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Question

A coordinate plane plots temperature in Celsius for different days. The x-axis labeled day, and the y-axis labeled temperature degree Celcius, both range from negative 5 to 5 in increments of 1. Six unlabeled points are plotted. Quadrant 1 is labeled with left parenthesis positive sign comma positive sign right parenthesis. Quadrant 2 is labeled with left parenthesis negative sign comma positive sign right parenthesis. Quadrant 3 is labeled with left parenthesis negative sign comma negative sign right parenthesis. Quadrant 4 is labeled with left parenthesis positive sign comma negative sign right parenthesis. Points are plotted at left parenthesis 2 comma 2 right parenthesis; left parenthesis negative 4 comma 4 right parenthesis, left parenthesis negative 3 comma 0 right parenthesis; left parenthesis negative 2 comma negative 2 right parenthesis; left parenthesis zero comma zero right parenthesis; left parenthesis 4 comma negative 1 right parenthesis.

Alex is tracking temperatures. Today the temperature was 0°
Celsius, which is represented by the point (0, 0)
. If the temperature yesterday was right in between today’s temperature and the temperature two days ago, what was the temperature yesterday?

(1 point)
Yesterday’s temperature was
°
Celsius.
Remaining Attempts : 3

Answer 1

To find yesterday's temperature, we first need to identify the temperatures for today and the day before yesterday.

1. Today's temperature is represented by the point (0, 0), which is 0° Celsius.
2. We need to determine the temperature two days ago. The temperatures are plotted at the following points:
   • (2, 2) → 2° Celsius
   • (-4, 4) → 4° Celsius
   • (-3, 0) → 0° Celsius
   • (-2, -2) → -2° Celsius
   • (0, 0) → 0° Celsius (today)
   • (4, -1) → -1° Celsius

Now, we assume the temperature two days ago could be from one of the points that's not today. Let's denote today's temperature (0°) as T_today, yesterday's temperature as T_yesterday, and the temperature two days ago as T_two_days_ago.

According to the problem, the temperature yesterday was right in between today’s temperature and the temperature two days ago. Mathematically, this can be expressed as: \[ T_yesterday = \frac{T_today + T_two_days_ago}{2} \]

We know that:

• T_today = 0°

To find T_two_days_ago, we can test the possible temperatures from the plotted points:

• If T_two_days_ago = 2° Celsius: \[ T_yesterday = \frac{0 + 2}{2} = 1° \text{ Celsius} \]
• If T_two_days_ago = 4° Celsius: \[ T_yesterday = \frac{0 + 4}{2} = 2° \text{ Celsius} \]
• If T_two_days_ago = 0° Celsius: (not valid since it's today)
• If T_two_days_ago = -2° Celsius: \[ T_yesterday = \frac{0 + (-2)}{2} = -1° \text{ Celsius} \]
• If T_two_days_ago = -1° Celsius: \[ T_yesterday = \frac{0 + (-1)}{2} = -0.5° \text{ Celsius} \]

Among the calculated values, the valid options for yesterday's temperature based on available points are:

• 1° Celsius (if yesterday was between 0° and 2°)
• 2° Celsius (if yesterday was between 0° and 4°)
• -1° Celsius (if yesterday was between 0° and -2°)

Since we are looking for a straightforward solution that adheres to the typical interpretation of "right in between" discrete values, and limiting the candidates based on the possibilities we have explored:

The best fit considering common temperatures would be that yesterday's temperature was 1° Celsius.

So, the final answer is: Yesterday’s temperature was 1° Celsius.