C it’s Just look it up bro
Answer:
0.5
Step-by-step explanation:
Answer:
The probability that there are 3 or less errors in 100 pages is 0.648.
Step-by-step explanation:
In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.
For the given Poisson distribution the mean is p = 0.03 errors per page.
We have to find the probability that there are three or less errors in n = 100 pages.
Let us denote the number of errors in the book by the variable x.
Since there are on an average 0.03 errors per page we can say that
the expected value is,
= E(x)
= n × p
= 100 × 0.03
= 3
Therefore the we find the probability that there are 3 or less errors on the page as
P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)
Using the formula for Poisson distribution for P(x = X ) = 
Therefore P( X ≤ 3) = 
= 0.05 + 0.15 + 0.224 + 0.224
= 0.648
The probability that there are 3 or less errors in 100 pages is 0.648.
Answer:
D. (17,-1)
Step-by-step explanation:
We'll start by canceling out x. To do so, multiple the first equation by -1
-1( x - 2y)=( 19 )(-1)
This gives us:
-x +2y = -19
x + 3y = 14
Add the equations together:
→ 5y = -15
→ y = -1
Plug in y = -1 into an equation:
→ x + 3(-1) = 14
→ x - 3 = 14
→ x = 17
(17,-1)
Hope this helps!
Answer:
x = 9
Step-by-step explanation:
If ΔMNO ≅ ΔPST, their corresponding sides must also be ≅(congruent).
NO is corresponds to TS, thus sides NO and side TS are ≅.
=> 20 = 3x - 7
=> 3x = 27
=> x = 9