The cdc estimates that each year in the united states, on average, 5% to 20% of the population gets the flu. using 20% as our best guess (p∗), find how large a sample would be needed to get a ±2 percentage point margin on a 95% confidence interval for the population proportion of americans who get the flu in a given year.
a. 1536
b. 1537
c. 2401
d. 2402
e. 40

b. 1537

Step-by-step explanation:

When prior estimate of population proportion is available , then the formula for sample size:  

, where p= prior estimate of population proportion

z*= critical-value.

E= Margin of sampling error.

Let p be the proportion of population gets the flu.

As per given , we have

p=20%=0.20

E= ± 0.02

Using z-table , the critical z-value corresponding to 95% confidence level = z*=1.960

Substitute all the values in the above formula , we get

Required sample size  :

[Rounded to next integer.]

Thus, the minimum sample size needed = 1537

Hence, the correct answer =  b. 1537

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0