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hichkok12 [17]
2 years ago
8

Suppose a 5-digit number is formed using the digits from 1 to 9 (without replacement). What is the probability that it will be a

n even number? ​
Mathematics
1 answer:
TiliK225 [7]2 years ago
8 0

By finding the probability that the unit digit is even, we will see that the probability that the 5-digit number will be even is P = 4/9.

So we have a 5-digit number, let's say that it is:

abcde

Now, we have 5 "slots" where we can put numbers from 1 to 9 (without replacement).

We want to find the probability that the number will be even, so we only care for the last digit, e (it must be even).

Let's assume that the first digit at which we assign its value is this, e, the last one.

For this digit we have 9 options, {1, 2, 3, 4, 5, 6, 7, 8, 9}

4 of these options are even numbers.

So there are 4 out of 9 cases where the <u>last digit is even</u>, then the probability that the 5-digit number will be even is just:

P = 4/9

Now the other four digits can be anything (even or odd), so we only care for the probability above.

If you want to learn more about probability, you can read:

brainly.com/question/23044118

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2 years ago
Simplify completely quantity 4 x squared minus 32 x plus 48 all over quantity 3 x squared minus 17 x minus 6. (1 point)
Galina-37 [17]

Given rational expression is:


\frac{(4x^2-32x+48)}{(3x^2-17x-6)}

Now we can simplify this by factoring as shown below:


=\frac{4(x^2-8x+12)}{(3x^2-17x-6)}


=\frac{4(x-2)\left(x-6\right)}{(3x^2-18x+1x-6)}


=\frac{4(x-2)\left(x-6\right)}{3x\left(x-6\right)+1\left(x-6\right)}


=\frac{4(x-2)\left(x-6\right)}{(3x+1)\left(x-6\right)}


=\frac{4(x-2)}{(3x+1)}



Hence final answer is \frac{4(x-2)}{(3x+1)}




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