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Mariulka [41]
2 years ago
5

Evaluate: 216 1/2 Question 2 options: A) 9 B) 1 C) 6 D) -3

Mathematics
2 answers:
Phantasy [73]2 years ago
8 0
B is the answerr okkkkkkk
zaharov [31]2 years ago
5 0
B, i just know, no context needed but whatever
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PLS HELP ME, PLS THIS IS DUE FIRST CORRECT ANSWER GETS BRAINLIEST​
zepelin [54]

Answer: C. 29 3/4

Ok done. Thank to me :>

8 0
2 years ago
Read 2 more answers
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 help on this.
tiny-mole [99]

Step-by-step explanation:

Given

22m - 33 = 11( 2m) - 11(3)

= 11 ( 2m - 3 )

Hope it helps :)

3 0
3 years ago
(HELP!)
denis-greek [22]

Let X be the random variable denoting the number of girls that are born. Then X has a binomial distribution on n=125 babies and probability of getting a girl p=0.5. The probability we want is

\mathbb P(X\ge60)=\displaystyle\sum_{x=60}^{125}\binom{125}x0.5^{125-x}0.5^x\approx0.704=70.4\%

7 0
3 years ago
Read 2 more answers
14. Which shows how to find 5 x 72 mentally by using the Distributive
Sholpan [36]
The answer to the question is B
Hope this helps :)
5 0
3 years ago
Read 2 more answers
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