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Brut [27]
3 years ago
14

What is the domain and range of the relation shown in the table?

Mathematics
1 answer:
TEA [102]3 years ago
6 0

Answer:

<u>domain: {10,15,19,32}</u>

<u>range:{5,9,-1}</u>

Step-by-step explanation:

  • As we know domain is the values of input and range is the values of output.
  • Here , x is the input and y is the output.
  • \left[\begin{array}{cc}x&y\\10&5\\15&9\\19&-1\\32&5\end{array}\right]
  • Thus the input values according to the given problem is : 10 ,15 , 19, and thus ,

⇒<em>The domain would accordingly be these four numbers : 10 , 15 , 19 , 32.</em>

  • <u>Note that we donot have any information regarding the other values of x.</u>
  • The range is : { 5,9,-1 } only as the 5 is repeated in two cases .
  • Range is unique and there must be not repetition. Thus the apt answer would be :

<em>domain: {10,15,19,32}</em>

<em>range:{5,9,-1}</em>

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A jar contains chocolate candies. There are 12 red, 17 blue, 5 green and 6 orange. What is the probability of picking either a g
nataly862011 [7]

Answer:

Step-by-step explanation:

If you sum everything, you'll get 50 candies. Now you know that 50 candies is 100%.

Now you'll calculate the probability:

12 on 50 : 24.00 %

17 on 50 : 34.00 %

5 on 50 : 10.00 %

6 on 50 : 12.00 %

Hope I was helpful :)

Edit :

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7 0
3 years ago
Please simplify this problem:
joja [24]

Answer:

-52/75

Step-by-step explanation:

5 0
3 years ago
Use the vertical method to multiply (4a3 – 2a + 3a2 + 1) and (3 - 2a + a²).
Igoryamba

Answer:

.a^3-2a+3a^2+1\right)and\left(3-2a+a 2\ri

Step-by-step explanation:

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6 0
2 years ago
How many distinct permutations are there of the letters of the word BOOKKEEPER?
emmainna [20.7K]
The letters of the word BOOKKEEPER can be arranged in 151,200 ways.

The group can be arranged in 8640 ways with all students of the same major together.

Explanation
There are 10 letters in the word bookkeeper.  There is 1 B; 2 O's; 2 K's; 3 E's; 1 P; and 1 R.

An arrangement of n total objects where n₁ is one kind, n₂ is another, etc. is given by:
\frac{n!}{n_1!\times n_2!\times ... \times n_k!}&#10;\\&#10;\\=\frac{10!}{1!2!2!3!1!1!} = \frac{10!}{2!2!3!} = 151,200

Keeping all of the students of each major together makes each one essentially a "unit."  With this in mind, there are 3 units, that can be arranged in 3!=6 ways.

Within the English unit, the students can be arranged in 3!=6 ways.
Within the anthropology unit, the students can be arranged in 2!=2 ways.
Within the history unit, the students can be arranged in 5!=120 ways.

This gives us 6(6*2*120) = 8640 
7 0
3 years ago
Read 2 more answers
-11 2/3 x (-4 1/5) pls help me​
sveta [45]

Answer:

49

Step-by-step explanation:

8 0
3 years ago
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