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Roman55 [17]
3 years ago
12

Divide. Write each quotient as either a whole number or a mixed number in lowest terms. Show your work.

Mathematics
1 answer:
Harman [31]3 years ago
7 0
12 times 8 is 96, 96+5=101    101/8
3 times 4 is 12, 12+1=13         13/4
101/8x4/13 because you switch signs from division to multiplication
=101/26= 3 23/26
26 x 3=78
101-78= 23
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742 hard boiled egg into cartons for a community farm fundraiser. Each carton holds 12 eggs.
ch4aika [34]

Answer:

I am assuming you want to know how many total cartons will be used which is about 61 cartons with a remainder of 10 eggs.

Step-by-step explanation:

To solve for this problem we have to divide 742 by 12:

742/12 = 61.833333...

This means that the "whole" number of cartons will be 61

If you want to find the remaining eggs not in decimal form, you have to mulitlpy 12 by 61 and take this number away from 742 to find the remainder:

**Note: 12 * 61 can also be written as 61 * 12

12 * 61 = 732 → 742 - 732 = 10 eggs


I hope this helps!

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Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected.
Leno4ka [110]

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The average, on the other hand, results from the sum of a list of values ​​divided by the amount of values ​​in the summed list.

Assume that the mean sought is x and consider that the list is composed of a single element a, in that case no random number can be selected, since the mean x must correspond to that number.

If the list were composed of two elements a and b, one of the two values ​​could be chosen randomly, and according to the chosen value the second should be the one whose sum with the previous one results in 2x, this given that the formula of the average \sum\limits_{i=1}^n \frac{ x_{i}}{n}.

With three values ​​a, b and c, it is possible to select two freely, since the thirteen must be the one that balances the sum of a+b, that is (a + b) + c = 3x.

Thus, in general, with n values, it is possible to select n-1 values ​​freely whose sum must be balanced by the last value so that the whole sum is nx.

Answer

In a list of \bf{n} values ​​you can assign \bf{n-1} values ​​freely, that is, you have n-1<em>degrees of freedom.</em>

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Answer:

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Step-by-step explanation:

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