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Sati [7]
3 years ago
13

One fourth of the candy bars were not chocolate. There were 45 chocolate candy bars. One-fifth of the chocolate candy bars conta

ined nuts. How many nonchocolate candy bars were there? How many chocolate candy bars contained nuts?
Mathematics
1 answer:
xeze [42]3 years ago
4 0
1/4 =non chocolate bar
3/4 = chocolate bar

45 ÷ 3 = 15
So 1/4 = 15 candy bars

And 1/5 candy bars contained nuts...

45 x 1/5 = 9 chocolate bars contained nuts

The answer is ...
15 non chocolate bars
And
9 chocolate candy bars contained nuts
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78 * 3/4 hour,
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The least-squares regression line is ________.
Murljashka [212]

Answer:

a. the line that passes through the most data points.

Step-by-step explanation:

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5 0
3 years ago
The coordinates of point P on a coordinate grid are (−5, −6). Point P is reflected across the y-axis to obtain point Q and acros
Scilla [17]
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5 0
4 years ago
If ef bisects angle ceb,angle cef=7x+31 and angle feb=10x-3
Nastasia [14]

Given : Angle  < CEB is bisected by EF.

< CEF = 7x +31.

< FEB = 10x-3.

We need to find the values of x and measure of < FEB, < CEF  and < CEB.

Solution: Angle  < CEB is bisected into two angles < FEB and < CEF.

Therefore,   < FEB = < CEF.

Substituting the values of < FEB and < CEF, we get

10x -3 = 7x +31

Adding 3 on both sides, we get

10x -3+3 = 7x +31+3.

10x = 7x + 34

Subtracting 7x from both sides, we get

10x-7x = 7x-7x +34.

3x = 34.

Dividing both sides by 3, we get

x= 11.33.

Plugging value of x=11.33 in < CEF = 7x +31.

We get

< CEF = 7(11.33) +31 =  79.33+31 = 110.33.

< FEB  = < CEF =  110.33 approximately

< CEB = < FEB +  < CEF  = 110.33 +110.33 = 220.66 approximately



7 0
3 years ago
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