Answer :
Sure! Let's break this down step by step.
### Question 1
To determine if a score of 77 would be within par for the course where the par is 72 and a golfer must score within 5 strokes of par:
1. Write an equation:
To express being within 5 strokes of par (72), we use the absolute value equation that captures the range:
[tex]$ |p - 72| = 5 $[/tex]
2. Solution for Minimum and Maximum Scores:
Using the equation [tex]\( |p - 72| = 5 \)[/tex]:
- When the expression inside the absolute value is 5:
[tex]$ p - 72 = 5 $[/tex]
[tex]$ p = 77 $[/tex]
- When the expression inside the absolute value is -5:
[tex]$ p - 72 = -5 $[/tex]
[tex]$ p = 67 $[/tex]
Thus, the acceptable score range is between 67 and 77, inclusive.
3. Check if 77 is within the range:
Since 77 falls within this range (67 to 77), a score of 77 is indeed within par for the course.
4. Evaluate the given options:
- [tex]\( |p + 72| = 5 \)[/tex]:
This does not correctly represent the scenario because it does not result in the proper range.
- [tex]\( |p + 5| = 72 \)[/tex]:
This also does not correctly represent the scenario because it does not compute the correct minimum or maximum scores.
- [tex]\( |p - 72| = 5 \)[/tex]:
This is the correct equation, and as stated, 77 is within the minimum and maximum scores to be within par.
- [tex]\( |p - 5| = 72 \)[/tex]:
This does not correctly represent the scenario because it does not compute the correct minimum or maximum scores.
So, the correct and acceptable equation is [tex]\( |p - 72| = 5 \)[/tex], and a score of 77 is within par for the course.
### Question 2
To determine the equation representing the maximum and minimum yards thrown on average for a professional quarterback who throws a football for an average of 65 yards plus or minus 10 yards:
1. Understand the range:
The quarterback's average throwing distance is 65 yards, but it can vary by plus or minus 10 yards. This means the range of the throwing distance is:
[tex]$ 65 \pm 10 $[/tex]
Therefore, the minimum and maximum throwing distances are:
[tex]\[ 65 - 10 = 55 \quad \text{(minimum)} \][/tex]
[tex]\[ 65 + 10 = 75 \quad \text{(maximum)} \][/tex]
2. Write the equation:
To represent this range using an absolute value equation, we set up the equation so that it can be written in the form:
[tex]$ |y - 65| = 10 $[/tex]
3. Evaluate the given options:
- [tex]\( |y + 10| = 65 \)[/tex]:
This does not provide the correct interpretation for the scenario described.
- [tex]\( |y - 10| = 65 \)[/tex]:
This also does not result in the proper representation because it suggests the average is 10 yards with a deviation of 65 yards.
- [tex]\( |y + 65| = 10 \)[/tex]:
This does not correctly represent the average and range given.
- [tex]\( |y - 65| = 10 \)[/tex]:
This is the correct equation, describing the range of throwing distances as being within 65 yards, plus or minus 10 yards.
Thus, the correct equation is [tex]\( |y - 65| = 10 \)[/tex].
Therefore, the answers to the questions are:
1. [tex]\( |p - 72| = 5 \)[/tex]; Yes, because 77 is within the minimum and maximum scores to be within par.
2. [tex]\( |y - 65| = 10 \)[/tex].
### Question 1
To determine if a score of 77 would be within par for the course where the par is 72 and a golfer must score within 5 strokes of par:
1. Write an equation:
To express being within 5 strokes of par (72), we use the absolute value equation that captures the range:
[tex]$ |p - 72| = 5 $[/tex]
2. Solution for Minimum and Maximum Scores:
Using the equation [tex]\( |p - 72| = 5 \)[/tex]:
- When the expression inside the absolute value is 5:
[tex]$ p - 72 = 5 $[/tex]
[tex]$ p = 77 $[/tex]
- When the expression inside the absolute value is -5:
[tex]$ p - 72 = -5 $[/tex]
[tex]$ p = 67 $[/tex]
Thus, the acceptable score range is between 67 and 77, inclusive.
3. Check if 77 is within the range:
Since 77 falls within this range (67 to 77), a score of 77 is indeed within par for the course.
4. Evaluate the given options:
- [tex]\( |p + 72| = 5 \)[/tex]:
This does not correctly represent the scenario because it does not result in the proper range.
- [tex]\( |p + 5| = 72 \)[/tex]:
This also does not correctly represent the scenario because it does not compute the correct minimum or maximum scores.
- [tex]\( |p - 72| = 5 \)[/tex]:
This is the correct equation, and as stated, 77 is within the minimum and maximum scores to be within par.
- [tex]\( |p - 5| = 72 \)[/tex]:
This does not correctly represent the scenario because it does not compute the correct minimum or maximum scores.
So, the correct and acceptable equation is [tex]\( |p - 72| = 5 \)[/tex], and a score of 77 is within par for the course.
### Question 2
To determine the equation representing the maximum and minimum yards thrown on average for a professional quarterback who throws a football for an average of 65 yards plus or minus 10 yards:
1. Understand the range:
The quarterback's average throwing distance is 65 yards, but it can vary by plus or minus 10 yards. This means the range of the throwing distance is:
[tex]$ 65 \pm 10 $[/tex]
Therefore, the minimum and maximum throwing distances are:
[tex]\[ 65 - 10 = 55 \quad \text{(minimum)} \][/tex]
[tex]\[ 65 + 10 = 75 \quad \text{(maximum)} \][/tex]
2. Write the equation:
To represent this range using an absolute value equation, we set up the equation so that it can be written in the form:
[tex]$ |y - 65| = 10 $[/tex]
3. Evaluate the given options:
- [tex]\( |y + 10| = 65 \)[/tex]:
This does not provide the correct interpretation for the scenario described.
- [tex]\( |y - 10| = 65 \)[/tex]:
This also does not result in the proper representation because it suggests the average is 10 yards with a deviation of 65 yards.
- [tex]\( |y + 65| = 10 \)[/tex]:
This does not correctly represent the average and range given.
- [tex]\( |y - 65| = 10 \)[/tex]:
This is the correct equation, describing the range of throwing distances as being within 65 yards, plus or minus 10 yards.
Thus, the correct equation is [tex]\( |y - 65| = 10 \)[/tex].
Therefore, the answers to the questions are:
1. [tex]\( |p - 72| = 5 \)[/tex]; Yes, because 77 is within the minimum and maximum scores to be within par.
2. [tex]\( |y - 65| = 10 \)[/tex].
