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kogti [31]
2 years ago
15

9_12-15-22-5_25-15-21translate the code ​

Mathematics
1 answer:
ipn [44]2 years ago
3 0

Answer:

this is sum u made up

Step-by-step explanation:

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Plz help with 6,7,8 and 9
Reil [10]
Grayson's mistake was that he multiplied 4 and 3 and then used the exponent he had to square 3 and then multiply it by 4.

Emily's mistake was that she added 2 to 36 instead of multiplying it by -2

Pat's mistake was that he forget to make y into -2 instead of 2

The right way to do this is 4(3^2)+2(-2)
(3^2)=9 9×4=36 2(-2)=-4 -4+9=5
3 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.

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
Marilyn uses a credit card with a 19.9% APR compounded monthly to pay for car repairs totaling $991.38. She can pay $410 per mon
Leokris [45]
Marilyn's finance charge at the end of the first month will be
  $991.38 × 0.199/12 = $16.44
The balance subject to the next month's finance charge will be
  $991.38 +16.44 -410.00 = $597.82

The finance charge at the end of the second month will be
  $597.82 × 0.199/12 = $9.91
The balance remaining after the second payment will be
  $597.82 +9.91 -410.00 = $197.73

The finance charge applied at the end of the third month is
  $197.73 × .199/12 = $3.28
so Marilyn can make one final payment of
  $197.73 +3.28 = $201.01
to pay off the balance.

In all, Marilyn has paid 2×$410.00 +201.01 = 
  $1021.01 . . . . . . . . corresponds to the first choice

_____
In real life, Marilyn's credit card may not accrue any finance charge until after the first statement on which the charge appears. Thus the total cost of the purchase may be only $1004.02. The attached spreadsheet shows the beginning balance and the finance charges for each month for the two different scenarios.

8 0
2 years ago
The highest common factor of 98 and 42
ohaa [14]

Answer:14

Step-by-step explanation:Start by writing like this below

98 =2×7×7

42 =2×3×7

H.C.F=2×7

=14

Take the number which is common like in 98 and 42, (2) and (7) are common

and then multiply the two numbers.

Hope it helps you!

5 0
2 years ago
Read 2 more answers
Four more than the product of x and y
Naddika [18.5K]
The answer to this one is<em> xy + 4</em>
5 0
3 years ago
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