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The average (arithmetic mean) weight of 5 crates is 250 pou
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05 Jun 2017, 02:18
05 Jun 2017, 02:18
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72% (02:16) correct
27% (01:41) wrong based on 174 sessions
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The average (arithmetic mean) weight of 5 crates is 250 pounds. The 2 lightest crates weigh between 200 and 205 pounds each, inclusive, and the 2 heaviest crates weigh between 300 and 310 pounds each, inclusive. If the weight of the fifth crate is x pounds, then x is expressed by which of the following?
A) 220 ≤ x ≤ 250
B) 230 ≤ x ≤ 260
C) 240 ≤ x ≤ 270
D) 250 ≤ x ≤ 270
E) 260 ≤ x ≤ 280
Re: The average (arithmetic mean) weight of 5 crates is 250 pou
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11 Oct 2017, 05:17
11 Oct 2017, 05:17
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We need to find an interval, thus we can proceed by first computing the sum of the crates weights as mean times number, i.e. 250*5 = 1250.
Now, we can derive x at the two extremes, i.e. when the lightest and heaviest crates have the higher weight and when they have the smaller weight.
In the first case, x = 1250-2*200-2*300= 250. This is the highest value of x because we have taken the minimum weight for the other crates.
In the second one, x = 1250-2*205-2*310= 220. This is the lowest value of x.
Thus the interval is 220-250. Answer A
Now, we can derive x at the two extremes, i.e. when the lightest and heaviest crates have the higher weight and when they have the smaller weight.
In the first case, x = 1250-2*200-2*300= 250. This is the highest value of x because we have taken the minimum weight for the other crates.
In the second one, x = 1250-2*205-2*310= 220. This is the lowest value of x.
Thus the interval is 220-250. Answer A
Re: The average (arithmetic mean) weight of 5 crates is 250 pou
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10 Feb 2019, 09:30
10 Feb 2019, 09:30
1
To calculate the greatest possible value of X: x = 1250-(2*200-2*300)= 250
To calculate the least possible value of x = x = 1250-2*205-2*310= 220.
So answer is A
To calculate the least possible value of x = x = 1250-2*205-2*310= 220.
So answer is A
