Answer: (751.05, 766.95)
Step-by-step explanation:
We know that the confidence interval for population mean is given by :-
,
where 
=population standard deviation.
= sample mean
n= sample size
z* = Two-tailed critical z-value.
Given : 
n= 42
We know that from z-table , the two-tailed critical value for 99% confidence interval : z* =2.576
Now, the 99% confidence interval around the true population mean viscosity :-
![759\pm (2.5760)\dfrac{20}{\sqrt{42}}\\\\=759\pm (2.5760)(3.086067)\\\\=759\pm7.9497=(759-7.9497,\ 759+7.9497)\]\\=(751.0503,\ 766.9497)\approx(751.05,\ 766.95)](https://tex.z-dn.net/?f=759%5Cpm%20%282.5760%29%5Cdfrac%7B20%7D%7B%5Csqrt%7B42%7D%7D%5C%5C%5C%5C%3D759%5Cpm%20%282.5760%29%283.086067%29%5C%5C%5C%5C%3D759%5Cpm7.9497%3D%28759-7.9497%2C%5C%20759%2B7.9497%29%5C%5D%5C%5C%3D%28751.0503%2C%5C%20766.9497%29%5Capprox%28751.05%2C%5C%20766.95%29)
∴ A 99% confidence interval around the true population mean viscosity : (751.05, 766.95)
Answer:
15
Step-by-step explanation:
Solve using the left-> right rule. Note that they all have addition signs and that they are all positive numbers:
3 + (8) + 4 = 3 + 8 + 4 = (3 + 8) + 4 = (11) + 4 = 15
15 is your answer.
~
Acceleration is the change of velocity per unit change of time. In this case, given the velocity equation, we can get the acceleration of the particle by taking the first derivative of the equation. v'(t)= t^2 (1/t+2) + 2t In(t+2). substituting t= 6, acceleration is equal to 29.45 m/s2.
