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3 years ago

How do you find the percent of a whole number?

2 answers:

7 0

A percentage figure shows that the quantity of parts per one hundred a segment sum compares to. For instance, "85 percent" is another method for saying "85 sections for each 100." To calculate a percentage, the entire sum must be known, in addition to the percentage or portion amount.

soldier1979 [14.2K]3 years ago

4 0

It is easy to determine the percentage of a whole number and this is the steps to do so:
First multiply the number by the percent for example 87 * 68 = 5916, Second divide the answer by 100, we will just move the decimal two places to the left (5916/100) = 59.16, and then round to the desired precision 59.16 rounded to the nearest whole number which is 59.

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Serggg [28]

Answer:45-67-3

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

uppose we want to build a rectangular storage container with open top whose volume is $$ cubic meters. Assume that the cost of m

Ainat [17]

Answer:

a = length of the base = 2.172 m

b = width of the base = 1.357 m

c = height = 4.072 m

Step-by-step explanation:

Suppose we want to build a rectangular storage container with open top whose volume is 12 cubic meters. Assume that the cost of materials for the base is 12 dollars per square meter, and the cost of materials for the sides is 8 dollars per square meter. The height of the box is three times the width of the base. What’s the least amount of money we can spend to build such a container?

lets call a = length of the base

b = width of the base

c = height

V = a.b.c = 12

Area without the top:

Area = ab + 2bc + 2ac

Cost = 12ab + 8.2bc + 8.2ac

Cost = 12ab + 16bc + 16ac

height = 3.width

c = 3b

Cost = 12ab + 16b.3b + 16a.3b = 12ab + 48b² + 48ab = 48b² + 60ab

abc = 12 → ab.3b = 12 → 3ab² = 12 → ab² = 4 → a = 4/b²

Cost = 48b² + 60ab = 48b² + 60b.4/b² = 48b² + 240/b

C(b) = 48b² + 240/b

C'(b) = 96b - 240/b²

Minimum cost: C'(b) = 0

96b - 240/b² = 0

(96b³ - 240)/b² = 0

96b³ - 240 = 0

96b³ = 240

b³ = 240/96

b³ = 2.5

b = 1.357m

c = 3b = 3*1.357 = 4.072m

a = 4/b² = 2.172m

6 0

3 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

Oksana_A [137]

Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.