11t=120(12)
11t=1440
t=1440/11
However t is in hours and we want to know days so:
d=(1440/11)/24
d=1440/264
d=60/11
d=5 5/11 days
Answer: D) The linear model shows a strong fit to the data
The actual strength of the relationship is unknown unless we have the actual values of each data point (so we can compute the correlation coefficient r), but the residuals are randomly scattered about both above and below the horizontal axis. This means we have a fairly good linear fit. If all of the points were above the line, or all below the line, or all residuals fit a certain pattern (eg: parabola), then it wouldn't be a good linear fit.
Answer:
I’m sorry but I’m not much help. I usually don’t answer questions I don’t know about, but I can do one thing.
The rate of change of a line is regarded as the slope/ gradient (m) of the line
The slope/ gradient (m) of the line moving through points (x₁,y₁) and (x₂,y₂) is given as
On substituting the above values in the formula above we will have the slope to be
Therefore,
The rate of change of the line through (2,3) and (3,8) is 5
Isolate x on one side of the equation. divide by k.
2p over k =x-t
add t
(2p/k)+t = x.
hope this helps and have a wonderful day!
