FVAD=8000[((1+0.10/4)^(4*4)-1)/(0.10/4)]*(1+0.10/4)
FVAD=158917.84
Answer:
15 percent
Step-by-step explanation:
- 30/200=15/100
- 15/100=0.15
- 0.15=15 percent
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
1) around 1300
2) around 30
3) a. about 500
b. about 2600
Hope this helps you!
Answer:
The answer is option A) <em>r</em><em> </em>and <em>s</em>