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Furkat [3]
3 years ago
14

An integer is chosen at random from first hundred natural numbers. The probability that the integer chosen the multiple of 5 is

Mathematics
1 answer:
Ede4ka [16]3 years ago
7 0

Answer: 0.2

Step-by-step explanation:

The first hundred natural numbers are 1 to 100.

Now, the numbers that are multiples of 5 are the numbers that end in 5 or 0, so we have:

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100

So in the range, we have 20 multiples of 5.

Now, when we select at random, each number has the same probability of being selected, then the probability of randomly selecting a multiple of 5 is equal to the number of multiples of 5 divided the total number of options in the set.

we have 20 multiples of 5, and in the set we have a total of 100 numbers, then the probability is:

P = 20/100 = 0.2

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BARSIC [14]

Hello, first of all we can notice that

5^2-3*5-10=25-15-10=0

and

2*5-10=0

and 0 divided by 0 is undetermined.

So, we need to factorise and simplify first.

x^2-3x-10=(x-5)(x+2)\\\\\text{Because we already known that 5 is a zero...}\\\\\text{... and the sum of the zeroes is 3 = 5 - 2 and the product is -10 = 5 * (-2).}

For x different from 5, we can write

\dfrac{x^2-3x-10}{2x-10}=\dfrac{(x-5)(x+2)}{2(x-5)}=\dfrac{x+2}{2}

So, the limit when x tends to 5 is

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Thank you

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