In a random survey of 226 college students, 20% reported being "only" children (no siblings). The standard deviation is 2%. Determine the confidence interval that is likely to contain the percent of all students nationwide who are only children.

The confidence interval that is likely to contain the percent of all students nationwide who are only children is (0.1505,0.2554)

Step-by-step explanation:

Total number of students = n = 226

We are given that 20% reported being "only" children (no siblings).

Number of students being "only" children (no siblings)

So, x = 46

Z at 95% is 1.96

Formula for confidence interval for sample proportion :

Substitute the values in the formula :

Hence  the confidence interval that is likely to contain the percent of all students nationwide who are only children is (0.1505,0.2554)

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

read more on -

the answer is 45 i just did it on oddesy ware