Answer: 0.665.
Step-by-step explanation:
60% of the bags are from supplier A, and 95% of them are undamaged.
40% of the bags are from supplier B, and 75% of them are undamaged.
Then, the number of undamaged bags from supplier A is:
0.6*0.95 = 0.57 or 57%
from B is:
0.4*0.75 = 0.3 or 30%
This means that the percentage of undamaged bags is:
57% + 30% = 87%.
Then, the probability that an undamaged bag is from supplier A is equal to the percentage of undamaged bags that came from supplier A (57%) and the total percentage of undamaged bags (87%) this is:
P = 57%/87% = 0.655 or 65.5%
Then the probability that an undamaged bag comes from supplier A is 0.665.
Answer:
im going so asume you mean a 6 sided dice since thats the most common
(i) 16 2/3%
(ii) 0%
(iii) 16 2/3
(iv) (2,3,5) 50% (1,2,3,5) 66 2/3%
(v) 50%
(vi) 50%
(vii) 33 1/3
Hope This Helps!!!
Answer:
A translation 1 unit left and 2 units up
Step-by-step explanation:
1) look at the point at the very bottom-we can call that point A
2) that point on figure C would be at (3,-5)
3) if you move it to the left one and up two, you get that same point but on figure D
So, the answer is the second option
SORRY IF IT DOESNT MAKE SENCE
Answer:
A) 0 , B) 0.0199 , C) 200
Step-by-step explanation:
Let X denotes no. of computers fail on day
X ~ Bin (n = 4 , p = 0.05)
A] P (X = x) = (n c x) p^x q^n-x
P (X = 4) = (4 c 4) (0.005)^4 (0.995)^0
(0.05)^4 ~ 0
B] P (X <u>></u> 1) = 1 - P (X < 1) = 1 - P (X = 0)
1 - { ( 4 c 0) (0.005)^0 (0.995)^ 4 }
1 - (0.995)^ 4 ~ 0.0199
C] Let Y depict mean number of days until a specific computer fails
Y ~ Geom (p = 0.005)
Mean = E (Y) = 1/p = 1/0.05 = 200