There should be 6 per number
ex.
1234
1243
1342
1324
1432
1423
14/23. You can’t simplify it anymore. It is also equivalent to about 61% (60.86% to be exact)
Answer:
Conversion and substitution
Step-by-step explanation:
One way to find out if the expression is true, we can convert all the percentages into their decimal equivalence.
60% = 0.6
10% = 0.1
6% = 0.06
Now let's substitute the values.
0.06 = 0.6 x 0.1
0.06 = 0.06
Since 0.06 is equal 0.06, this then proves that it is true.
Answer:
=21+11x
Step-by-step explanation:
Answer:

If we divide both sides by
we got:

And we can use the normal distribution table or excel to find the probabilites and we got:
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".
Solution to the problem
Let X the random variable that represent the area of a population, and for this case we know the distribution for X is given by:
Where
and 
We select a a sample of n =4 and since the distribution for X is normal then we know that the distribution for the sample mean
is given by:

And we want to find this probability:

If we divide both sides by
we got:

And we can use the normal distribution table or excel to find the probabilites and we got:
