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3 years ago

Please help click on the image below

Mathematics

2 answers:

MatroZZZ [7]3 years ago

4 0

I am pretty sure it's c

Arisa [49]3 years ago

3 0

I think it C
have a great day!!!

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The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10

Pavel [41]

Answer:

a) 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 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,

$N\sim Pois(0.2)$

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

where, n is the number of words in a page

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.

6 0

3 years ago

Help, please simplify.. thank you

Romashka [77]

$\bf \cfrac{3.7\times 10^{-12}(42)}{(.000000025)(4.0\times 10^{-15})}\implies 
\cfrac{37\times 10^{-13}(42)}{(25\times 10^9)(4\times 10^{-15})}$

$\bf \cfrac{37(42)\times 10^{-13}}{25(4)\times 10^910^{-15}}\implies 
\cfrac{1554\times 10^{-13}}{100\times 10^{9-15}}\implies 
\cfrac{1554\times 10^{-13}}{1\times 10^2\times 10^{9-15}}
\\\\\\
\cfrac{1554\times 10^{-13}}{10^{2+9-15}}\implies \cfrac{1554\times 10^{-13}}{10^{-4}}\implies 1554\times 10^{-13}\times 10^4
\\\\\\
1554\times 10^{-13+4}\implies 1554\times 10^{-9}\implies \cfrac{1554}{10^9}\implies 0.000001554$

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Given the quadratic function y=5x^2−2x identify a, b and c.

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Work shown above! Answer is
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