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3 years ago

The cost of g gallons of gasoline can be modeled by C(g) = 2.25g

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

5 0

Answer:

C(g) = 27

Step-by-step explanation:

2.25(12) = 27

You just plug in your g value and solve.

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What is the answer to q divided by 1/3= 1 1/2

katrin2010 [14]

The answer is q = 3/6.

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3 years ago

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83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

$\chi^2_{\alpha/2}=104.139$

$\chi^2_{1- \alpha/2}=62.132$

The confidence interval is given by:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

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3 years ago

Why is the opposite of a negative number a positive number?

andreev551 [17]

I don't really know how to explain but another example would be the opposite of multiplying is dividing.

6 0

4 years ago

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Linear or non linear

nevsk [136]

Answer:

Linear.

Step-by-step explanation:

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3 years ago

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Find the amount in the bank after 7 years if interest is compounded quarterly?

qwelly [4]

Answer:

a) amount in the bank after 7 years if interest is compounded quarterly is $6,605

b) amount in the bank after 7 years if interest is compounded quarterly is $6,612.57

Step-by-step explanation:

We are given:

Principal Amount P= 5000

Rate r= 4% = 0.04

time t = 7 years

The formula used is: $A=P(1+\frac{r}{n})^{nt}$

where A is future value, P is principal amount, r is rate, n is compounded value and t is time

a) Find the amount in the bank after 7 years if interest is compounded quarterly?

If interest is compounded quarterly then n = 4

Using values given in question and finding A

$A=P(1+\frac{r}{n})^{nt}\\A=5000(1+\frac{0.04}{4})^{4*7} \\A=5000(1+0.01)^{28}\\A=5000(1.01)^{28}\\A=5000(1.321)\\A=6,605$

So, amount in the bank after 7 years if interest is compounded quarterly is $6,605

b) Find the amount in the bank after 7 years if interest is compounded monthly?

If interest is compounded quarterly then n = 12

Using values given in question and finding A

$A=P(1+\frac{r}{n})^{nt}\\A=5000(1+\frac{0.04}{12})^{12*7} \\A=5000(1+0.003)^{84}\\A=5000(1.003)^{84}\\A=5000(1.322)\\A=6,612.57$

So, amount in the bank after 7 years if interest is compounded quarterly is $6,612.57

6 0

3 years ago