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topjm [15]
3 years ago
13

A library association plans to take an SRS of 150150150 adults from the population of adults in the US to see what proportion of

adults sampled believe that public libraries help them learn new things. Suppose that 76\vv, percent of adults in the US believe that public libraries help them learn new things. What are the mean and standard deviation of the sampling distribution of the proportion of US adults who believe that public libraries help them learn new things
Mathematics
1 answer:
Serga [27]3 years ago
3 0

Answer:

Mean: =0.76

SD= (squarerootof) 0.76(0.24)/150

Step-by-step explanation:

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If 5x+5=50 <br> then solve for x
pav-90 [236]

Answer:

9

Step-by-step explanation:

5 x 9 is 45 plus 5 gives you 50

6 0
2 years ago
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I need these correct answers nowww
faust18 [17]

Answer:

1 - 3/14

2 - 10 5/8

3 - 3/8

4 - 12 1/4

Step-by-step explanation:

numbers 1 and 3 are simple, just multiply the numerators and denominators together.

Numbers 2 and 4, you convert the mixed number to an improper fraction and multiply. Don't forget to switch the answer back to a mixed number if needed!

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3 years ago
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An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so
alexgriva [62]

Solution :

Let p_1 and p_2  represents the proportions of the seeds which germinate among the seeds planted in the soil containing 3\% and 5\% mushroom compost by weight respectively.

To test the null hypothesis H_0: p_1=p_2 against the alternate hypothesis  H_1:p_1 \neq p_2 .

Let \hat p_1, \hat p_2 denotes the respective sample proportions and the n_1, n_2 represents the sample size respectively.

$\hat p_1 = \frac{74}{155} = 0.477419

n_1=155

$p_2=\frac{86}{155}=0.554839

n_2=155

The test statistic can be written as :

$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}

which under H_0  follows the standard normal distribution.

We reject H_0 at 0.05 level of significance, if the P-value or if |z_{obs}|>Z_{0.025}

Now, the value of the test statistics = -1.368928

The critical value = \pm 1.959964

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

                                     $=2 \times 0.085667$

                                     = 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$, so we fail to reject H_0 at 0.05 level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the \text{proportion} of the seeds that \text{germinate differs} with the percent of the \text{mushroom compost} in the soil.

8 0
3 years ago
The weight of an object on a particular scale 155. 2lbs the measured weight May very from the actual weight by the most 0.4 lbs
zavuch27 [327]

Answer:

154.8-155.6

Step-by-step explanation:

155.2-.4=154.8 or maybe its more in that case the most it could be is 155.2+.4=155.6

7 0
4 years ago
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RSB [31]

Answer:

Step-by-step explanation:

3x² - 3x + 7 = 0

a = 3 ; b = -3 and c = 7

D = b² - 4ac = (-3)² - 4*3*7

D = 9 - 84 = - 75

D = 75i²

√D = √75i² = \sqrt{5*5*3 * i * i}= 5i\sqrt{3}

x = \dfrac{-b+\sqrt{D}}{2a}  \ ; \  x=\dfrac{-b-\sqrt{D}}{2a}\\\\ \\x=\dfrac{3+5i\sqrt{3}}{6} ; \ ; x = \dfrac{3-5i\sqrt{3}}{6}\\\\\\x =\dfrac{3}{6}+\dfrac{5i\sqrt{3}}{6} ; \ ; x=\dfrac{3}{6}-\dfrac{5i\sqrt{3}}{6}\\\\\\Solution:\\\\x=\dfrac{1}{2}+\dfrac{5\sqrt{3}}{6}i ; \ ; x=\dfrac{1}{2}-\dfrac{5\sqrt{3}}{6}i

5 0
2 years ago
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