According to an article in newsweek, the rate of water pollution in china is more than twice that measured in the us and more than three times the amount measured in japan. the mean emission of organic pollutants is 11.7 million pounds per day in china. assume the water pollution in china is normally distributed throughout the year with a standard deviation of 2.8 million pounds of organic emissions per day. a) what is the probability that on any given day the water pollution in china is at least 15 million pounds per day? b) what is the probability that on any given day the water pollution in china is between 6.2 and 9.3 million pounds per day?

(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: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

to determine the unit cost your would divide the total cost by the number sold.

7 bagels for $4.41       $4.41/7 = $.63 each at fresh side

5 bagels for $3.60         $3.60/5 = $.72 each at value mart