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inessss [21]
2 years ago
13

2 to the power of 3+7+6÷2 BDMAS PLEASE HELP GUYS SCHOOL IN 30 MINUTES​

Mathematics
1 answer:
chubhunter [2.5K]2 years ago
7 0

Answer:

10.5

Step-by-step explanation:

Break the problem down into small parts:

2 to the power of 3= 8

8+7+6=

15+6=21

21/2

21 divided by 2 is 10.5

Hope this helps!

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What is this math question answer 48=a+20
jeyben [28]

Answer:

a=28

Step-by-step explanation:

a+20=48

a=48-20

a=28

6 0
3 years ago
A rectangular prism has a length of 5/2 ft, width of 9/6 ft, and height of 2/3 ft. what is its volume?
ivanzaharov [21]
The formula of the volume of the rectangular prism:

V = length × width × height

We have:
l=\dfrac{5}{2}ft\\\\w=\dfrac{9}{6}ft\\\\h=\dfrac{2}{3}ft\\\\V=l\cdot w\cdot h=\dfrac{5}{2}\cdot\dfrac{9}{6}\cdot\dfrac{2}{3}\\\\=\dfrac{5}{2}\cdot\dfrac{1}{1}\cdot\dfrac{1}{1}=\dfrac{5}{2}=2.5ft.^3
6 0
3 years ago
432566 10.5.623467.14,5<br><br> 4. What is the 1s (lower) quartile of the data set?
egoroff_w [7]

Answer:

I feel sorry that looks hard. see how do you like it. waiting for an answer and then someone finally answers. and there being a smart Butt

Step-by-step explanation:

4 0
3 years ago
Special right triangles
Korolek [52]

Answer:

see down

Step-by-step explanation:

cos 60 = \frac{x}{6\sqrt{3} } \\x = 6\sqrt{3}  *  cos60\\x = 6\sqrt{3} * \frac{1}{2}  \\x = 3\sqrt{3}

5 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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