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:
1. This graph is not a function, because for every y there are more than one x.
2. This graph is not linear because it is not a straight line.
3. It took them 6 hours to get to their destination.
4. The destination was 325 miles away.
5. They might have stopped at a restaurant to eat.
Answer:

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

And replacing we got:


We can find the probability of interest using the normal standard table and with the following difference:

Step-by-step explanation:
Let X the random variable who represent the expense and we assume the following parameters:

And for this case we want to find the percent of his expense between 38.6 and 57.8 so we want this probability:

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

And replacing we got:


We can find the probability of interest using the normal standard table and with the following difference:

Domain x < or equal to 5
Range y < or equal to -1
Brainliest?