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3 years ago

How many solutions does 4(x-1)=4x+5

Mathematics

2 answers:

sergeinik [125]3 years ago

6 0

Answer:

NO SOLUTION

Step-by-step explanation:

To solve this problem, first you have to isolate it on one side of the equation.

First, expand.

$\displaystyle 4(x-1)=4x-4$

$\displaystyle 4x-4=4x+5$

Next, add 4 from both sides.

$\displaystyle 4x-4+4=4x+5+4$

Solve.

$\displaystyle 4+5=9$

$\displaystyle 4x=4x+9$

Subtract 4x from both sides.

$\displaystyle 4x-4x=4x+9-4x$

Solve.

$\displaystyle 0=9$

Therefore, there are no solution.

mihalych1998 [28]3 years ago

3 0

4(x - 1) = 4x + 5

Eliminate parentheses on the left:

4x - 4 = 4x + 5

Subtract 4x from each side:

-4 = +5

No value of x can make this a true statement. So no solution exists.

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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

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Solution :

Applying Poisson distribution,

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

where, n is the number of words in a page

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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,

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

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

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

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

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

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

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