Astandardized exam's scores are normally distributed. in a recent year, the mean test score was 14781478 and the standard deviation was 311311. the test scores of four students selected at random are 18701870, 12001200, 21802180, and 13801380. find the z-scores that correspond to each value and determine whether any of the values are unusual.

Z SCORE FOR 1870 = 1.26

Z SCORE FOR 1200 = -0.89

Z SCORE FOR 2180 = 2.25

Z SCORE FOR 1380 = -0.315

Step-by-step explanation:

given data:

test score = 1478

standard deviation = 311

z score can be calculated by using below relation

1) Z SCORE FOR 1870

= 1.26

2) Z SCORE FOR 1200

= -0.89

3) Z SCORE FOR 2180

= 2.25

4) Z SCORE FOR 1380

= -0.315

There are no unusual z value.

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

for this case we must apply distributive property to the following expression:

the distributive property states that:

then, when applying the property we have to multiply the -2 for each term within the parenthesis: