A class' test scores are normally d istributed with a mean of 48 points  and a standard  deviation of 3 points. What percent age of test scores are are between  47 and 52 
points? Round to the nearest hundre dth of a percent.

53.94%

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 48 points

σ is the population standard deviation = 3 points

For x = 47 points

z = 47 - 48/3

= -0.33333

Probability value from Z-Table:

P(x = 47) = 0.36944

For x = 52 points

z = 52 - 48/3

= 1.33333

Probability value from Z-Table:

P(x = 52) = 0.90879

Hence, the probability of test scores that between 47 and 52 points is calculated as:

P(x = 52) - P(x = 47)

= 0.90879 - 0.36944

= 0.53935

Therefore, the percentage of test scores that are between 47 and 52

points is given as:

0.53935 × 100

= 53.935%

Approximately to the nearest hundredth of a percent ≈ 53.94%

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>