The two-way table is attached.
There are 58 people. 31 do not play baseball; this means that 58-31=27 do play baseball.
16 people play football; this means 58-16=42 people do not play football.
20 people do not play football or baseball; 42-20 = 22 people play baseball but do not play football.
27-22=5 people play baseball and football.
16-5=11 people play football but do not play baseball.
Answer:
I want to answer 15%
Step-by-step explanation:
As for how to solve it I don't really know I would use process of elimination because the answer can't be so high that it's 45% so that's out. Personally I think that 35% is pretty high to so my best educated guess would be between 25% and 15% . Sorry this isn't more helpful.
here's a worksheet that might match what you're looking for
Answer
Multiply the divisor and dividend by
✔ 100
The new division problem becomes
✔ 4,860 ÷ 15
The quotient of 48.6 ÷ 0.15 is
✔ 324
Estimating to check, you get
✔ 5,000 ÷ 15 = 333.33
<u>Divide</u>
Answer:
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solutio to the problem
Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:
Where 
and 
For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.
The definition of sample mean is given by:
If we solve for the total T we got:
For this case then the expected value and variance are given by:
And the deviation is just:
So then the distribution for the total would be also normal and given by:
And we want this probability:
And we can use the z score formula given by:
