Answer:
AE = 15 cm; ED = 18 cm; AD = 15 cm (given)
Step-by-step explanation:
ΔBEC ~ ΔAED so ...
AD/BC = AE/BE = (BE+AB)/BE = 1 + AB/BE
Substituting given numbers (lengths in centimeters), we have ...
15/10 = 1 + 5/BE
1/2 = 5/BE
BE = 10
Similarly, ...
1/2 = 6/CE
CE = 12
Then the unknown sides are ...
AE = AB + BE = 5 + 10 = 15 . . . cm
ED = CE + CD = 12 + 6 = 18 . . . cm
I can use the angle and the length of JH to find the length of IJ.
To do this, I look at the relationship IJ and JH have to the 52 degree angle. JH is opposite to angle I, and IJ is adjacent to angle I. Because the two side lengths are opposite and adjacent, I use the tangent function to solve this.
Tangent of an angle = the length of the opposite side / the length of the adjacent side. This is just another way to say tan(x)=opposite/adjacent
Now I can fill in what I know...
tan(52)=4.2/x
Now, I want to isolate x.
tan(52) = 4.2/x
x(tan(52))=4.2
x=4.2/tan(52)
Now I put 4.2/tan(52) into a calculator and get x = 3.3 ft
Hope this helps!
Answer:
a. 0.689
b. 0.8
c. 0.427
Step-by-step explanation:
The given scenario indicates hyper-geometric experiment because because successive trials are dependent and probability of success changes on each trial.
The probability mass function for hyper-geometric distribution is
P(X=x)=kCx(N-k)C(n-x)/NCn
where N=4+3+3=10
n=2
k=4
a.
P(X>0)=1-P(X=0)
The probability mass function for hyper-geometric distribution is
P(X=x)=kCx(N-k)C(n-x)/NCn
P(X=0)=4C0(6C2)/10C2=15/45=0.311
P(X>0)=1-P(X=0)=1-0.311=0.689
P(X>0)=0.689
b.
The mean of hyper-geometric distribution is
μx=nk/N
μx=2*4/10=8/10=0.8
c.
The variance of hyper-geometric distribution is
σx²=nk(N-k).(N-n)/N²(N-1)
σx²=2*4(10-4).(10-2)/10²*9
σx²=8*6*8/900=384/900=0.427
M = (y-y)/(x-x)
= 9-1/-7-(-7)
= 8/0
It is a vertical line
He rounded it to the nearest hundred, not ten. Ten would be 560.
Hope this helps!
