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.

Result

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

Solution:

−6<=x<=2.4 (−6+2.4+a+b+c+d+e)/7=−2.8 s=a+b+c+d+e s=7⋅ (−2.8)+6−2.4=−16 p=s/5=(−16)/5=−165=−3.2 a=p−0.2=(−3.2)−0.2=−175=−3.4=−3.4∘C-6 <=x <=2.4 \ \\ (-6+2.4+a+b+c+d+e)/7=-2.8 \ \\ s=a+b+c+d+e \ \\ s=7 \cdot \ (-2.8)+6-2.4=-16 \ \\ p=s/5=(-16)/5=- \dfrac{ 16 }{ 5 }=-3.2 \ \\ a=p-0.2=(-3.2)-0.2=- \dfrac{ 17 }{ 5 }=-3.4=-3.4 ^\circ \text{C}
b=a+0.1=(−3.4)+0.1=−3310=−3.3=−3.3∘Cb=a+0.1=(-3.4)+0.1=- \dfrac{ 33 }{ 10 }=-3.3=-3.3 ^\circ \text{C}
c=b+0.1=(−3.3)+0.1=−165=−3.2=−3.2∘Cc=b+0.1=(-3.3)+0.1=- \dfrac{ 16 }{ 5 }=-3.2=-3.2 ^\circ \text{C}
d=c+0.1=(−3.2)+0.1=−3110=−3.1=−3.1∘Cd=c+0.1=(-3.2)+0.1=- \dfrac{ 31 }{ 10 }=-3.1=-3.1 ^\circ \text{C}
f=d+0.1=(−3.1)+0.1=−3 k=(−6+a+b+c+d+f+2.4)/7=(−6+(−3.4)+(−3.3)+(−3.2)+(−3.1)+(−3)+2.4)/7=−145=−2.8 f=d+0.1=(−3.1)+0.1=−3=−3∘Cf=d+0.1=(-3.1)+0.1=-3 \ \\ k=(-6+a+b+c+d+f+2.4)/7=(-6+(-3.4)+(-3.3)+(-3.2)+(-3.1)+(-3)+2.4)/7=- \dfrac{ 14 }{ 5 }=-2.8 \ \\ f=d+0.1=(-3.1)+0.1=-3=-3 ^\circ \text{C}

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