Answer:
x = 1 ; x = 8
Step-by-step explanation:
6x² - 5x + 10 = 5x² + 4x + 2
x² - 9x + 8 = 0
x² - 8x - x + 8 = 0
x(x - 8) - (x -8) = 0
(x - 8)(x - 1) = 0
x = 1, x = 8
Answer:
both the groups have the same value for the first quartile
Answer:

And we can solve this using the following z score formula:

And if we use this formula we got:

So we can find this probability equivalently like this:

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 heights of a population, and for this case we know the distribution for X is given by:
Where
and
We select n =100. Since the distribution for X is normal then we know that the distribution for the sample mean
is given by:
We want this probability:

And we can solve this using the following z score formula:

And if we use this formula we got:

So we can find this probability equivalently like this:

Answer:
just one solution
Step-by-step explanation:
4x + 8 = −2 − 2 + 5x
4x - 5x = -2 - 2 - 8
x = 12
Answer:
PPF, PFF
Step-by-step explanation:
There are several ways you can list all the possible combinations. A couple of my favorite are a) use a binary counting sequence; b) use a gray code counting sequence.
Using the first method, the binary numbers 000 to 111 can be listed in numerical order as 000, <em>001</em>, 010, <em>011</em>, 100, 101, 110, 111. Letting 0=P and 1=F, the ones missing from your list are the ones in italics in my list.
Using the second method, we change the right-most character, then the middle one, and finally the left-most character so there is one change at a time: 000, <em>001</em>, <em>011</em>, 010, 110, 111, 101, 100.
After you have a list of all possible combinations, it is a simple matter to compare the given list to the list of possibilities to see which are missing.