PLEASE HELP QUICKLY A scientist is studying the growth of a particular species of plant. He writes the following equation to sho
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The slope is 7/5 and the equation of the line is 5y - 7x = 0 if the points (10,14) and (35,49) form a proportional relationship.
<h3>What is the slope?</h3>The ratio that y increase as x increases is the slope of a line. The slope of a line reflects how steep it is, but how much y increases as x increases. Anywhere on the line, the slope stays unchanged (the same).
From the above formula, we can find the slope of the line:
m = (49-14)/(35-10)
m = 35/25
m = 7/5
The equation will be:
y = 7x/5 + c
here c is the y-intercept
Plug (10, 14) in the equation to find the value of c
14 = 7(10)/5 + c
c = 0
y = 7x/5 or
5y - 7x = 0
Thus, the slope is 7/5 and the equation of the line is 5y - 7x = 0 if the points (10,14) and (35,49) form a proportional relationship.
Learn more about the slope here:
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Answer:
a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.
c) This last case the histogram is not symmetrical and the data seems to be skewed.
Step-by-step explanation:
For this case we have the following data:
(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
We can use the following R code to get the histogram
> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
> hist(x1,main="Histogram a)")
The result is on the first figure attached.
For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> hist(x2,main="Histogram b)")
The result is on the first figure attached.
For this case the data is skewed to the left and we can't assume that we have the normality assumption.
(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> hist(x3,main="Histogram c)")
The result is on the first figure attached.
This last case the histogram is not symmetrical and the data seems to be skewed.
