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Fed [463]
3 years ago
6

Martin has 60 magazines and 42 books to put into stacks. Each stack will have 7 magazines and 4 books. How many complete stacks

of 7 magazines and 4 books can Ms. Martin make?
Mathematics
1 answer:
Delicious77 [7]3 years ago
7 0

Answer:

8 complete stacks

Step-by-step explanation:

Total magazine = 60

Total books = 42

Each stack will have 7 magazines and 4 books.

Each stack = 7 magazines + 4 books

How many complete stacks of 7 magazines and 4 books can Ms. Martin make?

Number of stack of 7 magazines possible = total magazines / magazines per stack

= 60/7

= 8 4/7

Complete stacks with 7 magazines = 8

Number of stack of 4 books possible = total books / books per stack

= 42/4

= 10 1/2

Complete stacks with 4 books = 10

If 7 magazines on each stacks can make 8 complete stacks

Then, 4 books on each stack can make 8 complete stacks

Note: after 8 stacks using 7 magazines each, magazines will not be available to complete other stacks

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First we need to find the slope of line k.  Using the slope formula, we fill in accordingly: m= \frac{4-1}{1-3}.  The slope is -3/2.  The slope of a line that is perpendicular to line k will have the opposite sign (so, positive) and will be the reciprocal of the line k's slope (so, 2/3).  You could also say, more "mathematically legal", that the product of the slopes of perpendicular lines is -1, but it's easier to just say that they are opposite reciprocals.  The slope you're looking for is 2/3.  Choice 3 above.
6 0
3 years ago
A coin will be tossed three times, and each toss will be recorded as heads (H) or tails (T). Give the sample space describing al
cricket20 [7]

Answer:

Sample space:

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Third toss is heads:

{HHT, HTT, THT, TTT}

First toss is heads, second tails and third heads:

{HTH}

Explanation:

In order to find the sample space we need to write all the possible outcomes from tossing the coin three times.You can start by writting all the possible outcomes for the first toss being a head and finish with all the possible outcomes for the first toss being Tails. The number of elements in the sample space can be found by multiplying the possible outcomes of each toss together, so 2*2*2=8

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

For the third toss being heads we will have one restriction this time. Which is that the third toss is Tails. You can take the elements from the sample space where the third element is a tails. In this case, the number of elements can be calculated by multiplying the possible outcomes for only the first two tosses 2*2=4

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

For first toss is heads, second tails and third heads there is only one possible way for you to get this outcome so there is only one element to this set.

{HTH}

4 0
3 years ago
There is something wrong with the reason. I don’t know what it is. Could someone help me. The picture of the figure is on the ne
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9514 1404 393

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  CPCTC

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The applicable reason for statement 4 is "Corresponding Parts of Congruent Triangles are Congruent (CPCTC)".

The reason shown in your problem statement is applicable only within one triangle. The segments of interest are in two different triangles.

8 0
3 years ago
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Answer:

what is it equal to

Step-by-step explanation:

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3 years ago
There are 11 women and 9 men in a certain club. If the club is to select a committee of 2 women and 2 men, how many different su
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We can tell it’s not a permutation because it’s picking two people. Order doesn’t matter.

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