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Arlecino [84]
4 years ago
15

Identify the range of the function shown in the graph.​

Mathematics
1 answer:
lys-0071 [83]4 years ago
7 0

Answer:

C. 0 ≤ y ≤ 8

Step-by-step explanation:

In any graph, the range represents the 'y' values covered by the graph.  The 'y' axis is the vertical (up and down) line on the graph.  Given the graph, it stops at the point (0, 8) on the top and at (-6,0) and (6,0) on the bottom.  This would be an overall range of 0 ≤ y ≤ 8.

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Watch help
babymother [125]

Answer:

m∠RPQ = 8°

Step-by-step explanation:

m∠QRS = 4x - 15

m∠RPQ = x + 1

m∠PQR = x - 2

m∠QRS is exterior angle and m∠RPQ and m∠PQr are opposite interior angles to m∠QRS

m∠QRS = m∠RPQ + m∠PQR      {Exterior angle property of triangle}

4x - 15 = x +1 + x - 2

4x - 15 = x + x + 1-2          {Combine like terms}

4x - 15 = 2x - 1        {Subtract 2x from both sides}

4x - 2x - 15 = - 1

2x - 15 = - 1                  {Add 15 to both sides}

2x = -1 + 15

2x = 14          {Divide both sides by 2}

x = 14/2

x = 7

m∠RPQ = x + 1 = 7 + 1 = 8°

4 0
3 years ago
Read 2 more answers
Which number line shows the solution to the inequality -2x-3> -11
Brut [27]

Answer:

-4

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
The population of a small town, P, as a function of time, t, in years past 1940 is given below.
romanna [79]

Answer: 1976

Steps:

Let t* be the year in which the population reached a count of 8920. Solve for t*:

P=200t+1720\\8920=200t^*+1720\\7200=200t^*\\t^* = 36

The population grew to the number 8920 in 36 years past 1940, i.e., 1976

3 0
3 years ago
Read 2 more answers
An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a d
anygoal [31]

Answer:

The probability of event <em>A</em> is \frac{7}{11}.

The probability of event <em>B</em> is \frac{6}{11}.

Step-by-step explanation:

The sample space of rolling a fair six-sided dice is as follows:

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

      (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

      (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

      (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

      (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

      (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Now consider the experiment of the computing the sum of the two rolls as follows:

X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Number of total outcomes, <em>N</em> = 11.

The probability of an event <em>E</em> is the ratio of the number of favorable outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

The event <em>A</em> is defined as the sum is greater than 5.

The sample space of <em>A</em> is:

A = {6, 7, 8, 9, 10, 11, 12}

n (A) = 7

Compute the probability of event <em>A</em> as follows:

P(A)=\frac{n(A)}{N}=\frac{7}{11}

Thus, the probability of event <em>A</em> is \frac{7}{11}.

The event <em>B</em> is defined as the sum is an even number.

The sample space of <em>B</em> is:

B = {2, 4, 6, 8, 10, 12}

n (B) = 6

Compute the probability of event <em>B</em> as follows:

P(B)=\frac{n(B)}{N}=\frac{6}{11}

Thus, the probability of event <em>B</em> is \frac{6}{11}.

7 0
3 years ago
What is 0.7 turned into a fraction
egoroff_w [7]

Answer:

Step-by-step explanation:

7/10

5 0
3 years ago
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