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crimeas [40]
3 years ago
10

The diagram shows a right-angled triangle and a parallelogram.

Mathematics
1 answer:
Aneli [31]3 years ago
3 0

Answer:

h = 12

Step-by-step explanation:

area of triangle = 16x9x0.5= 72 cm^2

area of parallelogram = 72 x 5 = 360 cm^2

area of parallelogram = base x height(h)

360 = 30h

divide both sides by 30

h = 12

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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,
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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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