Answer:
98 would be written as Nighty Eight in words!
Step-by-step explanation:
The data below is what was provided in the question and it is what I solved the question with
P(A1) = 0.23
P(A2) = 0.25
P(A3) = 0.29
P(A1 n A2 ) = 0.09
P(A1 n A3) = 0.11
P(A2 n A3) = 0.07
P(A1 n A2 n A3) = 0.02
a
P(A2|A1) = P(A1 n A2)/P(A1)
= 0.09/0.23
= 0.3913
We have 39.13% confidence that event A2 will occur given that event A1 already occured
b.)
P(A3 n A3|A1) = P(A2 n A3)n A1)/P(A1)
= 0.02/0.23
= 0.08695
We have about 8.7% chance of events A2 and A3 occuring given that A1 already occured.
C.
P(A2 u A3|A1)
= P(A1 n A2)u(A1 n A3)/P(A1)
= P( A1 n A2) + P(A1 n A3) - P(A1 n A2 n A3) / P(A1)
= (0.09+0.11-0.02)/0.23
= 0.18/0.23
= 0.7826
We have 78.26% chance of A2 or A3 happening given that A1 has already occured.
M(slope) = Δy/Δx = 8-(-6)/-4-(-3) = 14/-1 = -14
Answer: y= 1/2x - 2
Step-by-step explanation:
y= mx+b
m= 1/2 (plug it in)
b= -2 (plug it in)
x= ?
The minimum cost option can be obtained simply by multiplying the number of ordered printers by the cost of one printer and adding the costs of both types of printers. Considering the options:
69 x 237 + 51 x 122 = 22,575
40 x 237 + 80 x 122 = 19,240
51 x 237 + 69 x 122 = 20,505
80 x 237 + 40 x 122 = 23,840
Therefore, the lowest cost option is to buy 40 of printer A and 80 of printer B
The equation, x + 2y ≤ 1600 is satisfied only by options:
x = 400; y = 600
x = 1600
Substituting these into the profit equation:
14(400) + 22(600) - 900 = 17,900
14(1600) + 22(0) - 900 = 21,500
Therefore, the option (1,600 , 0) will produce greatest profit.
