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:
(1, 2 )
Step-by-step explanation:
Given the 2 equations
y = - 3x + 5 → (1)
y = 4x - 2 → (2)
Substitute y = 4x - 2 into (1)
4x - 2 = - 3x + 5 ( add 3x to both sides )
7x - 2 = 5 ( add 2 to both sides )
7x = 7 ( divide both sides by 7 )
x = 1
Substitute x = 1 into (2) for corresponding value of y
y = (4 × 1) - 2 = 4 - 2 = 2
Solution is (1, 2 )
We first need to pick the correct piece to use. In this case, we will use f(x) = -x-5 because -5<=-4.
Then, we use direct substitution:
f(-4) = -(-4) - 5
f(-4) = -1
Answer:
=3
=2
Step-by-step explanation:
12−5+6=0
using the Quadratic Formula where
a = 1, b = -5, and c = 6
=−±2−4‾‾‾‾‾‾‾‾√2
=−(−5)±(−5)2−4(1)(6)‾‾‾‾‾‾‾‾‾‾‾‾‾‾√2(1)
=5±25−24‾‾‾‾‾‾‾√2
=5±1‾√2
The discriminant 2−4>0
so, there are two real roots.
Simplify the Radical:
=5±12
=62=42
which becomes
=3
=2
hope this helps :)
11 pt. = 176 oz.
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