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Nadya [2.5K]
3 years ago
9

Determine the range of the graph above

Mathematics
1 answer:
kozerog [31]3 years ago
8 0

Answer:

The range of the graph is:

-3 ≤ y ≤ 3

Hence, option (B) is correct.

Step-by-step explanation:

We know that the range of a function is the set of values of the dependent variable 'y' for which a function is defined.

From the given graph, it is clear that the graph goes down at y=-3 and then goes up to y=3  and then goes down again.

In fact, this indicates that the range of the graph lies between y=-3 to y=3

Therefore, the range of the graph is:

-3 ≤ y ≤ 3

Hence, option (B) is correct.

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Select the correct answer from each drop-down menu.
kow [346]

Triangle ABC and triangle XYZ are similar based on angle-side-angle criterion.

Triangle ABC and triangle RQP are similar based on side-angle-side criterion.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Two triangles are said to be congruent if they have the same shape and their corresponding sides are congruent.

Triangle ABC and triangle XYZ are similar based on angle-side-angle criterion.

Triangle ABC and triangle RQP are similar based on side-angle-side criterion.

Find out more on equation at: brainly.com/question/2972832

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5 0
2 years ago
Sketch the graph of y (x +3)2-4 and identify the axis of symmetry.
masya89 [10]

The axis of symmetry is at x = -3.


This can be found by looking at the basic form of vertex form:


y = (x - h)^2 + k


In this basic form the vertex is (h, k). By looking at what is plugged into the equation, it is clear that h = -3 and k = -4. This means the vertex is at (-3, -4).


It is a fact that the axis of symmetry is a vertical line of x = (vertex value of x). So we can determine that the axis of symmetry is at x = -3

6 0
3 years ago
What is 6 plus 12 times 8-3
9966 [12]

Answer:

99

Step-by-step explanation: Use PEMDAS

12+8 = 96

96+6 = 102

102-3 = 96

Hope this helped!

4 0
3 years ago
Both sets of values have an average of 13. Is Set A's standard deviation smaller, larger, or about the same as Set B's? (Note: T
Softa [21]

Answer:

  Set A's standard deviation is larger than Set B's

Step-by-step explanation:

Standard deviation is a measure of variation. One way to judge the value of standard deviation is by looking at the range of the data. In general, a dataset with a smaller range will have a smaller standard deviation.

The range of data Set A is 25-1 = 24.

The range of data Set B is 18-8 = 10.

Set A's range is larger, so we expect its standard deviation to be larger.

__

The standard deviation is the root of the mean of the squares of the differences from the mean. In Set A, the differences are ±12, ±11, ±10. In Set B, the differences are ±5, ±3, ±1. We don't actually need to compute the RMS difference to see that it is larger for Set A.

Set A's standard deviation is larger than Set B's.

5 0
3 years ago
A population has the following characteristics. (a) A total of 75% of the population survives the first year. Of that 75%, 25% s
Artemon [7]

Answer:

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age  \leq 3}\end{array}\right

i.e after the first year ;

there 1344 members in the first age class

84 members for the second age class; and

28 members for the third age class

Step-by-step explanation:

We can deduce that the age distribution vector x represents the number of population members for each age class; Given that in each class of age there are 112 members present.

The current age distribution vector is as follows:

x = \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right] \left[\begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 0 \  \leq  age   \leq  2 }\\{0 \  \leq  age   \leq 3}\end{array}\right]

Also , the age transition matrix is as follows:

L = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]

After 1 year ; the age distribution vector will be :

x_2 =Lx_1 = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]  \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right]

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq 1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age   \leq  3}\end{array}\right

6 0
3 years ago
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