Daily temperature

The average of daily temperature measurements in one week every day at the same hour was -2.8 °C. All temperatures were measured in different days are different. The highest daily maximum temperature was 2.4 °C, the lowest -6 °C. Determine the options that temperature can be measured in the remaining five days.

Correct result:

a =  -3.4 °C
b =  -3.3 °C
c =  -3.2 °C
d =  -3.1 °C
f =  -3 °C

Solution:

$$\begin{gathered} -6<=x<=2.4 \\ (-6+2.4+a+b+c+d+e)\mathrm{/}7=-2.8 \\ s=a+b+c+d+e \\ s=7\cdot (-2.8)+6-2.4=-16 \\ p=s\mathrm{/}5=(-16)\mathrm{/}5=-{\displaystyle \frac{16}{5}}=-3.2 \\ a=p-0.2=(-3.2)-0.2=-{\displaystyle \frac{17}{5}}=-{{\displaystyle \frac{17}{5}}}^{\circ }\text{C}=-3.4^{\circ }\text{C} \end{gathered}$$

$b=a+0.1=(-3.4)+0.1=-{\displaystyle \frac{33}{10}}=-{{\displaystyle \frac{33}{10}}}^{\circ }\text{C}=-3.3^{\circ }\text{C}$

$c=b+0.1=(-3.3)+0.1=-{\displaystyle \frac{16}{5}}=-{{\displaystyle \frac{16}{5}}}^{\circ }\text{C}=-3.2^{\circ }\text{C}$

$d=c+0.1=(-3.2)+0.1=-{\displaystyle \frac{31}{10}}=-{{\displaystyle \frac{31}{10}}}^{\circ }\text{C}=-3.1^{\circ }\text{C}$

$$\begin{gathered} f=d+0.1=(-3.1)+0.1=-3 \\ k=(-6+a+b+c+d+f+2.4)\mathrm{/}7=(-6+(-3.4)+(-3.3)+(-3.2)+(-3.1)+(-3)+2.4)\mathrm{/}7=-{\displaystyle \frac{14}{5}}=-2.8 \\ f=d+0.1=(-3.1)+0.1=-3^{\circ }\text{C} \end{gathered}$$

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