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lys-0071 [83]
3 years ago
13

a 4.5-foot-long table and a 127-centimeter-long table side by side to form one long table.What is the combined length of the two

tables in inches?
Mathematics
1 answer:
dexar [7]3 years ago
7 0
4.5 feet is 54 inches, and 127cm is 50 inches, so 50+54=104inches
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The square root of 7 lies between what two numbers
ExtremeBDS [4]

The square root of 7 would like between 2 and 3. This is because 2 x 2 is 4's square root and 3 x 3 is nine's square root. Seven is between four and nine.

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Simplify (-2 + 3i) + 2(5 + 6i). Enter your answer in the form a + bi.
Reptile [31]

We will have the following:

(-2+3i)+2(5+6i)=-2+3i+10+12i

=(10-2)+(3i+12i)=8+15i

So, the solution would be:

8+15i

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How to find the formula for the sum of series that is not arithmetic nor geometric?please give examples
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It takes Maya 30 minutes to solve 5 logic puzzles, and it takes amy 28 minutes to solve 4 logic puzzles. use models, to show the
nalin [4]

Answer:

maya - 6 min per puzzle

amy - 7 min per puzzle

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Step-by-step explanation:

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2 years ago
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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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