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Naya [18.7K]
3 years ago
13

How many 3-digit numbers can you make such that they do not have any zero digit in any place?

Mathematics
2 answers:
Andre45 [30]3 years ago
7 0

Answer:

There are 648 numbers that are 3 digits numbers such that none of the digits are repeated. For the first digit, you have 9 choices (digits 1 -9) you don't want zero as the first digit.

Tanya [424]3 years ago
5 0

EDIT: There are 727 numbers that are 3 digits numbers such that none of the digits are repeated. For the first digit, you have 9 choices (digits 1 -9) you don't want zero as the first digit.------------------------------------ 727 numbers

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20 PTS!!!
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Answer: x = -2

<u>Step-by-step explanation:</u>

vertical line means it will be x = ___

Set the x-values equal to each other and solve for x.

3 - 2x = x + 9

<u>    +2x</u> <u>+2x     </u>

3        = 3x + 9

<u>-9      </u>    <u>      -9 </u>

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Is 5 1/2 yds greater than 192 ins?
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Answer:

no

Step-by-step explanation:

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3 years ago
If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers. Roun
AlekseyPX

Answer:

P = 0.008908

Step-by-step explanation:

The complete question is:

The table below describes the smoking habits of a group of asthma sufferers

               Nonsmokers      Light Smoker     Heavy smoker      Total

Men                        303                     35                           37        375

Women                   413                     31                            45        489

Total                        716                     66                           82        864

If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers.

The number of ways in which we can select x subjects from a group of n subject is given by the combination and it is calculated as:

nCx=\frac{n!}{x!(n-x)!}

Now, there are 82C2 ways to select subjects that are both heavy smokers. Because we are going to select 2 subjects from a group of 82 heavy smokers. So, it is calculated as:

82C2=\frac{82!}{2!(82-2)!}=3321

At the same way, there are 864C2 ways to select 2 different people from the 864 subjects. It is equal to:

864C2=\frac{864!}{2!(864-2)!}=372816

Then, the probability P that two different people from the 864 subjects are both heavy smokers is:

P=\frac{82C2}{864C2}=\frac{3321}{372816}=0.008908

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