From earlier studies, it is believed that the percentage of students favoring a four-day school week during May and June of every school year is approximately 85%. What is the minimum sample size that will create a margin of error of 2% with 90% confidence

The minimum sample size 'n' = 862.538

Step-by-step explanation:

Step(i):-

Given the percentage of students favoring a four-day school week during May and June of every school year is approximately 85%

Given the Proportion of students favoring a four-day school week during May and June of every school year

      p = 85% = 0.85

      q = 1-p = 1-0.85 = 0.15

Given Margin of error = 2 % or 0.02

Level of significance = 0.90 or 0.10

The critical value Z₀.₁₀ = 1.645

Step(ii):-

The Margin of error is determined by

Cross Multiplication, we get

         √n = 29.369

squaring on both sides, we get

         n = 862.538

Final -

Sample size 'n' = 862.538

step-by-step explanation:

107 units

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15