Answer:
Step-by-step explanation:
Given that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.
G = card drawn is green
Y = card drawn is yellow
E = card drawn is even-numbered
List:
Sample space = {G1, G2, G3, G4, G5, Y1, Y2, Y3}
2) P(G) = 5/8
3) P(G/E) = P(GE)/P(E)
GE = {G2, G4}
Hence P(G/E) = 2/5
4) GE = {G2, G4}
P(GE) = 2/8 = 1/4
5) P(G or E) = P(G)+P(E)-P(GE)
= 5/8 + 3/8-2/8 = 3/5
6) No there is common element as G2 and G4
Cannot be mutually exclusive
Step-by-step explanation:
2•4 =8
8/3 =2 2/3
answer : 3
This is a simultaneous equation
Rearrange the second to give
x - y = 1
Add the two equations
(x + y) + (x - y) = 5 + 1
2x = 6
x=3
Substitute 3 into either equation
3 - 1 = y
y = 2
Answer:
Area of ellipse is ≈ 77.78 ![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Step-by-step explanation:
Given the dimension of ellipse
a = 4.5 m
b = 5.5 m
Now, Area of ellipse = ![\pi ab](https://tex.z-dn.net/?f=%5Cpi%20ab)
= ![\frac{22}{7}\times 4.5\times 5.5](https://tex.z-dn.net/?f=%5Cfrac%7B22%7D%7B7%7D%5Ctimes%204.5%5Ctimes%205.5)
= ![\frac{5445}{70}](https://tex.z-dn.net/?f=%5Cfrac%7B5445%7D%7B70%7D)
= 77.78 ![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Hence Area of ellipse will be 77.78 ![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)