There are infinite equivalent expressions. Here are some:
1/5(m-100)
20(1/100m-1)
1/5m-(4•5)
If you expand any of these or any of the terms, you will get an equivalent expression.
Answer:
c. The sampling distribution of the sample means can be assumed to be approximately normal because the distribution of the sample data is not skewed
Step-by-step explanation:
From the given data, we have;
The category of the sample = Retired individuals
The number of participants in the sample = 20
The duration of program = six-weeks
The improvement seen by most participants = Little to no improvement
The improvement seen by few participants = Drastic improvement
Therefore, given that the participants are randomly selected and the majority of the participants make the same observation of improvement in the time to walk a mile, we have that, the majority of the outcomes show little difference in walk times after the program, therefore, the distribution of the sample data is not skewed and can be assumed to be approximately normal
Answer:
sorry I don't speak that language
Answer:
x=8
Step-by-step explanation:
2(x+4) - 10 =14
distribute the 2 2x +8 -10 +14
simplify 2x-2=14
add 2 2x=16
devide by 2 x=8
Answer:
see below
Step-by-step explanation:
The experimental probability of heads is
P( heads) = 28/50 = 14/25
The theoretical probability is
P (heads) =1/2
Getting a common denominator for the experimental
25/50
comparing the experimental and the theoretical
28/50 > 25/50
experimental > theoretical
