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Anastaziya [24]
4 years ago
8

Write an expression to represent the sum of three consecutive even numbers. Let C represent the first number in the series

Mathematics
2 answers:
Anna [14]4 years ago
7 0
Let, the first number = C
second number will be = C + 2   ['cause next will be odd so we can't include that]
Third number = C + 4

Now, their sum would be:
C + C+2 + C+4 = 3C + 6

In short, Your Answer would be 3C + 6

Hope this helps!


vladimir2022 [97]4 years ago
7 0
Consecutive even integers would differ by 2.

So if the first even integer is C,

The next is C + 2

The next after is C + 2 + 2 = C + 4

The sum of the three consecutive integers =  C + (C + 2) + (C + 4)

 C + C + 2 + C + 4

C + C + C + 2 + 4

3C + 6

Sum = 3C + 6

Hope this helps.
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andrey2020 [161]

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The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are E(1)= ^9C_2 places in which the other 2 red balls can be placed.

A red ball drawn third, there are E(3)= ^7C_2 places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are E(5)= ^5C_2 places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are E(7)= ^3C_2 places in which the other 2 red balls can be placed.

The total number of total event is S= ^{10}C_3

The probability that A selects the first red ball is

P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}

P(A \text{wins})=\frac{36+21+10+3}{120}

P(A \text{wins})=\frac{70}{120}

P(A \text{wins})=0.5833

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3 years ago
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