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djverab [1.8K]
2 years ago
9

In an archery competition, each archer shoots 3 arrows at a target. An archer’s score is calculated by measuring each arrow’s di

stance from the center of the target and finding the mean. One archer has a mean distance of 1.6 inches and a MAD of 1.3 inches in the first round. In the second round, this same archer’s arrows are farther from the center but more consistent. What values for the mean and MAD would fit this description for the second round?
Mathematics
1 answer:
balandron [24]2 years ago
8 0

Answer:

Mean: 3 Mad: 1/3

Step-by-step explanation:

First we need to find out the numbers, so to get a mean of 3 we have to get to 9 ebcause there are 3 arrows.

  • We have to make sure that the arrows are not too close to the center
  • You can have decimals

2nd: Once you find the numbers which are 2.5, 3, and 3.5 you add them together and divide it by 3 to get 3 the mean

3rd: Find the MAD but adding .5 and .5 to get 1 then divide it by 3 and you get 1/3 as the MAD.

Hope this helps!

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Roger can run one mile in 8 minutes. jeff can run one mile in 6 minutes. if jeff gives roger a 1 minute head​ start, how long wi
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Recall your d = rt, distance = rate * time.

now, if Roger can do 1 mile in 8 minutes, so in 1 minute, he has done then 1/8 of a mile, so his rate is 1/8 miles per minute.

if Jeff can do 1 mile in 6 minutes, he's faster, in 1 minute he has done 1/6 of a mile, so his rate is 1/6 miles per minute.

now, when Jeff catches up with Roger, the distance covered by both will be the same, say "d" miles, because, at that millisecond, Jeff will be neck and neck with Roger, and their covered distance will be the same.

now, Jeff is generous and let Roger roll on for 1 minute before him, so, by the time time Roger has covered "d" miles, he has been running for say "t" minutes.

however, since Jeff started later by 1 minute, he hasn't been running for "t" minutes, but for "t - 1" minutes.

\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mpm}{rate}&\stackrel{minutes}{time}\\
&------&------&------\\
Roger&d&\frac{1}{8}&t\\\\
Jeff&d&\frac{1}{6}&t-1
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=\frac{1}{8}t\\\\
d=\frac{1}{6}(t-1)\\
------\\
\boxed{\frac{1}{8}t}=\cfrac{t-1}{6}
\end{cases}
\\\\\\
\cfrac{t}{8}=\cfrac{t-1}{6}\implies 6t=8t-8\implies 8=2t\implies \cfrac{8}{2}=t\implies \boxed{\stackrel{mins}{4}=t}
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