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charle [14.2K]
3 years ago
13

the set $s = \$ contains the first $50$ positive integers. after the multiples of 2 and the multiples of 3 are removed, how many

integers remain in the set $s$?
Mathematics
1 answer:
Sophie [7]3 years ago
5 0

Clearly, |S| = 50.

Count the multiples of 2 between 1 and 50:

⌊50/2⌋ = ⌊25⌋ = 25

(where ⌊x⌋ denotes the "floor of x", or the largest integer that is smaller than or equal to x; in other words, round <u>down</u> to the nearest integer)

Count the multiples of 3 between 1 and 50:

⌊50/3⌋ ≈ ⌊16.667⌋ = 16

Since LCM(2, 3) = 6, the sets of multiples of 2 and multiples of 3 have some overlap. Count the multiples of 6 between 1 and 50:

⌊50/6⌋ ≈ ⌊8.333⌋ = 8

Then by the inclusion/exclusion principle, we remove from S

25 + 16 - 8 = 33

elements, so that the new set S contains 50 - 33 = 17 elements.

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5 0
3 years ago
Ken wants to buy some coffee mugs, m , that cost $5 each and some ground coffee, C, that costs $8 a pound. He has $25.
ValentinkaMS [17]

The inequality representation of the scenario is :

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  • Number of mugs = m
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  • Cost per ground coffee = $8
  • Total amount to spend = $25

  • (Number of mug × cost per mug) + (Number of ground coffee × cost of ground coffee) ≤total amount to spend

  • Representing as an inequality : 5m + 8c ≤ 25

Combination of m and c that makes the inequality true :

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Learn more : brainly.com/question/15748955

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