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If a basketball team scores an average (arithmetic mean) of x points p
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11 Aug 2021, 10:42
11 Aug 2021, 10:42
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If a basketball team scores an average (arithmetic mean) of x points per game for n games and then scores y points in its next game, what is the team's average score for the n+1 games?
(A) \(\frac{(nx + y)}{(n + 1)}\)
(B) \(x + \frac{y}{(n + 1)}\)
(C) \(x + \frac{y}{n}\)
(D) \(\frac{n(x + y)}{(n + 1)}\)
(E) \(\frac{(x + ny)}{(n + 1)}\)
(A) \(\frac{(nx + y)}{(n + 1)}\)
(B) \(x + \frac{y}{(n + 1)}\)
(C) \(x + \frac{y}{n}\)
(D) \(\frac{n(x + y)}{(n + 1)}\)
(E) \(\frac{(x + ny)}{(n + 1)}\)
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Re: If a basketball team scores an average (arithmetic mean) of x points p
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11 Aug 2021, 11:16
11 Aug 2021, 11:16
1
Carcass wrote:
If a basketball team scores an average (arithmetic mean) of x points per game for n games and then scores y points in its next game, what is the team's average score for the n+1 games?
(A) \(\frac{(nx + y)}{(n + 1)}\)
(B) \(x + \frac{y}{(n + 1)}\)
(C) \(x + \frac{y}{n}\)
(D) \(\frac{n(x + y)}{(n + 1)}\)
(E) \(\frac{(x + ny)}{(n + 1)}\)
(A) \(\frac{(nx + y)}{(n + 1)}\)
(B) \(x + \frac{y}{(n + 1)}\)
(C) \(x + \frac{y}{n}\)
(D) \(\frac{n(x + y)}{(n + 1)}\)
(E) \(\frac{(x + ny)}{(n + 1)}\)
Sum of points in \(n\) games = n(x)
Points in the next game = y
Total points = nx + y
Average = \(\frac{nx+ y }{n + 1}\)
Hence, option A
If a basketball team scores an average (arithmetic mean) of x points p
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11 Aug 2021, 11:29
11 Aug 2021, 11:29
1
x= mean of points scored in n games
xn= total number of points in n games
xn+y = total number of points in n+1 games
(xn+y)/(n+1) = new average
A
xn= total number of points in n games
xn+y = total number of points in n+1 games
(xn+y)/(n+1) = new average
A
