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Svetradugi [14.3K]
2 years ago
14

Any one smart and can help me catch up? ill give many points

Mathematics
1 answer:
Sergio [31]2 years ago
7 0

Answer:

4/15

Step-by-step explanation:

1) find common denominator, 15 in this problem

\frac{4}{5}= \frac{12}{15}

2) solve

\frac{12}{15}- \frac{8}{15} = \frac{4}{15}

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Simplify: -2 (8 - 6s + 11t - 3u)
kati45 [8]

Answer:

Step-by-step explanation:

-16 + 12s - 22t + 6u

5 0
3 years ago
Read 2 more answers
2 1.3.1 Study: Multiplying Binomials Drag the tiles to the blanks. Distribute. - 4(x - 2) =​
ehidna [41]

Answer:

I am not entirely sure what you are asking here but if I am right then the answer is -4(x-2) = 8 - 4x

Step-by-step explanation:

Multiplying binomials all you have to do is distribute the 4

Multiply  -4 by x to get: -4x

Muliply -4 by -2 to get: 8 (Because two negatives make a positive)

Now you have -4x + 8

Rewrite so it flows better and the answer is:

8 - 4x

6 0
2 years ago
Find the area of the figure. (Sides meet at right angles.) <br>​
ikadub [295]

Answer:

44ft^2

Step-by-step explanation:

Break the shape into 3 rectangles, two 3 by 2's and one 4 by 8. (Get 4 by subtracting 3 from 7 for the left and right most side)

3 x 2 = 6 ft^2

since there's two rectangles you end up 6 + 6 so you get 12 ft^2

then 4 x 8 = 32 ft^2

add 32 and 12, get 44 ft^2

8 0
1 year ago
I will give brainlest
AfilCa [17]
1. X + 5
2.15-15
3. C= ( 3x9.95)+(2x14.98)
4.?
5. 12,000+500x ?idk
6 0
2 years ago
Plz help me!!!!
sveta [45]
A.) For the Junior Varsity Team, mean would be the appropriate measure of center since the data is <span>symmetric or well-proportioned while we should use standard deviation for getting the measure of spread since it also measures the center and how far the values are from the mean.

b.) For the Varsity Team, the median would be the appropriate measure of the center since the data is skewed left and not evenly distributed so median could be used since it does not account for outliers while we use IQR or interquartile range in measuring the spread of data since IQR does not account for the data that is skewed. </span>
4 0
3 years ago
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