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2 years ago

What is the value of x+y? Eight grade math

Mathematics

1 answer:

atroni [7]2 years ago

6 0

Hello! The value of x+y is 12

Make sure to read an explanation so you can understand the solution better.

Any questions about the answer and explanation can be asked through comments.

Step-by-step explanation:

We can simply solve the system of equations by substituting x = 3y in the first equation

$\large{2x + 2y = 24} \\ \large{2(3y) + 2y = 24} \\ \large{6y + 2y = 24} \\ \large{8y = 24} \\ \large{y = 3}$

Therefore, the value of y is 3. Then we substitute the value of y in x = 3y.

$\large{x = 3y} \\ \large{x = 3(3)} \\ \large{x = 9}$

Therefore, the value of x is 9.

Since you want to find x+y, simply add 9 and 3 as we get 12.

Therefore x+y = 12.

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alexandr1967 [171]

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5 0

3 years ago

Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobil

o-na [289]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobile. A researcher is interested in finding out whether the variance of the annual repair costs also increases with the age of the automobile. A sample of 26 automobiles 4 years old showed a sample standard deviation for annual repair costs of $120 and a sample of 23 automobiles 2 years old showed a sample standard deviation for annual repair costs of $100. Let 4 year old automobiles be represented by population 1.

State the null and alternative versions of the research hypothesis that the variance in annual repair costs is larger for the older automobiles.

At a 0.01 level of significance, what is your conclusion? What is the p-value?

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.44

p-value = 0.1954

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

Step-by-step explanation:

Let σ₁² denotes the variance of 4 years old automobiles

Let σ₂² denotes the variance of 2 years old automobiles

State the null and alternative hypotheses:

The null hypothesis assumes that the variance in annual repair costs is smaller for older automobiles.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance in annual repair costs is larger for older automobiles.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic is given by

Test statistic = σ₁²/σ₂²

Test statistic = 120²/100²

Test statistic = 1.44

p-value:

The degree of freedom corresponding to 4 years old automobiles is given by

df₁ = n - 1

df₁ = 26 - 1

df₁ = 25

The degree of freedom corresponding to 2 years old automobiles is given by

df₂ = n - 1

df₂ = 23 - 1

df₂ = 22

Using Excel to find out the p-value,

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.44, 25, 22)

p-value = 0.1954

Conclusion:

When the p-value is less than the significance level then we reject the Null hypotheses

p-value < α (reject H₀)

But for the given case,

p-value > α

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

8 0

3 years ago

Can someone please help me on this

Karolina [17]

it is a because it is the line is in the spot we're it is ststop to go

8 0

2 years ago

Rochelle made three $25 withdrawals and then wrote a check for $100. If she started with $200 in her account, find the total amo

antoniya [11.8K]

$25. 25 x 3 is 75. Plus the 100 is 175. 200 - 175 is $25

4 0

3 years ago

Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a

tigry1 [53]

Answer:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

$p_v =P(z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

$\bar X =237$ represent the sample mean

$\sigma = 47$ represent the population deviation

$n =30$ represent the sample size selected

$\mu_0 = 247$ represent the value that we want to test.

The standard error for this case is given by:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

For the 90% confidence the value of the significance is given by $\alpha=1-0.9 = 0.1$ and $\alpha/2 = 0.05$ so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

$z_{\alpha/2}= 1.64$

And the margin of error would be:

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

The confidence interval for this case would be given by:

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

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".

Calculate the statistic

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

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

P-value

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

$p_v =P(z$

Conclusion

5 0

2 years ago