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svet-max [94.6K]
2 years ago
12

In one city, 75 percent of residents report that they regularly recycle. In a second city, 90 percent of residents report that t

hey regularly recycle. Simple random samples of 75 residents are selected from each city. Which of the following statements is correct about the approximate normality of the sampling distribution of the difference in sample proportions of residents who report that they regularly recycle?
a. The sampling distribution is approximately normal because both sample sizes are large enough.
b. The sampling distribution is not approximately normal because although the sample size for the first city is large enough, the sample size for the second city is not large enough.
c. The sampling distribution is not approximately normal because although the sample size for the second city is large enough, the sample size for the first city is not large enough.
d. The sampling distribution is not approximately normal because neither sample size is large enough.
e. There is not enough information provided to determine whether the sampling distribution is approximately normal.
Mathematics
1 answer:
Paraphin [41]2 years ago
6 0

can you show me a picture then maybe i can help you

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Obtain two numbers such their HCF is 20 and their LCM is 300....... By steps...<br><br>​
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Answer:

60 and 100

Step-by-step explanation:

Look for 2 multiples of 20 which are also factors of 300.

No need to go all Mathematician.

It's simple, no long explanation needed

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Factor 125x9 + 64. (5x3 – 4)(25x6 + 20x3 + 16) (5x3 – 4)(25x3 + 20x3 + 16) (5x3 + 4)(25x6 – 20x3 + 16) (5x3 + 4)(25x3 – 20x3 + 1
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(5x^3 + 4)(25x^6 - 20x3 +16)  remember factoring the cubes?
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2 years ago
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6.In a sample of 131 women with cosmetic dermatitis from using eye shadow, 12 were diagnosed with a nickel allergy. In a sample
aivan3 [116]

Answer:

a) 0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422

0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

b) \hat p = \frac{25}{250}= 0.1

0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628

0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

c) For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

Step-by-step explanation:

Part a

The estimated proportion of women with cosmetic dermatitis from using eye shadow is given by:

\hat p =\frac{12}{131}= 0.0916

The confidence  interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the proportion is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Replacing we got:

0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422

0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

Part b: A 95% confidence interval for the women with cosmetic dermatitis from using mascara

\hat p = \frac{25}{250}= 0.1

0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628

0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

Part c: Suppose you are informed that the true proportion with a nickel allergy for one of the two groups (eye shadow or mascara) is .12. Can you determine which group is referenced? Explain.

For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

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2 years ago
Lines AB and CD and are parallel. Find the measures of the three angles in triangle ABF.
Airida [17]

Answer:    ∠B = 42°   ∠A = 23°  ∠F = 115°

Step-by-step explanation:

∠B≅∠C  alternate interior angles  42°

  ∠CDF is supplementary to ∠CDE,

so m∠CDF = 23°  and ∠FAB≅∠CDF so m∠FAB= 23°

<em>also ∠A ≅ CDE  corresponding angles  157° ∠FAB is suplementary to ∠A </em>

<em>so 180 - 157 = 23 gives the m∠FAB</em>

The sum of the angles of a triangle is 180°, so m∠AFB = 180 -(23 +42)

180- 65 = 115 = m∠F

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2 years ago
If →u and →v are the vectors below, find the vector →w whose tail is at the point halfway from the tip of →v to the tip of →v−→u
S_A_V [24]
Û = (-1, -1, -1)
^v = (2, 3, -5)
^v - û = (2 + 1, 3 + 1, -5 + 1) = (3, 4, -4)
Half way from ^v to ^(v - u) = ((3 - 2)/2, (4 - 3)/2, (-4 + 5)/2) = (1/2, 1/2, 1/2)
Halfway from û to ^v = ((2 + 1)/2, (3 + 1)/2, (-5 + 1)/2) = (3/2, 2, -2)

The required vector ^w = ((3/2 - 1/2), (2 - 1/2), (-2 - 1/2)) = (1, 1/2, -5/2)
5 0
2 years ago
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