Auniversity newspaper is conducting a survey to determine what fraction of students support a $200 per year increase in fees to pay for a new football stadium. how big of a sample is required to ensure the margin of error is smaller than 0.04 using 95% confidence level?

The formula to calculate sample size is  Z * √((P(1-P)/n)

Z is the Z-score for the confidence level, in this problem the confidence level is given as 95%, so Z = 1.96

P is the population, which is unknown, but we want it to be the largest value possible, so in the equation above, P is 0.5.

n is the sample size.

Now we have: 1.96* √((0.5 * 0.5)/n)

They want this sample size to be less than 0.04

So now we have:

1.96* √((0.5 * 0.5)/n) < 0.04

Multiply both sides by 1/1.96:

√((0.5 * 0.5)/n) < 0.04  *1/1.96

Simplify:

0.5 / √n < 4/196

Take the reciprocal of both sides:

√n / 0.5 < 196/4

Simplify:

√n/0.5 < 49

Multiply both sides by 0.5:

√n = 49*0.5

√n = 24.5

to solve for n, raise both sides to the power of 2:

n = 24.5^2

n = 600.25

The sample needs to be about 600 people.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2