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stellarik [79]
3 years ago
15

Can someone give me some 4th grade math problems​

Mathematics
2 answers:
stich3 [128]3 years ago
8 0
1. What is 32×25?

2. What is 856×4?

3. What is 567×40?

Show your work. Sorry if these are too hard.
Leni [432]3 years ago
3 0

33×12=

45×23=

18÷_=9

14÷_=2

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Harvey is working on a weekly budget. He earns $459 per week. Harvey spends $200 per week on rent and utilities plus $85 per wee
Sedaia [141]

Answer:

$197

Step-by-step explanation:

459-200-85+23

8 0
2 years ago
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A backpacking tent in the shape of a triangular prism has sides that are 6 ft and a length of 7 ft. What is the volume of the te
Zanzabum
<span>the volume of the tent to the nearest whole number is 109.</span>
3 0
3 years ago
A + 2b = 1<br>2a + b = 8​
xxMikexx [17]

Answer:

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4 0
3 years ago
Each container has a base of 4 1/4 yds by 1 2/3 yds. What is the area covered by all the containers?
svet-max [94.6K]

Answer:

See Explanation

Step-by-step explanation:

Given

Base Dimension

Length = 4\frac{1}{4}yd

Width = 1\frac{2}{3}yd

Required

The base area of all containers

First, calculate the base area of 1 container.

This is calculated as:

Area = Length * Width

Area = 4\frac{1}{4}yd * 1\frac{2}{3}yd

Express as improper fraction

Area = \frac{17}{4}yd * \frac{5}{3}yd

So, we have:

Area = \frac{17*5}{4*3}yd^2

Area = \frac{85}{12}yd^2

The number of containers is not given. So, I will use 'n' as the number of containers.

So, we have:

Total = n * Area

Total= n * \frac{85}{12}yd^2

--------------------------------------------------------------------------------------------

Assume n is 3 (i.e. 3 containers)

The total area is:

Total= 3 * \frac{85}{12}yd^2

Total= \frac{85}{4}yd^2

Total= 21\frac{1}{4}yd^2

--------------------------------------------------------------------------------------------

6 0
3 years ago
How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each com
Mnenie [13.5K]

<span>If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

(15 choose 2) x (20 choose 2)


(15 choose 2) = 105

(20 choose 2) = 190

190*105 = 19950

Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.</span>

 

<span>If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

(15 choose 1) x (20 choose 3)


(15 choose 1) = 15

(20 choose 3) = 1140

15*1140 = 17100

Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.</span>

 

<span>We now find the total outcomes of having a group with 4 women.

We know this is the same as saying (20 choose 4) = 4845</span>

Therefore, there are 4845 ways to have a group of 4 with 4 women.

 

We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

 

19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee


5 0
3 years ago
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