Answer:
I think c 430 and if it is wrong I'm so sorry
Answer: 
Step-by-step explanation:
The confidence interval estimate for the population mean is given by :-
, where
is the sample mean and ME is the margin of error.
Given : Sample mean: 
The margin of error for a 98% confidence interval estimate for the population mean using the Student's t-distribution : 
Now, the confidence interval estimate for the population mean will be :-

Hence, the 98% confidence interval estimate for the population mean using the Student's t-distribution = 
AK = 640
Δ ABC and ΔFJK are similar. They are small triangles.
ΔCDF is the big triangle.
640 / 2 = 320 m= CF
AC & FK= 320/2 = 160 m each
2AC = CF = 2FK
2(160) = 320 = 2(160)
BG = 20 m
20/160 = x / 320
20*320 = 160x
6400 = 160x
6400/160 = x
40 = x
Area of CDF = (40 m * 320 m)/2 = 12,800 / 2= 6,400 m²
Answer:
198 is the variance for the number of defective parts made each week.
Step-by-step explanation:
We are given the following in the question:
Number of parts produced each week = 20,000
Percentage of defective parts = 1%
We have to calculate the variance for the number of defective parts made each week.
We treat defective part as a success.
P(Defective part) = 1% = 0.01

Then the number defective parts follows a binomial distribution
.
Formula for variance =

Thus, 198 is the variance for the number of defective parts made each week.