A. a researcher claims that 20% of students spend more than 10 hours a week on homework asignments and studying. a random sample of 53 students were surveyed and 18 of them said they spend more than 10 hours a week on homework assignments and studying. construct a 95% confidence interval to test the researcher's clain ie, is the value 20% within the confidence interval? you must show the value of e, p, and za/2 rounded to 4 significant digits.

The confidence interval is
, or 
(0.212, 0.4672); no, his claim is not accurate, as 20% is not in the interval.

To construct the interval we first find p:
18/53 = 0.3396

To calculate :
Convert 95% to a decimal:  95/100 = 0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 1.96.

We next calculate E:


Our confidence interval is then given by

Hello,

Let's assume CD=h, BD=x.

tan 24°=h/x ==>h=x*tan 24°

tan 16°=h/(7600+x)==>h=tan 16° *(7600+x)

==>x tan 24°=tan 16°*(7600+x)
==>x=(7600*tan 16°)/(tan 24°-tan 16°)

h=x*tan 24= (7600*tan 16°*tan 24°)/(tan 24°-tan16°)

b) CD=h=6122,2303...≈6122 (feet)

a) BC=h/sin 24°=15052,0746448...≈15052 (feet)

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