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Sagot :
Let's carefully analyze Corey’s approach and find where he went wrong in determining the outside temperature when there are 80 swimmers at City Pool given the regression equation [tex]\( y = 1.505x - 88.21 \)[/tex].
Corey's steps were:
1. He directly substituted 80 for [tex]\( x \)[/tex] in the equation: [tex]\( y = 1.505(80) - 88.21 \)[/tex]
2. He calculated [tex]\( y \)[/tex] as [tex]\( 1.505 \times 80 - 88.21 \)[/tex], resulting in:
[tex]\[ y = 120.4 - 88.21 = 32.19 \][/tex]
3. He concluded that when there are 80 swimmers, the temperature is [tex]\( 32.2^\circ F \)[/tex] (approximating [tex]\( 32.19 \)[/tex]).
Firstly, we need to identify the mistake:
- Corey used the value 80 for [tex]\( x \)[/tex] (temperature) rather than [tex]\( y \)[/tex] (number of swimmers).
- The problem states that there are 80 swimmers, so 80 should be assigned to [tex]\( y \)[/tex], not [tex]\( x \)[/tex].
Here is the correct step-by-step solution:
1. Given: [tex]\( y = 80 \)[/tex] (since 80 represents the number of swimmers).
2. Substitute [tex]\( y = 80 \)[/tex] into the equation [tex]\( y = 1.505x - 88.21 \)[/tex]:
[tex]\[ 80 = 1.505x - 88.21 \][/tex]
3. To isolate [tex]\( x \)[/tex], first add 88.21 to both sides of the equation:
[tex]\[ 80 + 88.21 = 1.505x \][/tex]
[tex]\[ 168.21 = 1.505x \][/tex]
4. Next, solve for [tex]\( x \)[/tex] by dividing both sides by 1.505:
[tex]\[ x = \frac{168.21}{1.505} \][/tex]
5. This yields:
[tex]\[ x \approx 111.77 \][/tex]
So, if there are 80 swimmers at the pool, the correct outside temperature is approximately [tex]\( 111.77^\circ F \)[/tex].
Corey’s mistake was indeed substituting 80 for [tex]\( x \)[/tex] (the temperature) instead of [tex]\( y \)[/tex] (the number of swimmers). He should have started by setting [tex]\( y \)[/tex] to 80 and solving the equation for [tex]\( x \)[/tex].
Thus, Corey's error was substituting 80 for [tex]\( x \)[/tex] rather than [tex]\( y \)[/tex]. If he had correctly substituted 80 for [tex]\( y \)[/tex], he would have found the temperature to be approximately [tex]\( 111.77^\circ F \)[/tex].
Corey's steps were:
1. He directly substituted 80 for [tex]\( x \)[/tex] in the equation: [tex]\( y = 1.505(80) - 88.21 \)[/tex]
2. He calculated [tex]\( y \)[/tex] as [tex]\( 1.505 \times 80 - 88.21 \)[/tex], resulting in:
[tex]\[ y = 120.4 - 88.21 = 32.19 \][/tex]
3. He concluded that when there are 80 swimmers, the temperature is [tex]\( 32.2^\circ F \)[/tex] (approximating [tex]\( 32.19 \)[/tex]).
Firstly, we need to identify the mistake:
- Corey used the value 80 for [tex]\( x \)[/tex] (temperature) rather than [tex]\( y \)[/tex] (number of swimmers).
- The problem states that there are 80 swimmers, so 80 should be assigned to [tex]\( y \)[/tex], not [tex]\( x \)[/tex].
Here is the correct step-by-step solution:
1. Given: [tex]\( y = 80 \)[/tex] (since 80 represents the number of swimmers).
2. Substitute [tex]\( y = 80 \)[/tex] into the equation [tex]\( y = 1.505x - 88.21 \)[/tex]:
[tex]\[ 80 = 1.505x - 88.21 \][/tex]
3. To isolate [tex]\( x \)[/tex], first add 88.21 to both sides of the equation:
[tex]\[ 80 + 88.21 = 1.505x \][/tex]
[tex]\[ 168.21 = 1.505x \][/tex]
4. Next, solve for [tex]\( x \)[/tex] by dividing both sides by 1.505:
[tex]\[ x = \frac{168.21}{1.505} \][/tex]
5. This yields:
[tex]\[ x \approx 111.77 \][/tex]
So, if there are 80 swimmers at the pool, the correct outside temperature is approximately [tex]\( 111.77^\circ F \)[/tex].
Corey’s mistake was indeed substituting 80 for [tex]\( x \)[/tex] (the temperature) instead of [tex]\( y \)[/tex] (the number of swimmers). He should have started by setting [tex]\( y \)[/tex] to 80 and solving the equation for [tex]\( x \)[/tex].
Thus, Corey's error was substituting 80 for [tex]\( x \)[/tex] rather than [tex]\( y \)[/tex]. If he had correctly substituted 80 for [tex]\( y \)[/tex], he would have found the temperature to be approximately [tex]\( 111.77^\circ F \)[/tex].
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