1/3+a=5/4 what do a equal

Solve the following equation for X: 1/2(5x+12)=2x-3 show your work

Debora [2.8K]

1/2(5x + 12) = 2x - 3
5/2 x + 6 = 2x - 3
5/2 x - 2x = -3 - 6 = -9
5x = 2 x -9 = -18
x = -18/5

5 0

2 years ago

Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average

aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let $\mu_{1}-\mu_{2}$ be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes $n_{1} = 40$ and $n_{2} = 35$ , the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$ , i.e., 699-682 = 17.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$ , i.e.,

$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{17}{9.8198} = 1.7312$ . Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for $\mu_{1}-\mu_{2}$ is given by $17\pm (z_{0.05/2})9.8198$ , i.e., $17\pm (z_{0.025})9.8198$ where $z_{0.025}$ is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0

3 years ago

I need to equation for the radius of a 400m running track I know the answer is 36.5 and I know the formula is =2<img src="https:

miskamm [114]

This formula only applies if the track is circular. If it is, then we write the equation:

2*r*pi=400

Dividing by 2pi, we see that:

r=200/pi

This is approximately 63.66.

6 0

3 years ago

Which statement describes a number with the smallest absolute value?

Snezhnost [94]

I believe C. The temperature is -23 degrees.

This is because, a negative (absolute value) becomes positive

Which would lead for |-23| to end up being |23|

Anytime a negative absolute value is placed, it becomes a positive and a positive stays positive.

So the smallest would be C then B later D and at last A

4 0

3 years ago

Using the formula , find four ordered pairs <br> F=1.8c+32

Crank

(5,41)
(-5,23)
(-10,14)
(-15,5)

4 0

2 years ago