Answer:
x < 9
Step-by-step explanation:
Move the -4 over as if there was an equal sign so that x < 9
Take the arrow that looks like this "<---O" that is not filled in and put it on the number 9.
Let me know if this helps <3
D and H are corresponding
G and C are corresponding
E and A are corresponding
B and F are corresponding
Answer:
-3
Step-by-step explanation:
5•-3=-15
4•-3=-12
-15+3=-12
8 = 8v - 4 (v + 8)
Distribute the 4 through the parentheses
8 = 8v - 4v - 32
combine like terms
8 = 4v - 32
add 32 to both sides
8 + 32 = 4v -32 + 32
combine like terms
40 = 4v
divide each side by 4
40/10 = v
4 = v
v= 4
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".
Solution to the problem
Let X the random variable that represent the grade points avergae of a population, and for this case we know the following properties
Where
and
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
So we can find the z score for the value of X=3.44 in order to see how many deviations above or belowe we are from the mean like this:

So the value of 3.44 is 2 deviations above from the mean, so then we know that the percentage between two deviations from the mean is 95% and on each tail we need to have (100-95)/2 = 2.5% , because the distribution is symmetrical, so based on this we can conclude that:
