alanforde800Maximus wrote:If a racehorse runs an average(arithmetic mean) of m miles per race for r races and then runs n miles in its next race, what is the average number of miles the horse has run for the r + 1 races?

a) (rm + n)/r+1
b) (m + n)/r+1
c) (m + n)/r
d) r(m + n)/r+1
e) (m + rn)/r+1

Please assist with above problem.

And here's how you solve it algebraically:

The average is defined as the total miles run divided by the number of races, so the total miles run is therefore the number of races multiplied by the average, in this case:

m*r

The one new race adds 1 to r and n to m*r, so the new average is:

(m*r+n)/(r+1)

alanforde800Maximus wrote:If a racehorse runs an average(arithmetic mean) of m miles per race for r races and then runs n miles in its next race, what is the average number of miles the horse has run for the r + 1 races?

a) (rm + n)/r+1
b) (m + n)/r+1
c) (m + n)/r
d) r(m + n)/r+1
e) (m + rn)/r+1

Please assist with above problem.

We are given that a racehorse runs an average (arithmetic mean) of m miles per race for r races and then runs n miles in its next race. We need to determine the average number of miles the horse has run for r + 1 races. Let's plug the given information into the average formula.

average = sum/quantity

Since the racehorse runs an average (arithmetic mean) of m miles per race for r races, the initial sum = mr and the initial quantity is r.

Since the racehorse then runs n miles in its next race, the new sum is mr + n and the new quantity is r + 1.

Thus, the average number of miles that the horse runs for r + 1 races is:

(mr + n)/(r + 1)

Answer: A

total miles covered in r number of races: m*r
mile covered in remaining 1 race = n

average = total miles/races
= (mr+n)(r+1)
SO A

If I run m miles per race for r races, I've run r*m miles.

If I run n miles in my next race, I've run another n miles, for a total of r*m + n.

To find the average, divide that mileage by the number of races: (r*m + n) / (r + 1).