Astudent measured the lead content of a paint sample 4 times. the standard deviation of the measurements was found to be 1.5% of the average. calculate the confidence interval at the 90% confidence level. confidence interval: can this student be 90% confident that the true value is within 1.8% of the measured average? o yes o no

Yes. There is 90% confidence the mean will be between ± 1.76% of the sample average.

Step-by-step explanation:

In this case, where the sample is little and the standard deviation of the population is unknown, the confidence interval (CI) can be expressed as

"s" is the standard deviation of the sample and t(n-1) is the critical t-value with (n-1) degrees of freedom. In this case, the degrees of freedom ar (4-1)=3.

The standard deviation is expressed as 1.5% or the mean, that is

Since we have two limits for the mean and a 90% CI, we have 5% of being outside the confidence. Looking up in the table for P(x>t)=0.05 and DF=3, we have

Then the CI is

There is 90% confidence the mean will be between ± 1.76% of the sample average.

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

you have to find the same variables -3n - 5n so this would be -3n +(-5n) which equals -9n