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o-na [289]
2 years ago
14

Last year, Ishiro Novato drove 21,986 miles and had these expenses: gas $1,187.25; maintenance and miscellaneous $118.80; parkin

g and tolls $125.00; tires $300.00; wash and wax $75; insurance $614.80; license $34.90; and depreciation $1,780.00
Cost per Mile = Annual Variable Cost + Fixed Cost
Number of Miles Driven
it would mean a lot to me if someone would answer this for me
Mathematics
1 answer:
Alisiya [41]2 years ago
3 0

Answer:

  $0.187 per mile

Step-by-step explanation:

The sum of the expenses for the year is ...

  $1,187.25 +$118.80 +$300.00 +$75 +$614.80 +$34.90 +$1,780.00

  = $4110.75

Then the cost per mile is ...

  cost/miles = $4110.75 / 21986 ≈ $0.18697

The cost per mile is about $0.187, or 18.7¢ per mile.

__

If you need whole cents, that's about $0.19 per mile.

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6.<br> 5. ¿Qué número tiene el mismo<br> valor que l centena, 7 decenas?
pashok25 [27]

Answer:

7 Decenas = 70 Unidades espero que te ayude

Step-by-step explanation:

5 0
3 years ago
Ayana has 220 coins in her piggy bank Of those 40percent are pennies. Out of the coins that are not pennies 25 percent are partn
Rzqust [24]
To find a percent of a number, multiply the percent by the number.

First we find the number of pennies.

40% of 220 =

= 40% * 220

= 0.4 * 220

= 88

88 coins are pennies.

220 - 88 = 132

132 coins are not pennies.

Now we find 25% of 132.

25% of 132 =

= 25% * 132

= 0.25 * 132

= 33

33 coins are partners.

(Did you mean quarters instead of partners?)
3 0
3 years ago
To multiply a monomial by a polynomial, use the Property. Using this property, the product of (2k2 – 7k + 3) and 4k is .
Daniel [21]

Answer:

4k(k-3)(2k-1)

Step-by-step explanation:

3 0
3 years ago
Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test
BaLLatris [955]

Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\  \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}

- Applying the formula, we have:

\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\  \\ t=\frac{5}{3.1752} \\  \\ t\approx1.5747 \end{gathered}

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis

7 0
1 year ago
Please help with this question.
katrin2010 [14]
The answer is 10 minutes
7 0
3 years ago
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