Answer:
$13.67
Step-by-step explanation:
Let's say that CD= c and DVDs= d. Four CDs and 4 DVDs cost $164, so the equation will be:
i. 4c + 4d= 164
The cost of the 4 CDs is half the cost of the 4 DVDs, to put into an equation it will be:
ii. 4c= 0.5(4d)
4c= 2d
d= 2c
Then we can substitute the second equation (ii) into the first equation (i). The calculation will be:
4c + 4d= 164
4c + 4(2c)= 164
4c+ 8c= 164
12c= 164
c=13.67
The cost for each CD is $13.67
Answer:
(-4, 2)
Step-by-step explanation:
4x+2y=-12
3x+y=-10
Start by dividing the first equation by 2 to simplify it...
2x+y=-6
Then, subtract -2x from both sides to isolate y...
y=-2x-6
Substitute -2x-6 for y in the second equation...
3x-2x-6=-10
Combine like terms...
x-6=-10
Add 6 to both sides
x-6+6=-10+6
x=-4
Plug -4 in for x to solve for y:
3(-4)+y=-10
-12+y=-10
Add 12 to both sides
-12+12+y=-10+12
y=2
(x,y)=(-4,2)
Answer:


As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.
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
Emilia case
Let X the random variable that represent the scores of a test, and we know that
Where
and
The z score formula is given by:
Since Emilia made 83 points we can find the z score like this:

Ferdinand case
Let X the random variable that represent the scores of a test, and we know that
Where
and
The z score formula is given by:
Since Ferdinand made 79 points we can find the z score like this:

As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.