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DedPeter [7]
3 years ago
15

The average of 5 numbers is 33. If the average of first 2 numbers is 35 and that of last 2 is 31, then find the 3rd number.

Mathematics
2 answers:
-BARSIC- [3]3 years ago
3 0
The answer is c. 33.
irinina [24]3 years ago
3 0

Answer:

the answer is c

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Do more than 50 percent of the children in a certain group have brown hair?
babymother [125]

Answer:

What we know is that 30% of the group have brown hair and are girls. It means that in the other 70% we have the boys with brown hair, the other boys and the other girls.

In the case all the girls have brown hair, the 70% remaining would be boys. So 70% of then would have brown hair. (0.7*0.7=0.49 -> 49% of the children would be brown hair) so 49% (boys) and 30% (girls) are more than 50%..

Now, let's suppouse that 90% of the group are girls. It means 10% are boys and 70% of that 10% (=7%) are the boys in the group that have brown hair. Now 7% + 30% of the children have brown hair, wich isn't more than the half.

We have explained with two extreme cases that the information is not enough for making a precise answer

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2 years ago
The formula F=9/5C+32 is used to convert Celsius to Fahrenheit temperature. If the temperature is 20 C what is it in Fahrenheit?
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3 years ago
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The perimeter of a rectangle is 36m. If its length is increased by 1m and the width is increased by 2m, its area will increase b
maw [93]

Answer:

  80 m^2

Step-by-step explanation:

The given information lets you write two equations involving length (x) and width (y).

  • 2(x +y) = 36 . . . . the perimeter is 36 m
  • (x+1)(y+2) -xy = 30 . . . . increasing the length and width increases area

The second of these equations simplifies to another linear equation, giving a system of linear equations easily solved.

  xy +y +2x + 2 -xy = 30

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Dividing the first equation by 2 gives

  x +y = 18

and subtracting this from the above equation gives ...

  (2x +y) -(x +y) = 28 -18

  x = 10

Then

  y = 18 -10 = 8

The area of the original rectangle is xy = 10·8 = 80 m^2.

4 0
3 years ago
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