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tankabanditka [31]
3 years ago
9

Joseph has a collection of stickers. When the stickers are aranged in piles of 3, there are 2 stickers left over. When the stick

ers are aranged in piles of 4, there are 2 stickers left over. When the stickers are arranged in piles of 12, how many stickers are left over?
Mathematics
1 answer:
Elodia [21]3 years ago
6 0

Answer:

2 stickers will be left over

Step-by-step explanation:

To answer the question, all you need to do is figure out at what value of stickers would the remaining stickers be equal to 2 if the piles are made up of either 3 to a set or 4 to a set. To do this, we need to figure out the size of the whole set of stickers.

Lets assume that the total number of stickers is 14. Lets test the 2 given conditions. If we divide it in piles of 3 then we will have a pile of 12 stickers (4 piles of 3 stickers each) with 2 stickers remaining.

Lets test the second condition. If we divide it in piles of 4, we will again have 12 stickers (3 piles of 4 stickers each) with 2 stickers remaining.

Now that both conditions have been satisfied, all we have to do is see how many stickers are left if we make a pile of 12 stickers. In this case, we will have just one pile of 12 stickers and, again, we will be left with 2 stickers.

Therefore, the answer is 2 stickers

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Problems Involving Fundamental Counting Principle
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The questions are illustrations of combination and probability

  • Combination involves the ways of choosing or selecting an item from a group of items
  • Probability involves the chance of choosing or selecting an item from a group of items

<h3>1. Number of systems to choose from</h3>

The given parameters are:

Monitors = 4

Keyboard = 2

CPU = 4

Printer = 3

The number of systems to choose from is:

n = 4 * 2 * 4 * 3 = 96

Hence, Patrick can choose from 96 systems

<h3>2. Combination of Ice cream</h3>

The given parameters are:

Containers = 2

Flavors = 4

The number of total combination to choose from is:

n = 2 * 4C3 = 8

Hence, the number of total combination is 8

<h3>3. Emma dressing up</h3>

The given parameters are:

Blouse = 5

Skirt = 3

Shoes = 4 pairc

The number of ways to dress up:

n = 5 * 3 * 4 = 60

Hence, Emma can dress up in 60 ways

<h3>4. Members of committee</h3>

The given parameters are:

Students = 6

Members of committee = 3

The number of ways to form a committee is:

n = 6C3 = 20

Hence, there are 20 ways to form the committee

<h3 /><h3>5. Probability of consonant</h3>

We have:

Letters = 11

Consonnats = 7

The probability of selecting a consonant is:

P = 7/11

P = 0.636

Hence, the probability of selecting a consonant is: 0.636

<h3>6. Probability of not selecting a grade 8 student</h3>

We have:

Total students = 34

Students not in grade 8 = 25

The probability that a student is not in grade 8 is:

P = 25/34

P = 0.735

Hence, the probability that a student is not in grade 8 is 0.735

<h3>7. Probability of a secondary color </h3>

Not enough details to solve

<h3>8. Probability of a prime number</h3>

There are 9 prime numbers in a total number of 31 numbers from 4 to 34.

So, the probability that a number is prime is:

P = 9/31

P = 0.290

Hence, the probability that a number is prime is 0.290

Read more about combination and probability at:

brainly.com/question/24756209

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