The principal randomly selected six students to take an aptitude test. their scores were: 81.6 72.0 81.1 86.4 70.2 83.1 determine a 90% confidence interval for the mean score for all students

84.38, 73.74

Step-by-step explanation:

score given 81.6, 72.0, 81.1, 86.4, 70.2, 83.1

sample size (n) = 6

mean = 79.06

standard deviation

σ = 6.47

level of significance (α) = 1 - 90% = 10%

confidence interval

=79.06 ± 5.32

= 84.38, 73.74

no this is a nonlinear negative

76%

step-by-step explanation:

24% are dogs and the remaining is 76% because 100-24=76