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Explanation: To determine if a number is divisible by 132, we need to check if it’s divisible by both 4 and 33 because 132 is the product of 4 and 33. First, we check divisibility by 4: A number is divisible by 4 if its last two digits form a number that is divisible by 4. Then, we check divisibility by 33: A number is divisible by 33 if the sum of its digits is divisible by 3 and the number formed by its last two digits is divisible by 11. Let’s check each of the given numbers: 264: Divisible by 4 (last two digits are 64) and 33 (sum of digits = 2 + 6 + 4 = 12, which is divisible by 3, and 64 is divisible by 11). 396: Divisible by 4 (last two digits are 96) and 33 (sum of digits = 3 + 9 + 6 = 18, which is divisible by 3, and 96 is divisible by 11). 462: Divisible by 4 (last two digits are 62) and 33 (sum of digits = 4 + 6 + 2 = 12, which is divisible by 3, and 62 is not divisible by 11). 792: Divisible by 4 (last two digits are 92) and 33 (sum of digits = 7 + 9 + 2 = 18, which is divisible by 3, and 92 is not divisible by 11). 968: Divisible by 4 (last two digits are 68) and 33 (sum of digits = 9 + 6 + 8 = 23, which is not divisible by 3). 2178: Not divisible by 4 (last two digits are 78). 5184: Divisible by 4 (last two digits are 84) and 33 (sum of digits = 5 + 1 + 8 + 4 = 18, which is divisible by 3, and 84 is divisible by 11). 6336: Divisible by 4 (last two digits are 36) and 33 (sum of digits = 6 + 3 + 3 + 6 = 18, which is divisible by 3, and 36 is divisible by 11). Out of the given numbers, 264, 396, 5184, and 6336 are divisible by 132. So, there are 4 numbers divisible by 132.

Explanation:

To determine if a number is divisible by 132, we need to check if it's divisible by both 4 and 33 because 132 is the product of 4 and 33.

First, we check divisibility by 4: A number is divisible by 4 if its last two digits form a number that is divisible by 4.

Then, we check divisibility by 33: A number is divisible by 33 if the sum of its digits is divisible by 3 and the number formed by its last two digits is divisible by 11.

Let's check each of the given numbers:

264: Divisible by 4 (last two digits are 64) and 33 (sum of digits = 2 + 6 + 4 = 12, which is divisible by 3, and 64 is divisible by 11).

396: Divisible by 4 (last two digits are 96) and 33 (sum of digits = 3 + 9 + 6 = 18, which is divisible by 3, and 96 is divisible by 11).

462: Divisible by 4 (last two digits are 62) and 33 (sum of digits = 4 + 6 + 2 = 12, which is divisible by 3, and 62 is not divisible by 11).

792: Divisible by 4 (last two digits are 92) and 33 (sum of digits = 7 + 9 + 2 = 18, which is divisible by 3, and 92 is not divisible by 11).

968: Divisible by 4 (last two digits are 68) and 33 (sum of digits = 9 + 6 + 8 = 23, which is not divisible by 3).

2178: Not divisible by 4 (last two digits are 78).

5184: Divisible by 4 (last two digits are 84) and 33 (sum of digits = 5 + 1 + 8 + 4 = 18, which is divisible by 3, and 84 is divisible by 11).

6336: Divisible by 4 (last two digits are 36) and 33 (sum of digits = 6 + 3 + 3 + 6 = 18, which is divisible by 3, and 36 is divisible by 11).

Out of the given numbers, 264, 396, 5184, and 6336 are divisible by 132. So, there are 4 numbers divisible by 132.

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- 1/2
- 1/3
- 3/8
**3/7**

**Explanation**

The probability that a red ball is drawn at random from the bag is 10/15.

Assuming the first ball was red, then the probability of picking a second red ball is now 9/14.

(You don’t need to worry about whether the first ball was blue, because if it was, both balls cannot be read, so picking a red ball first is the only probability you need to consider)

Multiply the two probabilities together and you have the combined probability of both balls being red.

10/15 * 9/14 = 90/ 210 = 3/7

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