25.8%
First, determine how many standard deviations from the norm that 3 tons are. So:
(3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%
So the probability that more than 3 tons will be dumped in a week is 25.8%
Answer:
I think it is c correct me if I'm wrong
I believe the equation is
![4 \sqrt[4]{2x} + 6 \sqrt[4]{2x}](https://tex.z-dn.net/?f=4%20%5Csqrt%5B4%5D%7B2x%7D%20%2B%206%20%20%5Csqrt%5B4%5D%7B2x%7D%20)
In this case, you would simplify it by adding them together.
=
![10 \sqrt[4]{2x}](https://tex.z-dn.net/?f=10%20%20%5Csqrt%5B4%5D%7B2x%7D%20)
And can even be changed to an exponential equation:
Answer:
5r + 2p + 6
Step-by-step explanation:
= 4r + 9 + r + 2p - 3
= 4r + r + 2p + 9 - 3
= 5r + 2p + 6
