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3 years ago

7

what is the probability that a red or green marble will be selected from a bag containing 9 red marbles, 7 green marbles, and 11

yellow marbles if one is selected randomly ? Help

1 answer:

SVEN [57.7K]3 years ago

3 0

P(red) = 9/(9 + 7 + 11) = 9/27 = 1/3
P(green) = 7/27

P(red or green) = 1/3 + 7/27 = 16/27

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The base of a garden box has an area of 27,404 square centimeters. It's height is 18 centimeters. What is the volume of the gard

taurus [48]

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What do AIA, SAG, AFP and AMA all stand for? Select four options.

American Institute of Architects

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3 years ago

The family of functions y = ce^{-2x} + e^{-x} is solution of the equation y' + 2y = e^{-x}. Find the constant c which defines th

Rufina [12.5K]

Answer:

$c = 1190.2$

Step-by-step explanation:

We have the following solution:

$y(x) = ce^{-2x} + e^{-x}$

We want to find the value of x that satisfies the following condition:

y(3) = 3.

This means that when x = 3, y = 3. So

$y(x) = ce^{-2x} + e^{-x}$

$y(3) = ce^{-2*3} + e^{-3}$

$3 = ce^{-6} + e^{-3}$

$3 = ce^{-6} + 0.0498$

$ce^{-6} = 2.9502$

$c = \frac{2.9502}{e^{-6}}$

$c = 1190.2$

4 0

3 years ago

Several years ago, 50% of parents who had children in grades K-12 were satisfied with the quality of education the students r

galina1969 [7]

Answer:

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

We have to see if 50% = 0.5 is part of the confidence interval.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 1095, \pi = \frac{478}{1095} = 0.4365$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 - 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4118$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 + 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4612$

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.

5 0

3 years ago

What is the value of x? Then determine the measure of each angle. will brainlist first correct answer

kherson [118]

Answer:

Step-by-step explanation:

I have to go with 9r-18= -27 is my answer for your math

4 0

3 years ago

The U.S. Census Bureau conducts annual surveys to obtain information on the percentage of the voting-age population that is regi

yan [13]

Answer:

We conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

Step-by-step explanation:

We are given that 513 employed persons and 604 unemployed persons are independently and randomly selected, and that 287 of the employed persons and 280 of the unemployed persons have registered to vote.

Let $p_1$ = <u><em>percentage of employed workers who have registered to vote.</em></u>

$p_2$ = <u><em>percentage of unemployed workers who have registered to vote.</em></u>

So, Null Hypothesis, $H_0$ : $p_1\leq p_2$ {means that the percentage of employed workers who have registered to vote does not exceeds the percentage of unemployed workers who have registered to vote}

Alternate Hypothesis, $H_A$ : $p_1>p_2$ {means that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote}

The test statistics that would be used here <u>Two-sample z test for proportions;</u>

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of employed workers who have registered to vote = $\frac{287}{513}$ = 0.56

$\hat p_2$ = sample proportion of unemployed workers who have registered to vote = $\frac{280}{604}$ = 0.46

$n_1$ = sample of employed persons = 513

$n_2$ = sample of unemployed persons = 604

So, <u><em>the test statistics</em></u> = $\frac{(0.56-0.46)-(0)}{\sqrt{\frac{0.56(1-0.56)}{513}+\frac{0.46(1-0.46)}{604} } }$

= 3.349

The value of z test statistics is 3.349.

<u>Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.</u>

Since our test statistic is more than the critical value of z as 3.349 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

5 0

3 years ago