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3 years ago

A function is defined by y = 2x + 1. What is the range of the function if the domain is {0, 2, 3}?

Mathematics

1 answer:

Eva8 [605]3 years ago

7 0

9514 1404 393

Answer:

{1, 5, 7}

Step-by-step explanation:

For x=0, y = 2·0 +1 = 1

For x=2, y = 2·2 +1 = 5

For x=3, y = 2·3 +1 = 7

The range is the list of y-values, so is {1, 5, 7}.

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5. SLOPE need help like, ASAP, tysm

Papessa [141]

Answer:

Since it's a horizontal line the slope would be 0

Step-by-step explanation:

Horizontal Lines A horizontal line is a line whose equation is of the form =, where is a real number. All points that lie on the line have a -coordinate of . The slope of any horizontal line is zero, 0.

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3 years ago

Drag a drop a number into each box to create pairs of corresponding angles

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Step-by-step explanation:

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4 years ago

Two friends compete with each other and five other, equally good, violinists for first and second chair in an orchestra, in a bl

Darina [25.2K]

Answer: 0.0476

Step-by-step explanation:

Given : Two friends and 5 other people compete with each other for first and second chair in an orchestra.

Total people in this competition= 2+5=7

By permutation , Number of ways to arrange 7 people= 7!

Also, number of ways for two friends end up as first and second chair together= 2 × 5! [ 2 ways to arrange friends on first and second chair and 5! ways to arrange others]

I.e. Required probability = $\dfrac{2\times5!}{7!}$

$=\dfrac{2!\times5!}{7\times6\times5!}\\\\=\dfrac{1}{7\times3}\\\\=\dfrac{1}{21}\\\\=0.0476$

Hence, the probability that the two friends end up as first and second chair together = 0.0476

4 0

4 years ago

Sets A and B are independent events and are mutually exclusive

Lynna [10]

The solutions are ${P}({A} \text { and }B})=0$ and $P}({A} \text { or } {B})=\frac{7}{12}$

Explanation:

Given that the sets A and B are independent events and are mutually exclusive events.

We need to determine the value of $P(A \text { and } B)$ and $P(A \text { or } B)$

The value of $P(A \text { and } B)$ :

Since, the sets A and B are mutually exclusive events, it is impossible for the two events to occur together.

Hence, we have,

${P}({A} \text { and }B})=0$

Thus, the value of $P(A \text { and } B)$ is 0.

The value of $P(A \text { or } B)$ :

For mutually exclusive events, the value of $P(A \text { or } B)$ is given by the formula,

$P(A \text { or } B)=P(A)+P(B)$

Substituting the values, we get,

$P(A \text { or } B)=\frac{1}{4}+\frac{1}{3}$

$P}({A} \text { or } {B})=\frac{7}{12}$

Hence, the value of $P(A \text { or } B)$ is $\frac{7}{12}$

Thus, the solutions are ${P}({A} \text { and }B})=0$ and $P}({A} \text { or } {B})=\frac{7}{12}$

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4 years ago

Natalie used 112 inches blue yarnas a border around her rectangular bulletin board. If the bulletin board is 36 inches wide, how

wel

3.1 with a repeating “1” (I think not sure)

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3 years ago