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motikmotik
3 years ago
11

The sum of three consecutive odd numbers is 63. Find the numbers.

Mathematics
2 answers:
bagirrra123 [75]3 years ago
7 0
<h2>Required Answer:</h2>

19,21,23

<h2>Question:</h2>

The sum of three consecutive odd numbers is 63. Find the numbers.

<h2>Given:</h2>
  • sum of three consecutive odd numbers = 63
<h2>Let:</h2>

Three consecutive odd numbers be:

  • x
  • x+2
  • x+4
<h2>A/Q</h2>

sum of three consecutive odd numbers is 63

.°. x+(x+2)+(x+4)=63

→x+x+x=63-2-4

→3x=63-6

→3x=57

→x=57/3

→x=19

<u>We get value of x → 19</u>

Now put values in each number

First odd number= x

.°.<u>First odd number=19</u>

Second odd number =x+2

.°. Second odd number= 19+2

<u>Second odd number = 21</u>

Third odd number =x+4

.°.Third odd number=19+4

<u>Third odd number=23</u>

hope it helps!♡

Westkost [7]3 years ago
5 0

Answer:

19, 21, 23

Step-by-step explanation:

Let the first odd number be n.

Then the second, consecutive odd number will be (n+2).

And the third will be (n+4).

We know that they sum to 63. Hence, we can write the following equation:

n+(n+2)+(n+4)=63

Solve for n. Combine like terms:

3n+6=63

Subtract 6 from both sides:

3n=57

Divide both sides by 3:

n=19

Hence, the first odd number is 19.

Therefore, our sequence is: 19, 21, 23.

Note: If we get an even or non-integer value for our n, then there are no three consecutive odd integers that exists that sum to 63.

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37. Socks. In your sock drawer you have 4 blue socks, 5 grey socks, and 3 black ones. Half asleep one morning, you grab 2 socks
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Answer:

a) \frac{1}{22}

b) \frac{7}{22}

c) \frac{5}{11}

d) \frac{35}{66}

Step-by-step explanation:

Given,

Number of blue socks = 4,

Grey socks = 5,

Black socks = 3,

Total socks = 4 + 5 + 3 = 12,

Ways of choosing any 2 socks = ^{12}C_2

=\frac{12!}{2!10!}

=\frac{12\times 11}{2}

=6\times 11

= 66,

a) Ways of choosing 2 blue socks = ^3C_2

= 3,

Since, \text{Probability}=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}

Thus, the probability of selecting 2 blue socks = \frac{3}{66}=\frac{1}{22}

b) Ways of choosing 2 shocks, non of them are grey socks = ^7C_2

=\frac{7!}{2!5!}

=7\times 3

=21

Thus, the probability of selecting no grey socks = \frac{21}{66}

=\frac{7}{22}

c) ways of selecting atleast 1 black sock = ways of selecting 1 black sock + ways of selecting 2 black socks

=^3C_1\times ^9C_1+^3C_2

=3\times 9 + 3

=27 + 3

= 30,

Thus, the probability of selecting at least 1 black sock = \frac{30}{66}

=\frac{5}{11}

d) Ways of selecting a green sock = ^5C_1\times ^7C_1

= 35,

Thus, the probability of selecting a green sock = \frac{35}{66}

3 0
3 years ago
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