A test is made of H subscript 0 colon space mu space equals space 43 space space v e r s u s space space H subscript 1 colon space mu space greater than space 43. A sample of size n equals 76 is drawn, and the sample mean is 47. The population standard deviation is 25. Compute the value of the test statistic and determine if the H subscript 0 is rejected at the alpha equals 0.05 level.

The calculated value Z = 1.39 <1.96 at 0.05 level of significance

The null hypothesis is accepted

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 43

Given that the mean of the sample = 47

Given that the standard deviation of the Population = 25

Given that the sample size 'n' = 76

Step(ii):-

Null hypothesis:H₀: μ = 43

Alternative Hypothesis: H₁: μ ≠ 43

Test statistic

          Z = 1.39

Level of significance = 0.05

The critical value = 1.96

Final -

The calculated value Z = 1.39 <1.96 at 0.05 level of significance

The null hypothesis is accepted

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .