D=event that chip selected is defective
d=event that chip selected is NOT defective
Four possible scenarios for the first two selections:
P(DDD)=15/100*14/99*13/98=13/4620
P(DdD)=15/100*85/99*14/98=17/924
P(dDD)=85/100*15/99*14/99=17/924
P(ddD)=85/100*84/99*15/98=17/154
Probability of third selection being defective is the sum of all cases,
P(XXD)=P(DDD)+P(DdD)+P(dDD)+P(ddD)
=3/20
The standard deviation of the frequency distribution is 5.54
<h3>How to determine standard deviation?</h3>
The table of values is given as:
x f(x)
0-3 13
4-7 13
8-11 10
12-15 11
16-19 0
20-23 3
Rewrite the table by calculating the class midpoints:
x f(x)
1.5 13
5.5 13
9.5 10
13.5 11
17.5 0
21.5 3
Start by calculating the mean using:

This gives

Evaluate

The standard deviation is then calculated as:

So, we have:

Evaluate

Solve

Hence, the standard deviation is 5.54
Read more about standard deviation at:
brainly.com/question/15858152
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Answer:
attached the drawing
Step-by-step explanation:
hope it works
GCF(24, 30, 42) = 6 I think it is.
Starting January 1, 2015, there are 364 remaining nights.
From 1,001 nights, it will be 637 nights left.
In 2016, there are 366 nights. Subtract it to 637 nights left, you will get 271 nights left.
In 2017, 271 is lesser than 365.
From January to August, the nights left is 28.
<span>
So the date will be September 28, 2017.</span>