The times for completing one circuit of a bicycle course are normally distributed with a mean of 72.5 minutes and a standard deviation of 6.5 minutes. An association wants to sponsor a race but only wants the top 25% of riders included. It a trial run, what should be the cutoff time?

76.881 minutes

Step-by-step explanation:

If we look for the value of 25% in the normal distribution table, it corresponds to z = - 0.674

I mean:

P( Z < - 0.674) = 0.25

Now, we know that the formula in this case would be

cut off time = m - z*sd

where m is the mean and sd is the standard deviation.

replacing these values we have left:

cut off time = 72.5 - (-0.674)*6.5

cut off time = 76.881

Therefore the time limit would be 76.881 minutes

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b