Kurt recorded the daily temperature highs for a science project:
[tex]\[ 27, 28, 28, 28, 29, 29, 30, 31, 31 \][/tex]

The temperature high on another day was included. The new mean is [tex]\[ 30^{\circ} \text{F} \][/tex]. What was this temperature?



Answer :

To solve the problem, we need to determine the temperature on the additional day that allows the new mean temperature to be 30°F. Here is a detailed step-by-step solution:

1. Understand the original data:
Kurt has the following recorded temperatures:
[tex]\[ 27, 28, 28, 28, 29, 29, 30, 31, 31 \][/tex]
Let's count the number of days for which temperatures were recorded:
[tex]\[ \text{Number of days} = 9 \][/tex]

2. Calculate the sum of the original temperatures:
[tex]\[ \text{Sum of original temperatures} = 27 + 28 + 28 + 28 + 29 + 29 + 30 + 31 + 31 \][/tex]
[tex]\[ = 27 + (3 \times 28) + (2 \times 29) + 30 + (2 \times 31) \][/tex]
[tex]\[ = 27 + 84 + 58 + 30 + 62 \][/tex]
[tex]\[ = 261 \][/tex]

3. Determine the new mean temperature
The new mean temperature after including the temperature of the new day is 30°F.

4. Set up the equation to find the new temperature:
Let's assume the temperature on the additional day is \( x \).

The equation for the new mean temperature is:
[tex]\[ \text{New mean} = 30 = \frac{\text{Sum of original temperatures} + x}{\text{Number of days} + 1} \][/tex]

Plugging the values into the equation:
[tex]\[ 30 = \frac{261 + x}{10} \][/tex]

5. Solve for \( x \):
Multiply both sides of the equation by 10 to isolate \( x \):
[tex]\[ 30 \times 10 = 261 + x \][/tex]
[tex]\[ 300 = 261 + x \][/tex]

Subtract 261 from both sides:
[tex]\[ x = 300 - 261 \][/tex]
[tex]\[ x = 39 \][/tex]

Thus, the temperature on the additional day is [tex]\( 39^\circ F \)[/tex].