Divide total full time employees by 20 to find how many groups of 20 there are, then multiply that number by 3 to find total part time employees:
250,000 / 20 = 12,500
12,500 x 3 = 37,500 part time employees.
34% of the scores lie between 433 and 523.
Solution:
Given data:
Mean (μ) = 433
Standard deviation (σ) = 90
<u>Empirical rule to determine the percent:</u>
(1) About 68% of all the values lie within 1 standard deviation of the mean.
(2) About 95% of all the values lie within 2 standard deviations of the mean.
(3) About 99.7% of all the values lie within 3 standard deviations of the mean.



Z lies between o and 1.
P(433 < x < 523) = P(0 < Z < 1)
μ = 433 and μ + σ = 433 + 90 = 523
Using empirical rule, about 68% of all the values lie within 1 standard deviation of the mean.
i. e. 
Here μ to μ + σ = 
Hence 34% of the scores lie between 433 and 523.
The other endpoint is (10, -8
A³ b² 4ab³
Rearrange order:
4 a³ a b² b³
Now add up the exponents from same base:
4 a³⁺¹ b²⁺³
4 a⁴ b⁵
Final answer: 4 a⁴ b⁵