Z-test should be used when the parent standard deviation is known and the sample size is greater than or equal to 30 (to satisfy the Central Limit Theorem). A T-test should be used when these two rules do not apply.
Answer:
1. 24
2. 24
3. -24
4. -24
I don`t know why some of the answers are 10, I think those are just trick ones. The answers are only in 24s
100%=all
all-change of wining=change of not winning
100%-11%=89% of not winning
probabilit is 89/100 that they dont win
$1.08
you would multiply your cost (1.00) by 8% which is equivalent to .08 and then add to to the original cost. or to skip the middle step just multiply it by 1.08 since you are adding the 8% to 100% of the original cost
No i believe but I’m unsure
