Answer:
The correlation coefficient 0.5 depicts that there is moderate positive correlation between y and x.
Step-by-step explanation:
The linear regression equation is
y= a+bx
The given linear equation is
y= 10x
Here, slope=b=10 and intercept=a=0.
From above equation,
ybar=10xbar
We are given that CVx=0.1 and CVy=0.2
where CV=(standard deviation/mean)*100
We know that
b=r(Sy/Sx)
Multiplying by xbar/ybar on both sides
(xbar/ybar)b=r(Sy/Sx)(xbar/ybar)
(xbar/ybar)b=r[(Sy/ybar)/(Sx/xbar))]
(xbar/ybar)b=r[CVy/CVx]
As CVy/CVx=(Sy/ybar)*100/(Sx/xbar)*100=(Sy/ybar)/(Sx/xbar).
By putting ybar=10xbar, b=10, CVx=0.1 and CVy=0.2.
(xbar/10xbar)10=r(0.2/0.1)
1=r(0.2/0.1)
1=r(2)
r=0.5
There is moderate positive correlation between y and x.
It's 12x. you just combine both numbers
1day=24 hours
10 days=240hours
60mins=1hr
30mins=1/2hr
total is 240hr+21hr+0.5hr=261.5hr
total is 261.5hr
Yes you can and you do that by multiplying $42×1.30=54.60 and then $54.60×1.25=68.25
Answer:
(a) 12.96 ft²
(b) 21.5 in²
Step-by-step explanation:
(a) For the first diagram
Area of the shaded region (A) = Area of Tripezium- area of circle
A = [1/2(a+b)h]-[πr²]............... Equation 1
Where a and b are the parallel side of the tripezium respectively, h = height of the tripezium, r = radius of the circle.
From the diagram,
Given: a = 15 ft, b = 6 ft, h = 12 ft, r = h/2 = 12/2 = 6 ft.
Constant: π = 3.14
Substitute these values into equation 1
A = [12(15+6)/2]-(3.14×6²)
A = 126-113.04
A = 12.96 ft²
(b) For the second diagram,
Area of the shaded region (A') = Area of square- area of circle
A' = (L²)-(πr²)............. Equation 2
Where L = lenght of one side of the square, r = radius of the circle
From the diagram,
Given: L = 2r = (2×5) = 10 in, r = 5 in
Substitute these values into equation 2
A' = (10²)-(3.14×5²)
A' = 100-78.5
A = 21.5 in²
