It is generally believed that the mean value of the district's total SAT score distribution is equal to 1200 (the null hypothesis). The principal from one of the schools claims that the mean value of the district's total SAT score distribution is not 1200. Perform the appropriate test with a 5% margin of error. What is the corresponding p-value?

The corresponding p-value, is p = 1

Step-by-step explanation:

The maximum score SAT score, n =  1,600

The mean of the district's total SAT score distribution = 1,200

The claim of one of the districts principal, is the that mean of the district's total SAT score distribution ≠ 1,200

Using proportions, we have;

p = 1,200/1,600 = 0.75

q = 1 - p = 0.25

The margin of error, E = Z√(p·q/n)

∴ E = 5% = Z×√((0.75 × 0.25)/1,600)

z = 0.05/(√((0.75 × 0.25)/1,600)) ≈ 4.61880

Therefore, the corresponding p-value, p = 1

a) 7x=7

step-by-step explanation:

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