−5/6e−2/3e=−24 solve for e

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

2 years ago

You are riding in a taxi that charges a $10 flat fee, plus an additional $2 per

irakobra [83]

Answer:

C= 2M + 10

Step-by-step explanation:

In this equation, we will calculate the total cost. So, we'll start off with "C =" Then, we know tat the charge is $2 per mile so we'll add it and make C = 2M. Then, it states that the flat fee is 10 dollars which makes the equation

C = 2M +10.

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The radius, r, of a sphere is increasing at 0.03 meters per second. The rate of increase is constant.

Vladimir79 [104]

A)
Given: dr/dt = .03 m/s and r = 18 m
Volume of a sphere = (4/3)*PI*r^3
Take the derivative of both sides, so
dv/dt = 4*PI*r^2*dr/dt
Plug in the givens and you have the rate of increase of the volume
B)
Given: V = 288 m^3 and I'm going to assume that r and dr/dt are the same as in part A
Surface Area = 4*PI*r^2
Take the derivative on both sides, so
dA/dt = 8*PI*r*dr/dt
Again, plug in the givens and you'll have the rate of increase in the surface area
C)
This part, it's been awhile since I've done related rates, so it may take me awhile, but perhaps the next person can answer it before I finish.

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2 years ago

Todd begins piano practice at 4:15pm.and ends at 5:50pm.How long dose he practice

xxTIMURxx [149]

Answer:

If He began at 4:15 and ended at 5:50 then that would be 95 minutes

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There are (49)5 ⋅ 40 books at the library. What is the total number of books at the library?

Romashka-Z-Leto [24]

it is 9,800 total books

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3 years ago

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