Your answer is 8 + 5/100 + 9/1000 but in simplest form it would be 22/1100
Answer:
Step-by-step explanation:
Given is a sample of contents of a random sample of 10 containers as follows:
10.1 9.7, 10.5, 10.4, 9.8, 9.7, 9.4, 9.9, 10.7, and 9.7 liters.
To check whether the mean is 10.1 litres.
Set up hypothesis as
![H_0: \bar x = 10.1\\H_a:\bar x \neq 10.1](https://tex.z-dn.net/?f=H_0%3A%20%5Cbar%20x%20%3D%2010.1%5C%5CH_a%3A%5Cbar%20x%20%5Cneq%2010.1)
(two tailed test at 10% level of significance)
Sample mean =9.9
Sample std dev=0.420185
Std error =0.1329
n = 10
df = 9
Mean difference = 9.9-10.1 =-0.2
Test statistic t = -0.2/0.1329 =-1.505
p value= 0.1666
Since p value is >0.10 we accept null hypothesis
Conclusion:
the average content of containers of a particular lubricant is 10.1 liters can be taken as true at 90% level of confidence
0.25% bc he is charged 5% per year so just multiple 5% times 5=0.25%
Answer:
ITS 32 m^3
Step-by-step explanation:
Option C
just trust me homie... :)