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ad-work [718]
3 years ago
10

The ratio of the lengths of an ellipse is 3:2. If it’s area is 150 cm2, what are the lengths of The major and minor semiaxes res

pectively?
Mathematics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

the length of major semiaxes a = 8.45 cm  and minor semiaxes b= 5.63 cm.

Step-by-step explanation:

Area of ellipse = π*a*b

Let major semiaxes = a and minor semiaxes = b

then a:b = 3:2

a/b = 3/2

=> a = 3/2 b

Putting in the value of formula

150 = π * (3/2)b * b

150 = 3.14 * 1.5b * b

150= 4.71 b^2

150/4.71 = b^2

=> b^2 = 31.8

b= 5.63

a = 3/2 * b

a = 1.5 * 5.63

a = 8.45

So, the length of major semiaxes a = 8.45 cm  and minor semiaxes b= 5.63 cm.

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Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is \hat{p}=26/42=0.6190. The estimated standard deviation is given by \sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749. Because we have a large sample, the 90% confidence interval for p is given by 0.6190\pm z_{0.05}0.0749 where z_{0.05}=1.6448 is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is 0.6190\pm (1.6448)(0.0749), i.e., (0.4958, 0.7422).

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