A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compression strength of more than 1300 KN/m^2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with \sigma =60. Let \mu denote the true average compressive strength. What are the appropriate null and alternative hypothesis?

The null hypothesis is: u >= 1300 KN/m^2

And the alternative hypothesis is:

u < 1300 KN/m^2.

Step-by-step explanation:

The null hypothesis describes hat there is no difference or assiciation between variables of a specific population or that it is a particular average value while the alternative hypothesis states otherwise.

In this case study, the null hypothesis is: u >= 1300 KN/m^2

And the alternative hypothesis is:

u < 1300 KN/m^2.

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

their is no graph showing

step-by-step explanation: