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:
Step-by-step explanation:
Given that acceleration of an object is

is the solution to the differential equation
Since v(0) =7
we get ln 7 = C
Hence 
since velocity is rate of change of distance s we have
![v=\frac{ds}{dt} =7e^{-2t}\\s= [tex]s(t) =\frac{-7}{2} (e^{-2t})+C)[](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bds%7D%7Bdt%7D%20%3D7e%5E%7B-2t%7D%5C%5Cs%3D%20%5Btex%5Ds%28t%29%20%3D%5Cfrac%7B-7%7D%7B2%7D%20%28e%5E%7B-2t%7D%29%2BC%29%5B)
substitute t=0 and s=0

So solution for distance is

Answer: HAPPY BIRTHDAY :)))
the answer is C
Step-by-step explanation:
Recall d = rt, distance = rate * time.
now, he has a speed rate of 56 mph for a distance of 504

so, at that speed the whole driving time is 9 hours then. Ok, he drove 2 hours today, that means he drove 7 yesterday, 9 - 2.
how many miles is it for 7 hours at 56mph? d = rt ---> d = 56 * 7
that many miles he drove yesterday.
Answer:
x = 15.3°
Step-by-step explanation:
Trig identities:

where
- x = angle
- O = side opposite the angle
- A = side adjacent the angle
- H = hypotenuse
