Need help ASAP!! Will help you back.
1 answer:
1) y=0.94x+12.5
for x=40 months
y=0.94*40+12.5=50.1------------------ > *1000=$50100
the answer is $50100
2) the companys sales increase per months is equal to the slope of the equation of the line
y=0.94x+12.5--------------------- > y=mx+b
the slope m=0.94
094*1000=940 $/month
the answer is 940 $/month
3) the equation of the trend line is
k=mj+b
<span>I take two points of the graph
</span>point A(2,24) point B(10,8)
m=(k2-k1)/(j2-j1)------------- >(8-24)/(10-2)=-2
m=-2
then
k=mj+b--- >for point B(10,8)
8=-2*(10)+b------ > b=28
k=-2j+28
the answer is k=-2j+28
4) the line which is a model lineal for the data is the one shown in the attached figure (because <span>that line shows the estimated values closest to the real values)</span>
for x=40 months
y=0.94*40+12.5=50.1------------------ > *1000=$50100
the answer is $50100
2) the companys sales increase per months is equal to the slope of the equation of the line
y=0.94x+12.5--------------------- > y=mx+b
the slope m=0.94
094*1000=940 $/month
the answer is 940 $/month
3) the equation of the trend line is
k=mj+b
<span>I take two points of the graph
</span>point A(2,24) point B(10,8)
m=(k2-k1)/(j2-j1)------------- >(8-24)/(10-2)=-2
m=-2
then
k=mj+b--- >for point B(10,8)
8=-2*(10)+b------ > b=28
k=-2j+28
the answer is k=-2j+28
4) the line which is a model lineal for the data is the one shown in the attached figure (because <span>that line shows the estimated values closest to the real values)</span>
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