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3 years ago

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Lily needed to get her computer fixed. She took it to the repair store. The technician at the store worked on the computer for 3

.5 hours and charged her $159 for parts. The total was $456.50. Write and solve an equation which can be used to determine x x, the cost of the labor per hour.

2 answers:

iragen [17]3 years ago

6 0

Answer
456.50=159 + 3.5x
OR
3.5x=456.50-159

Reason
456.50 is the total
159 is for the parts
3.5 is the number of hours

So to find the cost you have to find out how much it cost for the labor by subtracting the total cost from the cost of the parts. To find the cost of labor for each hour you have to divide the total cost of labor by the number of hours used to work on the computer

Ber [7]3 years ago

5 0

Answer:

456.50 = 3.5x + 159, $85 per hour for labor

Step-by-step explanation:

456.50 = 3.5x + 159

subtract 159 from both sides

297.50 = 3.5x

divide by 3.5 on both sides to isolate x

x = 85

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Mia wants to purchase a pair of jeans. The first time she saw the jeans, they were priced at $36. When she goes back to make the

jasenka [17]

Answer: D

Step-by-step explanation:

1. Subtract 36 to 28.80

2. Divide the answer to the first price (36.00)- 7.5 divided by 36.00 or 36

3. When you calculate and find the answer as a decimal, then make it into a percentage

So if... you divide 7.5 divided by 36, you get the decimal of 0.2, and when you move the decimals two places only, you get your answer of 20%.

Hopefully it helped you!. Have a good day :-).

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3 years ago

Suppose that a random sample of 10 newborns had an average weight of 7.25 pounds and sample standard deviation of 2 pounds. a. T

frosja888 [35]

Answer:

$z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565$

$p_v =P(Z$

a) If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.

b) $\chi^2 =\frac{10-1}{1.96} 4 =18.367$

$p_v =P(\chi^2 >18.367)=0.0311$

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation

$\bar X=7.25$ represent the sample mean

$s=1.2$ represent the sample standard deviation

$\sigma=1.4$ represent the population standard deviation

$n=10$ sample size

$\mu_o =7.5$ represent the value that we want to test

$\alpha=0.05,0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

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

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:

Null hypothesis: $\mu \geq 7.5$

Alternative hypothesis: $\mu < 7.5$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

3) Calculate the statistic

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

$z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565$

4)P-value

Since is a left tailed test the p value would be:

$p_v =P(Z$

5) Conclusion

Part a

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.

Part b

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=10$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s^2 =4$ represent the sample variance obtained

$\sigma^2_0 =1.96$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \leq 1.96$

Alternative hypothesis: $\sigma^2 >1.96$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{10-1}{1.96} 4 =18.367$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2 >18.367)=0.0311$

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(18.367,9,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

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Order from least to greatest<br> 45<br> 11, 45, 520, 54

pentagon [3]

Answer: 11, 45, 45, 54, 520

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agasfer [191]

Step-by-step explanation:

step 1. 1-12/20 = 1.6

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Sonja [21]

Answer:

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Step-by-step explanation:

1. the formula of the required area is:

$A=\frac{1}{2} *side_1*side_2*sin45;$

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