It has been conjecture by the U. S. Census Bureau that "approximately 60% of foreign-born people who live in the U. S. are not naturalized citizens". In a national random sample of 70 foreign-born people who live in the U. S., on average, how many people would you expect to get that are not naturalized citizens?

a. 28 people
b. 42 people
c. 4.10 people
d. None of these.

b. 42 people

Step-by-step explanation:

For each foreigner living in the U.S., there are only two possible outcomes. Either they are naturalized citizens, or they are not. The probability of a foreigner living in the U.S. not being a naturalized citizen is independent from other foreigners living in the U.S. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

"approximately 60% of foreign-born people who live in the U.S. are not naturalized citizens"

This means that

70 foreign-born people who live in the U.S

This means that

How many people would you expect to get that are not naturalized citizens?

So the correct answer is:

b. 42 people

4

step-by-step explanation:

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)