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Firlakuza [10]
3 years ago
14

You randomly select k integers between 1 and 100, inclusive. What is the smallest k that guarantees that at least one pair of th

e selected integers will sum to 101? Prove your answer.
Mathematics
1 answer:
vredina [299]3 years ago
6 0

Answer:

51

Step-by-step explanation:

The possible components to sum up to 101 can only be divided into 2 groups, 1 is larger than 50 and the other is less than or equals to 50

For example

50 + 51 = 101

52 + 49 = 101

60 + 41 = 101

...

99 + 2 = 101

100 + 1 = 101

Therefore, the worst case scenario is to pick all numbers from only 1 group, either all number less than or equal to 50, which there are 50 of them from 1 to 50, or greater than 50, which there are 50 of them from 51 to 100.

So k has to be at least 51 to guarantee that at least one pair of the selected integers will sum to 101

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Mendel crossed tall pea plants with short pea plants. The offspring were all tall plants.
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An expression is shown below: 2x3y + 18xy − 10x2y − 90y Part A: Rewrite the expression by factoring out the greatest common fact
SCORPION-xisa [38]

Answer:

Part A : 2y( x³ + 9x - 5x² - 45 ), Part B : 2y( x - 5 )( x² + 9 )

Step-by-step explanation:

Part A : Let's break every term down here to their " prime factors ", and see what is common among them,

2x³y + 18xy − 10x²y − 90y -

2x³y  = 2 * x³ * y,

18xy = 2 * 3 * 3 * x * y,

− 10x²y = 2 * - 5 * x² * y, - so as you can see for this example I purposely broke down - 10 into 2 and - 5. I could have placed the negative on the 2, but as that value was must likely common among all the terms, I decided to place it on the 5. The same goes for " − 90y. " I placed the negative there on the 5 once more.

− 90y = 2 * - 5 * 3 * 3 * y

The terms common among each term are 2 and y. Therefore, the GCF ( greatest common factor ) is 2x. Let's now factor the expression using this value.

2y( x³ + 9x - 5x² - 45 )

Part B : Let's simply factor this entire expression. Of course starting with the " factored " expression : 2y( x³ + 9x - 5x² - 45 ),

2y\left(x^3+9x-5x^2-45\right) - Factor out " (x^3+9x-5x^2-45\right)) " by grouping,

\left(x^3-5x^2\right)+\left(9x-45\right) - Factor 9 from 9x - 45 and x² from x³ - 5x²,

9\left(x-5\right)+x^2\left(x-5\right) - Factor out common term x - 5,

\left(x-5\right)\left(x^2+9\right) - And our solution is thus 2y( x - 5 )( x² + 9 )

3 0
3 years ago
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