Write and solve proportion answer the question.

Rewrite in simplest radical form! Show your work.

vazorg [7]

$x^{\frac{1}{\frac{-3}{6} }}$

First, let's deal with the fraction in the denominator of the exponent. Multiply the top and bottom of the exponent by 6.

$x^{\frac{6}{-3} }$

Now that the fraction in the denominator is taken care of, we can reduce the denominator.

$x^{-2}$ . Some professors might accept this as simplest form, but others might ask you to get rid of the negative.

$x^{-2} = \frac{1}{x^{2} }$

7 0

3 years ago

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-9b-6=-3b +48<br> Step by step

Travka [436]

Answer:

b = -9

Step-by-step explanation:

1. Move the variables to the left side. Be sure to change the terms (add/subtract)

-9b-6 = -3b+48

-9b+3b-6 = 48

2. Combine like terms.

-9b+3b-6 = 48

-6b = 48 + 6

3. Divide both sides by -6.

-6b = 54

b = -9

7 0

3 years ago

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A sample of 100 workers located in Atlanta has an average daily work time of 6.5 hours with a standard deviation of 0.5 hours. A

Readme [11.4K]

Answer:

$t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398$

$df = n_1 +n_2 -2= 100+110-2= 208$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{208}$

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=6.5$ represent the sample mean for Atlanta

$\bar X_{2}=6.7$ represent the sample mean for Chicago

$s_{1}=0.5$ represent the sample deviation for Atlanta

$s_{2}=0.7$ represent the sample standard deviation for Chicago

$n_{1}=100$ sample size for the group Atlanta

$n_{2}=110$ sample size for the group Chicago

t would represent the statistic (variable of interest)

$\alpha=0.01$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the meanfor atlanta is different from the mean of Chicago, the system of hypothesis would be:

Null hypothesis: $\mu_{1}=\mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398$

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

$df = n_1 +n_2 -2= 100+110-2= 208$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{208}$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

3 0

3 years ago

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Could someone help me for this?

lina2011 [118]

Answer:

x = - $\frac{5}{3}$ , x = $\frac{5}{2}$