Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. from past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known. the data obtained are as follows:

n_1 = 10 and x_1 = 87.6 sigma_1 = 1 and n_2 = 12 x^2 = 74.5 sigma_2 = 1.5. if mu_1 and mu_2 denote the true mean tensile strengths for the two grades of spars. construct a 90 percentage confidence interval on the .difference in mean strength.

(12.1409, 14.0591

Step-by-step explanation:

Given that Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known.

Group   Group One     Group Two  

Mean87.60074.500

SD1.0001.500

SEM0.3160.433

N10      12    

The mean of Group One minus Group Two equals 13.100

standard error of difference = 0.556

 90% confidence interval of this difference:  

  t = 23.5520

 df = 20

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

$649.60

step-by-step explanation: