10 = (1/400) x^2 - 20 x ??? you did not use parenthesis so I assume this
4000 = x^2 - 8000 x
x^2 - 8000 x - 4000 = 0
x = [ 8000 +/- sqrt(64^10^6+16000) ]/2
x =
nah, impossible you must mean
f(x) = x^2 /(400 - 20 x)
10 = x^2 / (400 - 20 x)
4000 -200 x = x^2
x^2 + 200 x - 4000 = 0
x = [ -200 +/- sqrt(40000+16000) ]/2
= [ -200 +/- 236 ]/2
= 18, -218 ignore -218
0 </= x </= 18
at a single ticket booth, customers arrive randomly at a rate of x per hour.
the average line length is given by
F(x)= x^2/400-20x
Where 0<= x <20. To keep the wait in line reasonable, it is required that the average line length should not exceed 10 customers.
Determine the range of rates at which customers can arrive before a second attendant is needed. Express your answer in interval form.
2 answers
thank you bottom is the correct way.