Answer: a) P = 0.5, b) P = 0.07
Step-by-step explanation:
Hi!
Lets call X₁ the time at which you arrive, and X₂ the time at which Bob arrives. Both are random variables with uniform density in the interval [0, 60] (in minutes). Their joint distribuition is uniform over the square in the image, with value P = 1/(60*60) = 1/3600.
a) For you to get more cake than Bob, you should arrive earlier. This event is A = { X₁ < X₂ }, the shaded triangle in the figure.The area of this event (set) is half the total area of the square, so P(A) = 0.5.
It makes sense, beacuse its equally probable for you or Bob to arrive earlier, as both have uniform density over the time interval.
b) In this case you arrive later than Bob, but less than 5 minutes later. So the event is B = { X₂ < X₁ < (X₂ + 5) } . This is the gray shaded area in b) part of the image. Its area is the difference two triangles (half square - blue triangle), then the probability is:

Answer:
true
Step-by-step explanation:
yes the statement is correct
By definition we have that the average rate of change is given by:
AVR = (f (x2) - f (x1)) / (x2 - x1)
Substituting the values we have:
AVR = (204 - (-6)) / (10 - 0)
Rewriting we have:
AVR = (204 + 6) / (10 - 0)
AVR = 210/10
AVR = 21
Answer:
the average rate of change for f (x) from x = 0 to x = 10 is:
AVR = 21
412-217
195 units
Just subtract them and that would be the distance I’m pretty sure
PLS HELP ASAP I DONT HAVE TIME AND IT ALSO DETECT IF ITS RIGHT OR WRONG! PLS HELP ASAP I DONT HAVE TIME AND IT ALSO DETECT IF ITS RIGHT OR WRONG!