A possible solution of the system of inequalities is 5 dimes and, 8 nickels
<h3>How to determine a possible solution?</h3>
From the question, the given parameters are:
Number of coins = Not less than 12
Worth = At most $0.95
Represent dime with x and nickels with y
So, we have the following inequalities
x + y >= 12
0.1x +0.05y <= 0.95
Next, we plot the graph of the inequalities (see attachment)
From the graph, we have a possible solution to be (5, 8)
This is so because this point is in the shaded region
Read more about inequalities at
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Pretty sure the mean is 8
Answer:
(5,-6)
Step-by-step explanation:
ONE WAY:
If , then .
Let's simplify that.
Distribute with :
Combine the end like terms :
Use identity for :
Combine like terms and :
We are given .
So we have that .
The vertex happens at .
Compare to to determine .
Let's plug it in.
So the coordinate is 5.
Let's find the corresponding coordinate by evaluating our expression named at :
So the ordered pair of the vertex is (5,-6).
ANOTHER WAY:
The vertex form of a quadratic is where the vertex is .
Let's put into this form.
We are given .
We will need to complete the square.
I like to use the identity .
So If you add something in, you will have to take it out (and vice versa).
So we have in vertex form is:
.
The vertex is (3,-6).
So if we are dealing with the function .
This means we are going to move the vertex of right 2 units to figure out the vertex of which puts us at (3+2,-6)=(5,-6).
The coordinate was not effected here because we were only moving horizontally not up/down.
Answer:
20
Step-by-step explanation:
The lower quartile Q₁ is the value at the left side of the box, that is
Q₁ = 20
Answer: The most appropriate measure of center is the median.
Step-by-step explanation:
- The measure of center generally represented by 1) mean 2) median 3) mode.
- Mean is the best measure to represent the center of the data but when data have outliers it gets affected badly.
- In that case, we use the Median as the best measure of center.
The given data is 12 10,9,68,12 .
As it can be seen that, the data have extreme value 68 ( called outliers).
In this case, we will use the Median as the best measure of center.
So, the most appropriate measure of center is the median.