Very interesting question!
Let's try to get something that would work for just 5 for now.
Any of these numbers would leave a remainder of 1 when divided by just 5:
6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, ...
Notice that they all look like 5*(something) + 1. Basically, they are all multiples of 5 plus 1.
Let's see what would work for just 8:
9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, ...
Are there any matches... AHA! It's 81. This is the lowest positive integer greater than 1 that follows the rule for both 5 and 8.
Answer: (13.4996, 32.5004)
Step-by-step explanation:
The confidence interval for mean difference is given by :-
,
where 
= sample mean difference .
= critical t value (two-tailed)
SE= Standard error.
Given : 
SE= 4.2
Now, the confidence interval for mean difference will be :-
Hence, the required confidence interval : (13.4996, 32.5004)
Answer:
80 degrees
Step-by-step explanation:
Answer:
the third one
Step-by-step explanation:
y+8= 4(x-3)
y+8= 4x-12
y= 4x-20
