Suppose the mean percentage in Algebra 2B is 70% and the standard deviation is 8%. What percentage of students receive between a 70% and 94% Enter the
value of the percentage without the percent sign.
HELP PLEASE!!

49.87%

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 = 70%

σ is the population standard deviation = 8%

a) For x = 70%

z = 70% - 70%/8%

z = 0

Probability value from Z-Table:

P(x = 70) = 0.5

b) For x = 70%

z = 94% - 70%/8%

z = 3

Probability value from Z-Table:

P(x = 94) = 0.99865

The probability of students that receive between a 70% and 94%

P(x = 94) - P(x = 70)

0.99865 - 0.5

0.49865

Therefore, the percentage of students that receive between a 70% and 94% is

0.49865 × 100

= 49.865%

Approximately = 49.87%

3.76%

step-by-step explanation:

the answer is 9.8975

explanation:

when you multiply the "39.59" with the 25% it reduces it and shows its final discounted price. hope this !

step-by-step explanation: