Answer: y=+2
Step-by-step explanation:
<em>This is y intercept form</em>
First you find the slope
Then you find the y intercept and plug it in t the equation y=mx+b
How do you solve for tan-1 or tan theta without using a calculator? I don't want to rely on the calculator to solve problems tha
1 answer:
You might be interested in
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.
Answer:
Step-by-step explanation:
Given A = 5i + 11j – 2k and B = 4i + 7k, the vector projection of B unto a is expressed as
b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)
note that i.i = j.j = k.k =1
b.a = 5(4)+11(0)-2(7)
b.a = 20-14
b.a = 6
||a|| = √5²+11²+(-2)²
||a|| = √25+121+4
||a|| = √130
square both sides
||a||² = (√130)
||a||² = 130
<em>Hence the projection of b unto a is expressed as </em><em></em>