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2 years ago

What is the range of the function shown on the graph above?

Mathematics

1 answer:

nikklg [1K]2 years ago

4 0

Answer:

D: 0 < y < 7

Step-by-step explanation:

The domain is the input (i.e. the x values)

The range is the output (i.e. the y values)

So for the range, look at the y-coordinates of the end points of the line: (-6, 0) and (9, 7) ⇒ 0 < y < 7

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Plz help and thank you!

hoa [83]

6 and 7 because if you solve for r the answer would be r > 6

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3 years ago

A store sells 3 ears of corn for $1 they round prices to the nearest cent.Tell whether you would describe the relationship betwe

ziro4ka [17]

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3 years ago

Find the length of the missing side. The triangle is not drawn to scale.

timofeeve [1]

Answer:

Step-by-step explanation:

a^2 + 8^2 = 10^2

a^2 + 64 = 100

a^2 = 36

sq root of a = sq root of 36

a = 6

8 0

3 years ago

Read 2 more answers

Review: To simplify an expression, combine like terms by adding or

gavmur [86]

Answer:

10b + 2

Step-by-step explanation:

To simplify this expression. 10b + 7 - 3 - 2 use the sample you demonstrated above to simplify it

10b + 7 -3 -2

combine the like terms

10b + 7 - 5

10b + 2

4 0

3 years ago

Lizzie rolls two dice. What is the probability that the sum of the dice is:

zhenek [66]

Answer:

$A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\$

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A) = \dfrac{12}{36} = \dfrac{1}{3}$

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)} Count 1

Possible cases for sum 3: {(1,2), (2,1)} Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)} Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)} Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

$P(B) = \dfrac{15}{36} = \dfrac{5}{12}$

C. Divisible by 3 and less than 7:

$P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}$

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

$P(A \cap B) = \dfrac{7}{\text{36}}$

5 0

3 years ago