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.


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


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)+62.4=16 p=s/5=(16)/5=165=3.2 a=p0.2=(3.2)0.2=175=3.4=3.4C-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=3.3Cb=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=3.2Cc=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=3.1Cd=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=3Cf=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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