The probability that a high school student in in the marching band is 0.47
so, 
The probability that a student plays a varsity sport is 0.32
so, 
the probability that a student is in the marching band and plays a varsity sport is 0.24
so, we get
P(M∩V)=0.24
a student plays a varsity sport if we know she is in the marching band
so,
P(V|M)=(P(V∩M))/(P(M))
now, we can plug values
and we get


So,
the probability that a student plays a varsity sport if we know she is in the marching band is 0.51063......Answer
By first principles, the derivative is

Use the binomial theorem to expand the numerator:


where

The first term is eliminated, and the limit is

A power of
in every term of the numerator cancels with
in the denominator:

Finally, each term containing
approaches 0 as
, and the derivative is

Answer: 105 miles
Step-by-step explanation:
You can write an equation for this problem and solve it.
32 miles=1/3x-3, where x is equal to the total length of the highway.
35=1/3x, and then you multiply by 3, to get x=105
Equations of straight lines are in the form y = mx + c (m and c are numbers). m is the gradient of the line and c is the y-intercept (where the graph crosses the y-axis).