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alekssr [168]
3 years ago
11

What is 2 2/5 times 3 1/4

Mathematics
2 answers:
Naddik [55]3 years ago
5 0

Answer:

7 4/5

Step-by-step explanation:

2 2/5 = 12/5

3 1/4 =  13/4

12/5 x 13/4 = 156/20 = 7 16/20

7 16/20 / 4 = 7 4/5

babymother [125]3 years ago
3 0
7 4/5

Steps:
12/5x13/4=156/20
Step 2:
Simplify 156/20
Answer: 7 4/5
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The variable in a relation who’s value depends on the value of the independent variable is called what?
Viefleur [7K]
That is called the dependent variable. It depends upon the value of the independent variable.

f(x)=y

So we can see by inspection that the dependent variable y value depends upon the input value of the independent variable x

Any questions please feel free to ask. Thanks
3 0
3 years ago
A store is having a going -out-of buisness sale.a television set that orginally cost 3000 has been marked down by30% what do you
allsm [11]

2,247

Step-by-step explanation:

take 3000 and minus that by 30 percent, then after that, add 7 percent to get your total.

7 0
2 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.

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
Which is greater 0.4 or 0.137
ElenaW [278]
<h2>0.4 is greater </h2>

hope that helps :)

4 0
3 years ago
Read 2 more answers
Solve the function.<br> h(x)= 6x - 10<br> ————-<br> X + 3 <br><br><br> Help please
Rus_ich [418]

Answer:

X+3=0

X=-3

H(x)=6*-3 - 10

H(x) =-18 - 10

H(x) =-28

8 0
2 years ago
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