Answer:
The enrollment after 5 years is 10,724
Step-by-step explanation:
Generally, we can have the depreciation formula written as follows;
A = P(1 - r)^t
A is the number of enrollment in after a certain number of years t
P is the initial population which is 13,500
r is the rate of depreciation which is 4.5% = 4.5/100 = 0.045
t = 5 years
Substituting these values, we have it that;
A = 13,500(1-0.045)^5
A = 10,723.84
Answer:
the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)
Step-by-step explanation:
since the variance S² of the batch follows a normal distribution , then for a sample n of 20 distributions , then the random variable Z:
Z= S²*(n-1)/σ²
follows a χ² ( chi-squared) distribution with (n-1) degrees of freedom
since
S² > 3.10 , σ²= 1.75 , n= 20
thus
Z > 33.65
then from χ² distribution tables:
P(Z > 33.65) = 0.02020
therefore the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)
The equation is -2x-4. u can use y2-y1/x2-x1.
Answer:216 is your answer
Step-by-step explanation: