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Zolol [24]
3 years ago
10

Victor baked 50 cookies. He put 70% of them on a plate to serve at a family party. The rest of the cookies he placed in a box to

take to his soccer team. How many cookies did he put on a plate for the family party?
Mathematics
1 answer:
Ne4ueva [31]3 years ago
6 0

Answer:

Victor put 35 cookies on a plate for the family party.

Step-by-step explanation:

The information provided is:

  • Victor baked 50 cookies.
  • He put 70% of them on a plate to serve at a family party.
  • The rest of the cookies he placed in a box to take to his soccer team.

The total number of cookies Victor baked is, <em>N</em> = 50.

Compute 70% of 50 as follows:

70\%\ \text{of}\ 50=\frac{70}{100}\times 50=35

Thus, Victor put 35 cookies on a plate for the family party.

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The correct answer is option (C) median.

The median will help Emily to find the number that appears in the middle of the 25 numbers that are arranged in ascending order.

<h3>What is the mean, median and mode? </h3>

The mean, median, and mode are the three most commonly used measures of central tendency for populations that do not have much data, that is, they do not need to be grouped.

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The number that appears most frequently in a set of numbers is called the mode.

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4 0
1 year ago
What is the percentage equivalent of 1.48?Select one of the options below as your answer: A. 148% B. 1.48% C. 1480% D. 14.8%
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3 0
3 years ago
Explain your work please
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7 0
3 years ago
Use the principle of inclusion and exclusion to find the number of positive integers less than 1,000,000 that are not divisable
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This is a simple problem based on combinatorics which can be easily tackled by using inclusion-exclusion principle.
We are asked to find number of positive integers less than 1,000,000 that are not divisible by 6 or 4.
let n be the number of positive integers.
∴ 1≤n≤999,999
Let c₁ be the set of numbers divisible by 6 and c₂ be the set of numbers divisible by 4.
Let N(c₁) be the number of elements in set c₁ and N(c₂) be the number of elements in set c₂.

∴N(c₁) = \frac{999,999}{6} = 166666
N(c₂) = \frac{999,999}{4} = 250000
∴N(c₁c₂) = \frac{999,999}{24} = 41667
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N(c₁`c₂`) = 999,999 - (166666+250000) + 41667 = 625000
Therefore, 625000 integers are not divisible by 6 or 4
8 0
4 years ago
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