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.

1,150 give brainliest

step-by-step explanation

2r= 2300

r =1150

it's 100

step-by-step explanation:

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

  n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the  14th  row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100

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