A sample of 18001800 computer chips revealed that 52R% of the chips do not fail in the first 10001000 hours of their use. The company's promotional literature states that 53S% of the chips do not fail in the first 10001000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Find the value of the test statistic. Round your answer to two decimal places.

The test statistic is -0.83.

Step-by-step explanation:

Test statistic (z) = (p' - p) ÷ sqrt[p(1-p) ÷ n]

p' is sample proportion = 0.52

p is population proportion = 0.53

n is sample size = 1800

z = (0.52 - 0.53) ÷ sqrt[0.53(1 - 0.53) ÷ 1800] = -0.01 ÷ 0.012 = -0.83 (to two decimal places)

1. 34

2. 268

3. x=14.4

step-by-step explanation:

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