Answer:
a) Figure attached
b) For this case we want this probability:

And we can use the complement rule:

And we can use the cumulative distribution function given by:

And replacing we got:

Step-by-step explanation:
For this case we define the random variable X= depth (in centimeters) of the bioturbation layer in sediment for a certain region, and we know the distribution for X, given by:

Part a
For this case we can see the figure attached.
Part b
For this case we want this probability:

And we can use the complement rule:

And we can use the cumulative distribution function given by:

And replacing we got:

The answer would be z=wx/6
Answer:
2 x 2 x 13
Step-by-step explanation:
2 times 26 = 52
You can break 26 up into 2 times 13.
So, you get 2 times 2 times 13.
25 mph the train is accelerating at a constant rate<span />