Choose the correct simplification of the expression (7ab^3)^2

4) The cost of membership at local country club is $125 per family member per month. Each family also has to pay $75 in service

lyudmila [28]

C(x)=125x+75
You multiply 125 by how many family members and and the 75

6 0

4 years ago

Melissa bought 2 shirts and 3 skirts and spent between $45 and $60. Each shirt costs the same amount. Each skirt costs the same

sukhopar [10]

Answer:

Least is 7

Most is 12

Step-by-step explanation:

So 2 shirts for 12.

So 24

Take 24 off of both

Left with 21 to 36

21÷3 is 7

36÷3 is 12

Hope this helps

6 0

3 years ago

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True or false. <br> -5+ ( -3 +( -8) = -5 + ( - 3 + 8 )

ivanzaharov [21]

False:

-5 + -3 + (-8) = -16

-5 + -3 + 8 = 0

hope this helps

8 0

4 years ago

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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

4 years ago

Isaak is writing an explicit formula to represent the sequence.

gulaghasi [49]

In a geometric progression:
a, b, c, d...
The common ratio can be obtained using:
b/a = c/b = d/c
b/a = 112 / 64 = 1.75
c/b = 196 / 112 = 1.75
d/c = 343 / 196 = 1.75
The common ratio = 1.75

7 0

3 years ago

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