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2 years ago

Y = 5x - 13 4x + 3y = 18

1 answer:

3 0

<span>replace Y = 5x - 13
into
4x + 3y = 18

</span>

4x + 3 (5x - 13) = 18
4x + 15 - 39 = 18
4x -24 = 18
4x = 18 +24
4x = 42
x = 42/4
x = 10.5

Y = 5(10.5) - 13
= 52.5 - 13
= 39.5

answer: x = 10.5 and y = 39.5

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Answer:

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b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

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$P(N=r)=\frac{e^{-np}(np)^r}{r!}$

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and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

$P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}$

$P(N=0)=\frac{e^{-0.2}}{1}$

$P(N=0)=e^{-0.2}$

$P(N=0)=0.8187$

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute $r\geq 2$ in formula,

$P(N\geq 2)=1-P(N$

$P(N\geq 2)=1-[P(N=0)+P(N=1)]$

$P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]$

$P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]$

$P(N\geq 2)=1-[0.8187+0.1637]$

$P(N\geq 2)=1-0.9825$

$P(N\geq 2)=0.0175$

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

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