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3 years ago

13

Tirzah wants to put a fence around her garden. She has 22 yards Of fence material. Does she have enough to go all the way around

the garden? 4 8/12 6 9/12

1 answer:

CaHeK987 [17]3 years ago

3 0

No. She needs at least 31 yards of material.

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Ray opened a savings account with $600 at 4.5% interest compounded quarterly and kept it in the account for 2 years. How much wa

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600(1 +0.045/4)^8

Step-by-step explanation:

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21.39 + 9=x ok this seems easy but I'm stuck it is supposed to= 30.39 how do you get that answer

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Answer:

30.39

Step-by-step explanation:

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3. Your friend has eight sweatshirts. Three sweatshirts are green, one is white, and

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3 years ago

The function Q(x)=2x^2-kx+18. For what value of k does Q(x) have one distinct real solution? (NEEDS TO BE DONE WITHIN 2 HOURS!)

Aliun [14]

Answer:

k = 12

Step-by-step explanation:

Given:

The equation $Q(x)=2x^2-kx+18$

To find:

Value of $k = ?$ for which the given equation has one distinct real solution.

Solution:

The given equation is a quadratic equation.

There are always two solutions of a quadratic equation.

For the equation: $ax^{2} +bx+c=0$ to have one distinct solution:

$b^2 - 4ac = 0$

Here,

a = 2,

b = -k and

c = 18

Putting the values, we get:

$(-k)^2 - 4\times 2\times 18 = 0\\\Rightarrow k^2 = 18\times 8\\\Rightarrow k^2 =144\\\Rightarrow k = 12$

The equation becomes:

$Q(x)=2x^2-12x+18$

And the one root is:

$2(x^2-6x+9 ) = 0\\\Rightarrow 2(x-3)^2=0\\\Rightarrow x = 3$

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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.

4 0

3 years ago