Question

*plz help it’s due tomorrow:)

The graph shows the temperature,y, in a freezer x minutues after it was turned on. Five minutes after being turned on, the temperature was actually three degrees from what the trend line shows. What values could the actual temperature be after the freezer was on for five minutes?

Answer 1

The values that the actual temperature could be after the freezer was on for five minutes are 12°F or 18°F.

A trendline is sometimes referred to as a line of best fit and it is a statistical tool which is commonly used in conjunction with a scatter plot, in order to determine whether or not there's any form of correlation (either positive or negative) between a given data.

By critically observing the scatter plot, we can logically deduce that the trend line passes through the point (5, 15), which implies that the temperature is 15°F, 5 minutes after the freezer was turned on.

Since the actual temperature deviates by 3°F from what the trend line shows, we have:

Actual temperature = 15 ± 3

Actual temperature = 15 - 3 = 12°F

Actual temperature = 15 + 3 = 18°F

Answer 2

• The actual temperature after the freezer was on for five minutes could be 18ºF or 12ºF

Step-by-step explanation:

First, you must find the y-value returned by the line in the graph for a time (x) of five minutes, and then use the deviation of 3º to conclude the range of the real temperature.

1. y- value at x = 5 min.

The trend line (blue line in the graph) passes through the point (5,15), which means that it predicts a temperature of 15ºF when the time (x) is 5 minutes, or five minutes after the freezer was turned on.

2. Actual temperature value

It is stated that the temperature was actually three degrees from what the trend line shows, that means that the actual temperature could be 3ºF more or 3ºF less than the predicted temperature of 15ºF.

Mathematically:

• Temperature = 15ºF ± 3ºF

• Temparature = 15ºF + 3ºF or 15ºF - 3ºF

• Temperature = 18ºF or 12ºF ← answer.