Suppose that 20% of voters are in favor of a certain legislation. A large number n of voters are polled and a relative frequency estimate fA(n) for the above proportion is estimated. How many voters should be polled so that the probability is at least 0.95 that fA(n) differs from 0.20 by less than 0.02?

8000 voters

Step-by-step explanation:

E is the difference between probability that is real and those which are estimated

Sd² = p(1-p)

When solve this, using the formula in the attachment,

0.95 = 1-0.2(0.2)²/n(0.02)

0.95 = 1 - 0.16/n(0.02)²

1-0.95 = 0.16/n(0.02)²

0.05 = 0.16/0.0004n

We cross multiply here

0.05 x 0.0004n = 0.16

0.00002n = 0.16

Divide through by 0.00002 to get n

n = 8000

This is the number of voters that should be polled

B and B

Step-by-step explanation:

The first one is B because all of the slopes are negative except B. And so I think that's the correct answer

The second one I know is B for sure, I remember learning it in 8th grade 2 years ago.

It's really just the first one I am second guessing myself on but B and B make sense to me :)