Q1=3 Q2[median]=12 Q3=14 and the IQR is 11
Y=Acos(p)+m, A=amplitude, p=period, m=midline, in this case:
A=1/2, p=360(t/12)=30t, m=(10-9)/2+9=9.5 so
h(t)=(1/2)cos(30t)+9.5
For finding the value of b, we must consider that Line MN passes through points M(4, 3) and N(7, 12). With this condition y = mx + b, can be written 3=4m+ b (because line passes through M(4,3) ) and 12=7m+b, b ( because line passes through M(7,12)).
We have a system of equation
4m+ b=3
7m+b=12
For solving this, 4m+b- (7m+b)= 3-12, it is equivalent to -3m= -9 and then m=3, if m=3 so
4x3 +b =3 implies b= 3 -12= -9, so the value of b= -9
<span>
</span>
