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2 years ago

What is 12,595,000 as a percent

Mathematics

2 answers:

Degger [83]2 years ago

8 0

Step-by-step explanation:

12595000 * 100 = 1259500000%

marysya [2.9K]2 years ago

7 0

Step-by-step explanation:

To write 12595000 as a percent have to remember that 1 equal 100% and that what you need to do is just to multiply the number by 100 and add at the end symbol % .

12595000 * 100 = 1259500000%

<h2>And finally we have:</h2>

12595000 as a percent equals 1259500000%.

I Hope Its Helpfull.:)

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A natural number is _____ a rational number.

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2 years ago

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PLZ HELP! Complete this proof. Given: a | | b Prove: ΔMOP ~ ΔRON

SashulF [63]

As shown in the given figure
line a // line b ⇒ given
MR and NP are transversals
So, for ΔMOP and ΔRON , we can conclude the following
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3 years ago

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.

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3 years ago

What is the 100th term if the sequence? 5 ,1 ,-3,-7

wel

5 ,1 ,-3,-7 ? what the dickens is going on?

notice, from 5 to 1, is really -4 units,

from 1 to -3, is -4 units as well, 1 -4 = -3 and so on.

so, to get the next term's value, you simply subtract 4 from the current one, namely -4 is the "common difference".

and we also know the first term is 5, ok... then

 $\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=-4\\
a_1=5\\
n=100
\end{cases}
\\\\\\
a_{100}=5+(100-1)(-4)\implies a_{100}=5+(99)(-4)
\\\\\\
a_{100}=5-396\implies a_{100}=-391$

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3 years ago