (VLBW group), of size n1=244 has a proportion of p1=0.77049.

p_1=X_1/n_1=188/244=0.77049

The sample 2 (control group), of size n2=247 has a proportion of p2=0.79757.

p_2=X_2/n_2=197/247=0.79757

The difference between proportions is (p1-p2)=-0.02708.

p_d=p_1-p_2=0.77049-0.79757=-0.02708

The pooled proportion, needed to calculate the standard error, is:

p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{188+197}{244+247}=\dfrac{385}{491}=0.988

The estimated standard error of the difference between means is computed using the formula:

s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.988*0.012}{244}+\dfrac{0.988*0.012}{247}}\\\\\\s_{p1-p2}=\sqrt{0.00005+0.00005}=\sqrt{0.0001}=0.0098

Then, we can calculate the z-statistic as:

z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.02708-0}{0.0098}=\dfrac{-0.02708}{0.0098}=-2.76

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

P-value=P(t

As the P-value (0.0030) is smaller than the significance level (0.05), the effect issignificant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.

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