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27 Sep 2023, 01:30
Kudos
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D
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Difficulty:
75%
(hard)
Question Stats:
60% (02:27) correct
40%
(02:18)
wrong
based on 279
sessions
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The Mindojo College basketball team has tryouts for the position of point guard. The coaching staff decides to consider only players whose height is within 2 inches of 6 feet and 3 inches, the average height for point guards in the NBA. (1 foot = 12 inches)
If x is the player's height in feet, which of the following inequalities can the coaching staff use to find the players with suitable heights?
A. \(|x-2| \leq 6.3\)
B. \(|x-6.3|\leq 2\)
C. \(|x -\frac{1}{6}| \leq \frac{25}{4}\)
D. \(| x -\frac{25}{4} | \leq \frac{1}{6}\)
E. \(| x - \frac{25}{4} | \geq \frac{1}{6}\)
If x is the player's height in feet, which of the following inequalities can the coaching staff use to find the players with suitable heights?
A. \(|x-2| \leq 6.3\)
B. \(|x-6.3|\leq 2\)
C. \(|x -\frac{1}{6}| \leq \frac{25}{4}\)
D. \(| x -\frac{25}{4} | \leq \frac{1}{6}\)
E. \(| x - \frac{25}{4} | \geq \frac{1}{6}\)
27 Sep 2023, 12:00
Kudos
Bookmarks
Bunuel
6 ft + 3 in = (6 × 12) in + 3 in = 75 in = 75/12 ft
According to the coach’s rule, a height x, in feet, is suitable if it is between (75 – 2)/12 = 73/12 ft and (75 + 2)/12 = 77/12 ft, inclusive.
So, we have:
73/12 ≤ x ≤ 77/12 [Subtract 75/12, the midpoint of the interval, from both sides.]
-2/12 ≤ x – 75/12 ≤ 2/12
-1/6 ≤ x – 25/4 ≤ 1/6
|x – 25/4| ≤ 1/6
Answer: D
General Discussion
08 May 2025, 19:30
Kudos
Bookmarks
The coaching staff decides to consider only players whose height is within 2 inches of 6 feet and 3 inches, the average height for point guards in the NBA. (1 foot = 12 inches)
3 inches = \(\frac{3}{12}\) foot = \(\frac{1}{4}\) foot
If x is the player's height in feet, which of the following inequalities can the coaching staff use to find the players with suitable heights?
x has to be within 2 inches from 6 foot + \(\frac{1}{4}\) foot = \(\frac{25}{4}\) foot
2 inch = \(\frac{2}{12}\) foot = \(\frac{1}{6}\) foot
=> \(\frac{25}{4}\) - \(\frac{1}{6}\) ≤ x ≤ \(\frac{25}{4}\) + \(\frac{1}{6}\)
Subtracting \(\frac{25}{4}\) from all the sides we get
=> \(\frac{25}{4}\) - \(\frac{1}{6}\) - \(\frac{25}{4}\) ≤ x - \(\frac{25}{4}\) ≤ \(\frac{25}{4}\) + \(\frac{1}{6}\) - \(\frac{25}{4}\)
=> -\(\frac{1}{6}\) ≤ x - \(\frac{25}{4}\) ≤ \(\frac{1}{6}\)
This is of the form -a ≤ X ≤ a
=> |X| ≤ a
=> |x - \(\frac{25}{4}\)| ≤ \(\frac{1}{6}\)
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Inequality + Absolute value Problems
3 inches = \(\frac{3}{12}\) foot = \(\frac{1}{4}\) foot
If x is the player's height in feet, which of the following inequalities can the coaching staff use to find the players with suitable heights?
x has to be within 2 inches from 6 foot + \(\frac{1}{4}\) foot = \(\frac{25}{4}\) foot
2 inch = \(\frac{2}{12}\) foot = \(\frac{1}{6}\) foot
=> \(\frac{25}{4}\) - \(\frac{1}{6}\) ≤ x ≤ \(\frac{25}{4}\) + \(\frac{1}{6}\)
Subtracting \(\frac{25}{4}\) from all the sides we get
=> \(\frac{25}{4}\) - \(\frac{1}{6}\) - \(\frac{25}{4}\) ≤ x - \(\frac{25}{4}\) ≤ \(\frac{25}{4}\) + \(\frac{1}{6}\) - \(\frac{25}{4}\)
=> -\(\frac{1}{6}\) ≤ x - \(\frac{25}{4}\) ≤ \(\frac{1}{6}\)
This is of the form -a ≤ X ≤ a
=> |X| ≤ a
=> |x - \(\frac{25}{4}\)| ≤ \(\frac{1}{6}\)
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Inequality + Absolute value Problems
