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hammer [34]
2 years ago
14

Solve the following using the elimination method and write the value of x.

Mathematics
1 answer:
yawa3891 [41]2 years ago
4 0

Answer:

the first one is ×=-2-3/5

the second one is ×=1-3y/2

hope it helps

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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
3 years ago
What is the value of d and w?
Anettt [7]

Answer:

d = 65

w = 55

Step-by-step explanation:

w + 90 + 35 = 180

w + 125 = 180

w = 55

35 + x = 60 (Vertical angle thm)

x = 25

d + x = 90

d + 25 = 90

d = 65

3 0
3 years ago
85:245 <br> Gente preciso da conta armada,pra agr pfv
goblinko [34]

Answer:

BOYBOYOBOYOBYOBYOYOBYOBYOBYOBYOBOBYOBYOBY

Step-by-step explanation:

4 0
3 years ago
I have a Brainliest please help
Basile [38]
1) Preparing questions about the company
2) Accountant
3) The end result of an employees work behavior.
4) A

Hope this helps :)
8 0
3 years ago
Dylan has 765 cards in his baseball card collection. He sells 153 of the cards. What is the percent of decrease in the number of
AlladinOne [14]
Simply divide 153 by 765 to get the percent. This equals .2

ANSWER 20%
6 0
3 years ago
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