The average weight of 12 men decreases by 2 kg when one of them weighing 91 kg was replaced by a new man. Find the weight of the new man.
(x + y + z) / 3
Area of a square = side length squared (x+y)2 (x+y)(x+y) x2+xy+yx+y2 Area = x2+2xy+y2
12/25 = 2/y so y = 2*25/12 = 25/6 = 4.166... units.
To find ( x ), we set up the equation for the average: ( \frac{3 + x}{2} = 5 ). Solving this gives ( 3 + x = 10 ), so ( x = 7 ). For ( y ), we set up ( \frac{5 + y}{2} = 7 ), leading to ( 5 + y = 14 ), so ( y = 9 ). The average of ( x ) and ( y ) is ( \frac{7 + 9}{2} = 8 ).
The average weight of 12 men decreases by 2 kg when one of them weighing 91 kg was replaced by a new man. Find the weight of the new man.
3y - 3
(x+y+z)/3The mean average is (x+y+z)/3 miles.
Add the mileage for each day and divide by the number of days to get the average: (x + y + z) ÷ 3 = Average
multiply the x-length and the y-length and 2
(x + y + z) / 3
Area of a square = side length squared (x+y)2 (x+y)(x+y) x2+xy+yx+y2 Area = x2+2xy+y2
If the average of 3 and x is 5, and the average of 5 and y is 7, what is the average of x and y?
"Yo y mi hombre" translates to "Me and my man" in English.
12/25 = 2/y so y = 2*25/12 = 25/6 = 4.166... units.
To find ( x ), we set up the equation for the average: ( \frac{3 + x}{2} = 5 ). Solving this gives ( 3 + x = 10 ), so ( x = 7 ). For ( y ), we set up ( \frac{5 + y}{2} = 7 ), leading to ( 5 + y = 14 ), so ( y = 9 ). The average of ( x ) and ( y ) is ( \frac{7 + 9}{2} = 8 ).
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