H = 60M
I think this is what u need
We are given with
random samples = 300 batteries
defective samples = 9 batteries
level of significance = 0.1
Using the P-table values and using the given values, we see that the number of defective batteries are really less than 5%
Answer:
15
Step-by-step explanation:
Answer: The answer here is x=12.49
Step-by-step explanation:
The idea is tho remove the brackets on the right hand side by multiplying by 4.9
4.90= 10(x-12)
4.90=10x-120
x=12.49
