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Elodia [21]
3 years ago
9

A key code must contain 6 numerals. There are 10 numerals available. Using these numerals, how many different key codes may be c

reated?
Mathematics
1 answer:
sattari [20]3 years ago
4 0
Imagine tha the each numeral of the code is written over each line:

__  __  __  __  __  __
1      2    3    4    5   6

Every position may have any of the 10 different numerals.

So over the first line you can put 10 different numerals and the same is true for each of the six positions.

So you can create 10x10x10x10x10x10 = 1,000,000 different codes.
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20 points to whoever answers this first
taurus [48]
Answer: 
71˚

Explanation:
To find the <em>mean</em> of a sum of numbers, you add all the numbers together, and then divide by how many numbers there are. 
71 + 72 + 68 + 67 + 74 + 72 + 73 = 497
497 ÷ 7 = 71
4 0
3 years ago
Read 2 more answers
What is 4.45 repeating as a fraction.
KiRa [710]

This would be so easy for you to solve, in about 3 minutes with a calculator.

A fraction means division ... (the top number) divided by (the bottom number).

Take the fraction part of each choice, do the division for each one,
and see which one gives you  0.4545... repeating.

A).  227/500  =  0.454     That's not 0.45 repeating.

B).  9/20  =  0.45             That's not 0.45 repeating.

C).  5/11  =  0.4545...      THAT's  0.4545... repeating !

D).  9/2  =  4.5                 That's not 0.45 repeating.

6 0
3 years ago
The quotient of 10 and 2
Irina18 [472]

Answer:

5

Step-by-step explanation:

10 divided by 2 is 5

4 0
3 years ago
Read 2 more answers
The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability th
maksim [4K]

Answer:

The probability that there are 3 or less errors in 100 pages is 0.648.        

Step-by-step explanation:

In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.

For the given Poisson distribution the mean is p = 0.03 errors per page.

We have to find the probability that there are three or less errors in n = 100 pages.

Let us denote the number of errors in the book by the variable x.

Since there are on an average 0.03 errors per page we can say that

the expected value is, \lambda = E(x)

                                       = n × p

                                       = 100 × 0.03

                                       = 3

Therefore the we find the probability that there are 3 or less errors on the page as

     P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)

                 

   Using the formula for Poisson distribution for P(x = X ) = \frac{e^{-\lambda}\lambda^X}{X!}

Therefore P( X ≤ 3) = \frac{e^{-3} 3^0}{0!} + \frac{e^{-3} 3^1}{1!} + \frac{e^{-3} 3^2}{2!} + \frac{e^{-3} 3^3}{3!}

                                 = 0.05 + 0.15 + 0.224 + 0.224

                                 = 0.648

 The probability that there are 3 or less errors in 100 pages is 0.648.                                

7 0
3 years ago
Read 2 more answers
Sara is collecting signatures to pu on her towns ballot. So far, she has 432 signatures. This is 36% of the number of signatures
In-s [12.5K]

Sara needs to collect 1200 signatures.

Step-by-step explanation:

Let,

x be the total number of signatures she needs to collect

Given percentage = 36%

Signature collected = 432

According of statement;

36% of x = 432

\frac{36}{100}*x=432\\

\frac{36x}{100}=432\\Multiplying\ both\ sides\ by 100\\100*\frac{36x}{100}=432*100\\36x=43200\\Dividing\ both\ sides\ by\ 36\\\frac{36x}{36}=\frac{43200}{36}\\x=1200

Sara needs to collect 1200 signatures.

Keywords: Percentage, Division

Learn more about percentages at:

  • brainly.com/question/10879401
  • brainly.com/question/10940255

#LearnwithBrainly

6 0
3 years ago
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