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Sergio039 [100]
3 years ago
11

Kenji walks 44 feet in 10 seconds. At this rate, how many miles does Kenji walk in an hour? show your work. (1 mile = 5,280 feet

)
Mathematics
2 answers:
Andreas93 [3]3 years ago
7 0

Answer:

Kenji walks 3 miles in an hour.

Step-by-step explanation:

1 hour = 3600 seconds

3600/10 = 360

So there are 360 increments of 10 in an hour. Multiply 44 by 360 to find the amount of feet in an hour:

(44)(360) = 15840

Now we have to divide 15,840 by the amount of feet in a mile (5,280):

15840/5280 = 3

So Kenji walks 3 miles in an hour.

expeople1 [14]3 years ago
4 0

Answer:

3 miles

Step-by-step explanation:

44 in 10 secs

264 in one min

15840 in an hour

15840 / 5280 = 3

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Answer:

The width W is :

W=\frac{3}{5}

Step-by-step explanation:

<em>Let A be the area of the rectangle </em>

<em>Let L be the length of the rectangle</em>

<em>Let W be the width of the rectangle </em>

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Formula:

A = L × W

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\begin{aligned}A &=L \times W \\\Rightarrow W &=\frac{A}{L} \\\Rightarrow W &=\frac{\frac{3}{8}}{\frac{5}{8}} \\&=\frac{3}{8} \times \frac{8}{5} \\&=\frac{3 \times 8}{8 \times 5} \\&=\frac{3}{5}\end{aligned}

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What is the answer to number 18?
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Answer: 3n ×9

Step-by-step explanation:

The product of 3 times a N number equals 3n or 3 ×N but you should write 3n then times it by 9

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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.

The total amount of possibilities for this case is

4 * {n \choose 2}

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

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

I hope that works for you!

4 0
3 years ago
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