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Anit [1.1K]
3 years ago
7

Calvin packs boxes of greeting cards into a larger box. The volume of each box of greeting cards is 90 cubic inches. what is the

approximate volume of the large box?

Mathematics
1 answer:
kaheart [24]3 years ago
4 0
Well if you count there are 24 greeting card boxes and the voulme of each is 90 in^3 so it's 2160 in^3.
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Jenny distributes 33 candies and 37 chocolates to each student in the class on her birthday. What is the ratio of the number of
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suffice iris8crib8do Amo stop FM Durham

3 0
3 years ago
Help me pls! giving out brainliest
Nana76 [90]

Answer:

1) 3/2

2) 13/10

3) 1/3

4) 9/7

5) 13/20

Step-by-step explanation:

6 0
2 years ago
Solve each system using the substitution method.
san4es73 [151]
-2x+3(x-1) = -6
-2x + 3x - 3=-6
x-3 = -6
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x= -3
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7 0
4 years ago
Can someone please explain to me please
sveta [45]

Answer:

  120

Step-by-step explanation:

nCk or C(n, k) are ways to write the "combinations" function. That function tells you how many combinations of n things there are when they are taken k at a time.

Here, you have 10 vegetables, and you want to count the number of ways you can take 3 at a time. You are being asked to find the value of 10C3, or C(10,3). You do this by putting 10 where n is in the formula, and 3 where k is. Then do the arithmetic.

  10C3 = (10!)/(3!(10-3)!) = 10·9·8/(3·2·1) = 5·3·8 = 120

120 3-vegetable plates may be made.

__

Read more on Brainly.com - brainly.com/question/14297983#readmore

3 0
3 years ago
For this problem, assume that three students are chosen from 5 freshmen, 4 sophomores, and 4 juniors.
Sati [7]

The formula C(n, r)= \frac{n!}{r!(n-r)!}, where r! is 1*2*3*...r

is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.

for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.


<span>The sample space S in our problem is as follows: </span>

triples selected from 5 freshmen, 4 sophomores, and 4 juniors, so that at least 1 freshman is selected.

The selection of 1 freshman can be done in 5 ways. The selection of the remaining 2 out of 4 freshmen (one is already chosen), 4 sophomores, and 4 juniors can be done in C(12, 2) many ways.

C(12, 2)=\displaystyle{  \frac{12!}{2!10!}= \frac{12\cdot11\cdot10!}{2!10!}=66

So there are in total, 5*66= 330 possible triples.

n(S)=330


The event E "selecting 2 freshmen and 1 junior" can occur in :

C(5, 2)\cdot 4=\displaystyle{ \frac{5!}{2!3!}\cdot4=40 many ways.


so n(E)=40


P(E)=n(E)/n(S)=40/330=0.12



Answer: 0.12


 

4 0
4 years ago
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