The answer is d. Thank you
Answer:
(a) 0.119
(b) 0.1699
Solution:
As per the question:
Mean of the emission,
million ponds/day
Standard deviation,
million ponds/day
Now,
(a) The probability for the water pollution to be at least 15 million pounds/day:


= 1 - P(Z < 1.178)
Using the Z score table:
= 1 - 0.881 = 0.119
The required probability is 0.119
(b) The probability when the water pollution is in between 6.2 and 9.3 million pounds/day:



P(Z < - 0.86) - P(Z < - 1.96)
Now, using teh Z score table:
0.1949 - 0.025 = 0.1699
Answer:
(4x-3)(x+2)
Step-by-step explanation:
i hope it will help you
The answer is b I believe
Answer:
8/20, 5/20
Step-by-step explanation:
2/5 - 4/10 6/15 8/20
1/4 - 2/8 3/12 4/16 5/20