45/43 as a percentage rounded to the nearest tenth?

Given z value let x be a normal random variable with standard deviation (d) = 1.65 (i) find the margin of error for a 90 percent confidence interval for mean corresponding to a sample size of 12 (3/100) (ii) interpret the result of question 3( (4/100) /- [cpt115] -5 (ii) with reference to question 3(d)(i), find an approximate 90 percent confidence interval for mean if the 12 sample values of x are as follows: 95 103 107 98 90 110 92 108 90 94 105 100 (7/100) interpret the result of question 3(d)(ii) (iv)