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3 years ago

CAN SOMEONE HELP ME WITH THIS PLEASE ??

Mathematics

1 answer:

masya89 [10]3 years ago

3 0

Remark
The very first thing you should do is graph the function, which I have done. It is at the bottom of this answer.

Step One
Where is the hole and what kind is it? This graph has a hole at x = 2. At that point the function is 0/0. A hole is defined as the point created by a zero in one of the factors that is the same in the numerator and denominator. Both = 0.

Step 2
Find the vertical asymptote.
That occurs when denominator alone has a factor that is = 0. In this case when x = 3 the whole equation blows up (or down) and you have a vertical asymptote. The graph is discontinuous at this point.

Step three
Find the horizontal Asymptote.
Horizontal assymptotes occur when (in this case) the power of the factors in the denominator exceeds the power of the factors in the numerator. In this case that is true and the horizontal assymptote occurs around the line y = 0.

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A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the

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Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

Step-by-step explanation:

(a)

Let X = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

X = 610

n = 1000

Confidence level = 90%

The (1 - α)% confidence interval for population proportion is:

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

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{610}{1000}=0.61$

The critical value of z for a 90% confidence level is:

$z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645$

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)$

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

Interpretation:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let X = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

X = 440

n = 1000

Confidence level = 95%

The (1 - α)% confidence interval for population proportion is:

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

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{440}{1000}=0.44$

The critical value of z for a 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)$

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend is (0.42, 0.46).

Interpretation:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

The confidence level
Sample size
Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of z in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

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