Let μ1 and μ2 denote true average densities for two different types of brick. assuming normality of the two density distributions, test: h0: μ1 − μ2 = 0 versus ha: μ1 − μ2 ≠ 0 use the following data: n1 = 6, x1 = 23.72, s1 = 0.55, n2 = 4, x2 = 20.88, and s2 = 0.530. the conservative degrees of freedom, min(n1-1, n2-1), are:

dont think so

step-by-step explanation:

i can't solve unless i know what the line is

step-by-step explanation:

welcome to/p>

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.

answer:   2 multiplied by x =24

step-by-step explanation: