it's plural so axes*
and in pilar coordinates, the axes are circular so they never intersect but the centre point is still called origin.
Answer:
D.3.5
Step-by-step explanation:
4x+2y=10(given)
2x+y=5
y=5-2x-----(1)
y=2x-1------(2)(given)
Since the left side of the equations are the the same,
5-2x=2x-1
4x=6
x=1.5
Sub x=1.5 into eq(1),
y = 5-2(1.5) = 5-3 = 2
so, x+y = 1.5+2 = 3.5
first is 86
we know because of the sum of the interior angles of a triangle being 180, and that 5 + 6 =180
Remark
I think you want us to do both 3 and 4
Three
QR is 4 points going left from the y axis + 2 points going right from the y axis.
QR = 6 units long.
RT is 4 units above the x axis and 3 units below the x axis
RT = 7 units long
TS = 3 units. For this one you just go from T to S. The graph really helps you. You just need to count.
Problem 4
You need to find the area of the combined figure. You could do it as a trapezoid, which might be the easiest way. I'll do that first.
Area = (b1 + b2)*h/2
Givens
B1 = QR = 6
B2 = UT + TS = 6 + 3 = 9
h = RT = 7
Area
Area = (6 + 9)*7/2
Area = 15 * 7 / 2
Area = 52.5
Comment
You could break this up into a rectangle + a triangle
Find the area of QRTU and add the Area of triangle RTS
<em>Area of the rectangle </em>= L * W
L = RT = 7
W = QR = 6
Area = 7 *6 = 42
<em>Area of the triangle</em> = 1/2 * B * H
B = TS = 3
H = RT =7
Area = 1/2 * 3 * 7
Area = 1/2 * 21
Area = 10.5
Total Area = 42 + 10.5 = 52.5 Both answers agree.
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
