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
Answer:
1. A(1;5), B(10;23), slope of 2
2. A(10;9), B(4;0), slope of 1.5
3. A(-35/4;6), B(9;-35/6), slope of -2/3
4. A(5;18), B(25,22), slope of 0.2
5. A(2/9;1/3), B(2/3,-1/3), slope of -1.5
Step-by-step explanation:
Substitute values with correct variable and slopemis coefficient of x when x and y-int. are on 1 side, y is on other side and has coefficient of 1. Hope it works!
Answer:
244
Step-by-step explanation:
74+42=116
360-116=244.
<u>Explanation:</u>
a) First, note that the Type I error refers to a situation where the null hypothesis is rejected when it is actually true. Hence, her null hypothesis would be H0: mean daily demand of her clothes in this region should be greater than or equal to 100.
The implication of Type I error in this case is that Mary <u>rejects</u> that the mean daily demand of her clothes in this region is greater than or equal to 100 when it is actually true.
b) While, the Type II error, in this case, is a situation where Mary accepts the null hypothesis when it is actually false. That is, Mary <u>accepts</u> that the mean daily demand of her clothes in this region is greater than or equal to 100 when it is actually false.
c) The Type I error would be important to Mary because it shows that she'll be having a greater demand (which = more sales) for her products despite erroneously thinking otherwise.
