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DerKrebs [107]
3 years ago
12

What is the value of (Negative three-fourths) Superscript negative 4?

Mathematics
2 answers:
Goryan [66]3 years ago
8 0

Answer:

-3.16049382716

Step-by-step explanation:

-3/4 = -0.75

-0.75 ^ -4 = -3.16049382716

ZombieBoy
2 years ago
Ah yes, Pi is the answer
svetoff [14.1K]3 years ago
5 0

D) 256/81

I just took the quiz on edge

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A random sample of n 1 = 249 people who live in a city were selected and 87 identified as a democrat. a random sample of n 2 = 1
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Answer:

CI=\{-0.2941,-0.0337\}

Step-by-step explanation:

Assuming conditions are met, the formula for a confidence interval (CI) for the difference between two population proportions is \displaystyle CI=(\hat{p}_1-\hat{p}_2)\pm z^*\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2} where \hat{p}_1 and n_1 are the sample proportion and sample size of the first sample, and \hat{p}_2 and n_2 are the sample proportion and sample size of the second sample.

We see that \hat{p}_1=\frac{87}{249}\approx0.3494 and \hat{p}_2=\frac{58}{113}\approx0.5133. We also know that a 98% confidence level corresponds to a critical value of z^*=2.33, so we can plug these values into the formula to get our desired confidence interval:

\displaystyle CI=(\hat{p}_1-\hat{p}_2)\pm z^*\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\\\\CI=\biggr(\frac{87}{249}-\frac{58}{113}\biggr)\pm 2.33\sqrt{\frac{\frac{87}{249}(1-\frac{87}{249})}{249}+\frac{\frac{58}{113}(1-\frac{58}{113})}{113}}\\\\CI=\{-0.2941,-0.0337\}

Hence, we are 98% confident that the true difference in the proportion of people that live in a city who identify as a democrat and the proportion of people that live in a rural area who identify as a democrat is contained within the interval {-0.2941,-0.0337}

The 98% confidence interval also suggests that it may be more likely that identified democrats in a rural area have a greater proportion than identified democrats in a city since the differences in the interval are less than 0.

5 0
2 years ago
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