A health club annually surveys its members. Last year, 33% of the members said they use the treadmill at least four times a week. How large a sample should be selected this year to estimate the percentage of members who use the treadmill at least four times a week? The estimate is desired to have a margin of error of 5% with a 95% level of confidence.

The large  sample n = 174.24

Step-by-step explanation:

Given 33% of the members said they use the treadmill at least four times a week.

Given the margin of error is 5% = 0.05

we know that the margin of error at 95% 0f level of confidence  =

Given standard deviation = 33% = 0.33

cross multiplication , we get

√n = 13.2

squaring on both sides, we get

n = 174.24

Conclusion:-

The large  sample n = 174.24

answer: ur mum is gay

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