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When Jayden left his house in the morning, his cell phone battery was partially charged. Let B represent the charge remaining in Jayden's battery, as a percentage, t hours since Jayden left his house. The table below has select values showing the linear relationship between t and B. Determine how many hours after leaving his house it would take until the phone's battery level got down to 31.5%. t B 1.5 38.25 6 18. 8. 9

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Final answer:

To determine how many hours after leaving his house it would take until the phone's battery level got down to 31.5%, we can use the linear relationship between time and battery charge remaining percentage.

Step-by-step explanation:

To determine how many hours it would take until the phone's battery level reaches 31.5%, we can use the linear relationship between t (hours) and B (battery charge remaining as a percentage).

  • t = 1.5, B = 38.25
  • t = 6, B = 18
  • t = 8, B = 9

We can see that the battery charge is decreasing linearly over time. To find the equation of this linear relationship, we can use the two points: (1.5, 38.25) and (8, 9). Using the formula for the equation of a line, y = mx + b, where m is the slope and b is the y-intercept, we can find the values of m and b.

Using the two points (1.5, 38.25) and (8, 9), we can calculate the slope m: m = (9 - 38.25) / (8 - 1.5) = -29.25 / 6.5 = -4.5

Then, we can substitute one of the points into the equation y = mx + b to find b. Let's use the point (1.5, 38.25): 38.25 = -4.5 * 1.5 + b. Solving for b, we get b = 45.75

Now that we have the equation of the line, y = -4.5x + 45.75, we can set y (battery charge as a percentage) to 31.5 and solve for x (hours): 31.5 = -4.5x + 45.75.

Subtracting 45.75 from both sides gives -14.25 = -4.5x. Dividing both sides by -4.5 gives x = 3.

Therefore, it would take 3 hours after leaving his house for the phone's battery level to reach 31.5%.

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