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3 years ago

Find the measure of b. (Explain how you found it please!)

Mathematics

1 answer:

Allisa [31]3 years ago

5 0

Answer:

125 deg

Step-by-step explanation:

Keep these three rules in mind:

1) A central angle (vertex is the center of the circle) has the same measure as the arc it intercepts.

2) The measure of an inscribed angle (vertex is point on circle) is half the measure of the intercepted arc.

3) Opposite angles of a rectangle inscribed in a circle are supplementary.

110 deg is a central angle.

By rule 1), the arc intercepted by the central angle 110 deg also measures 110 deg.

a is an inscribed angle that intercepts an arc of 110 deg.

By rule 2), the measure of an inscribed angle is half the measure of the intercepted arc.

angle a measures 55 deg.

Rule 3) Angles a and b are supplementary.

a + b = 180

55 + b = 180

b = 125

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A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

The standard error is given by the following formula:

$SE=\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$

And replacing we have:

$SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

Part c: What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

Confidence interval

Now we have everything in order to replace into formula (1):

$3.92-1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=2.009$

$3.92+1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=5.830$

So on this case the 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

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