Answer :
To determine when William's phone will lose its charge completely, we can use the given equation of the line of best fit, which is [tex]\( y = -10x + 100 \)[/tex].
Here, [tex]\( y \)[/tex] represents the percentage of charge left, and [tex]\( x \)[/tex] represents the time in hours. We need to find out when the phone's charge percentage [tex]\( y \)[/tex] becomes [tex]\( 0 \)[/tex].
Given the equation:
[tex]\[ y = -10x + 100 \][/tex]
Set [tex]\( y \)[/tex] to [tex]\( 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = -10x + 100 \][/tex]
To isolate [tex]\( x \)[/tex], we start by moving [tex]\( -10x \)[/tex] to the other side of the equation:
[tex]\[ 10x = 100 \][/tex]
Next, divide both sides of the equation by 10:
[tex]\[ x = \frac{100}{10} \][/tex]
Thus, solving the equation gives:
[tex]\[ x = 10 \][/tex]
So, it will take about 10 hours for the phone to lose its charge completely.
The correct answer is:
B. about 10 hours
Here, [tex]\( y \)[/tex] represents the percentage of charge left, and [tex]\( x \)[/tex] represents the time in hours. We need to find out when the phone's charge percentage [tex]\( y \)[/tex] becomes [tex]\( 0 \)[/tex].
Given the equation:
[tex]\[ y = -10x + 100 \][/tex]
Set [tex]\( y \)[/tex] to [tex]\( 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = -10x + 100 \][/tex]
To isolate [tex]\( x \)[/tex], we start by moving [tex]\( -10x \)[/tex] to the other side of the equation:
[tex]\[ 10x = 100 \][/tex]
Next, divide both sides of the equation by 10:
[tex]\[ x = \frac{100}{10} \][/tex]
Thus, solving the equation gives:
[tex]\[ x = 10 \][/tex]
So, it will take about 10 hours for the phone to lose its charge completely.
The correct answer is:
B. about 10 hours