Consider a value to be significantly low if its z score less than or equal to minus2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent year, the mean test score was 21.9 and the standard deviation was 4.9. Identify the test scores that are significantly low or significantly high.

Step-by-step explanation:

Since we are dealing with z scores, then the distribution is a normal distribution. The formula for determining the z score is expressed as

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 21.9

σ = 4.9

For significantly low values, z = - 2

Therefore,

- 2 = (x - 21.9)/4.9

- 2 × 4.9 = x - 21.9

- 9.8 = x - 21.9

x = - 9.8 + 21.9

x = 12.1

Significantly low test score = 12.1

For significantly high values, z = 2

Therefore,

2 = (x - 21.9)/4.9

2 × 4.9 = x - 21.9

9.8 = x - 21.9

x = 9.8 + 21.9

x = 31.7

Significantly high score = 31.7

a. 74 degrees

step-by-step explanation:

because the lines through the side and the box in the corners shows that these are congruent triangles and since they are mirrored the 37 degrees is also mirrored so the answer is 37 * 2 which is 74.