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when madison left her house in the morning, her cell phone battery was …

Question

when madison left her house in the morning, her cell phone battery was partially charged. let b represent the charge remaining in madisons battery, as a percentage, t hours since madison left her house. the table below has select values showing the linear relationship between t and b. determine the percentage charge that madisons phone loses each hour.

tb
630
8.511.25

Explanation:

Step1: Find change in time and battery charge

The change in time $\Delta t$ from $t = 1$ to $t = 6$ is $\Delta t=6 - 1=5$ hours. The change in battery - charge $\Delta B$ is $\Delta B=30 - 67.5=- 37.5$.

Step2: Calculate rate of battery - charge loss per hour

The rate of change (slope) $m$ of the linear relationship between $t$ and $B$ is given by the formula $m=\frac{\Delta B}{\Delta t}$. Substituting the values of $\Delta B$ and $\Delta t$, we have $m=\frac{30 - 67.5}{6 - 1}=\frac{-37.5}{5}=-7.5$.

Answer:

7.5%