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UNO [17]
3 years ago
5

an office supply company is shipping a case of paper to a school.There are 10 reams of paper in the case.If each ream of paper w

eighs 32 ounces,what is the weight,in pounds, of the case of paper?
Mathematics
2 answers:
AnnZ [28]3 years ago
7 0
If there are 10 reams, and each ream is 32 ounces, then there is 320 ounces total im the case! Then, if you know that there are 16 ounces in a pound, then you can do 320/16, and get a total of 20 pounds in the case!

Hope this helped!
vodka [1.7K]3 years ago
5 0
16 ounces = 1 pound 1 ream = 32 ounces= 2pounds 10 reams x 2 pounds= 20 case weighs 20 pounds hope this helps
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Greg has enough framing to enclose a rectangular photograph with a perimeter of 160 centimeters. If the width of his photograph
kotegsom [21]

Answer:

32cm

Step-by-step explanation:

Let the width be w. Now we know that the width is 16cm less than the length, this means that the length will thus have a total length of w + 16

Now, we know that the perimeter of a rectangle is 2(l + b)

This means 2(w + w + 16) = 160

4w + 32 = 160

4w = 128

w = 128/4 = 32cm

8 0
3 years ago
Type the ordered pair that is the solution to these questions. <br> 3x-y=13
Dvinal [7]

Answer:

Here's 3 ordered pairs:

(0,-13)(1,-10)(2,-7)

Step-by-step explanation:

They should all be correct. It doesn't matter which one you choose.

3 0
3 years ago
What is the length of the hypotenuse of the triangle?
GarryVolchara [31]
15^2 + 8^2 = 225 + 64 = 289

square root 289 = 17

answer

<span>length of the hypotenuse of the triangle = 17 cm (3rd choice)</span>
6 0
3 years ago
Read 2 more answers
One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R
marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

Urn\ 1 = \{B_1, R_1, R_2, R_3\}

Urn\ 2 = \{R_4, R_5, B_2, B_3\}

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

If urn 2 is selected, the following selection exists:

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

<em>See attachment for possibility tree</em>

Solving (b): The total number of outcome

<u>For urn 1</u>

There are 4 balls in urn 1

n = \{B_1,R_1,R_2,R_3\}

Each of the balls has 3 subsets. i.e.

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

So, the selection is:

Urn\ 1 = 4 * 3

Urn\ 1 = 12

<u>For urn 2</u>

There are 4 balls in urn 2

n = \{B_2,B_3,R_4,R_5\}

Each of the balls has 3 subsets. i.e.

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

So, the selection is:

Urn\ 2 = 4 * 3

Urn\ 2 = 12

Total number of outcomes is:

Total = Urn\ 1 + Urn\ 2

Total = 12 + 12

Total = 24

5 0
2 years ago
32/8+2*12 what is the answer?
lidiya [134]

Answer:

The correct answer is 28

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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