How many four-letter codes can be made if no letter can be used twice?
1 answer:
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Step-by-step explanation:
You need to translate all the points to the right 3 and up 6
Therefore, you are going to use this formula:
(x,y) ⇾ (x + 3, y + 6)
This is the same format as the previous problem, if you have noticed.
Using this, plug in each coordinate, starting with P (5, -1)
(5, -1) ⇾ ( 5 + 3, -1 + 6)
(5, -1) ⇾ ( 8, 5 )
P = (8, 5)
Now point Q, (0, 8)
(0, 8) ⇾ (0 + 3, 8 + 6)
(0, 8) ⇾ ( 3, 14 )
Q = (3, 14)
And last but not least, the point R, (7, 5)
(7, 5) ⇾ (7 + 3, 5 + 6)
(7, 5) ⇾ ( 10, 11 )
R = (10, 11)
Therefore, P = (8, 5), Q
= (3, 14), R
= (10, 11) is your answer. This is the 4th option or D.
Hope this for you to understand this a bit more! =D
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.