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olya-2409 [2.1K]
3 years ago
7

If the student is a domain/inout and the hair color is the range/output how is these two set of relation not a function?

Mathematics
1 answer:
IRINA_888 [86]3 years ago
6 0

Answer:

A function is a relation that maps inputs from a set called the domain, into outputs from a set called the range.

Such that each input can be mapped into only one output.

So for example, if we have a relation that maps the input 2 into two different values:

f(2) = 4

f(2) = 8

Then this is not a function.

In the case of the problem, we have a student as the input, and the hair color as the output.

So we will have something like:

f(student) = blond

And if this student decides to change his/her hair color to red?

Then the function becomes:

f(student) = red

So for the same input, we had two different outputs, which means that this is not a function.

We also could have the case where a given student has two colors (Californian for example)

Where again, we would see two different outputs for one single input.

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A. 5(b+2)

Step-by-step explanation:

Five times the sum of b and two

The sum of "b" and "2" means

b + 2

Five times the sum means,

5(b+2)

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3 years ago
The function y = -3x + 5 is transformed by reflecting it over the y axis. What is the equation of the new function? *
SpyIntel [72]

Answer:

x= −1/3 y+5/3

Step-by-step explanation:

Let's solve for x.

y=−3x+5

Step 1: Flip the equation.

−3x+5=y

Step 2: Add -5 to both sides.

−3x+5+−5=y+−5

−3x=y−5

Step 3: Divide both sides by -3.

−3x

−3

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4 0
3 years ago
9 is added to a number and multiplied by 7. The result is 84.​
bearhunter [10]

Answer:

<h2> 12 because 9+3 = 12 12 • 7 = 84</h2>

Step-by-step explanation:

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3 years ago
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A positive integer from one to six is to be chosen by casting a die. Thus the elements c of the sample space C are 1, 2, 3, 4, 5
Alex_Xolod [135]

Answer with Step-by-step explanation:

We are given that six integers 1,2,3,4,5 and 6.

We are given that sample space

C={1,2,3,4,5,6}

Probability of each element=\frac{1}{6}

We have to find that P(C_1),P(C_2),P(C_1\cap C_2) \;and\; P(C_1\cup C_2)

Total number of elements=6

C_1={1,2,3,4}

Number of elements in C_1=4

P(E)=\frac{number\;of\;favorable \;cases}{Total;number \;of\;cases}

Using the formula

P(C_1)=\frac{4}{6}=\frac{2}{3}

C_2={3,4,5,6}

Number of elements in C_2=4

P(C_2)=\frac{4}{6}=\frac{2}{3}

C_1\cap C_2={3,4}

Number of elements in (C_1\cap C_2)=2

P(C_1\cap C_2)=\frac{2}{6}=\frac{1}{3}

C_1\cup C_2={1,2,3,4,5,6}

P(C_1\cup C_2)=\frac{6}{6}=1

4 0
3 years ago
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