A tennis shop uses a machine to string tennis racquets to specific tensions requested by their customers. There's variability in the results, so when the machine is set to string at a tension of 60 pounds, the resulting racquets have a mean tension of 60 pounds with a standard deviation of 0.5 pounds. Suppose that we took random samples of 36 racquets and calculated the sample mean tension from each sample. We can assume that the racquets in each sample are independent.

What would be the shape of the sampling distribution of the sample mean tension?

Approximately Normal

Step-by-step explanation:

Even though we don't know the shape of the population distribution of tensions, the sampling distribution of the sample mean will be approximately normal since the sample size is large (n=36≥30).

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

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !