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Vinil7 [7]
3 years ago
9

Im just wasting points :) Ignore this

Mathematics
1 answer:
ollegr [7]3 years ago
5 0
Why? how much points do you have to spare?
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⚠️⚠️What should I change ? And why ? ⚠️⚠️
aksik [14]

Answer/Step-by-step explanation:

Change the 3rd and 4th input. ALL the input values must be different even if they have the same output values in order to be a function. NONE of the input values should be the same, it just has to be something completely different for it to be correct. It may not be linear, but It will be correct.

8 0
3 years ago
Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordl
Karo-lina-s [1.5K]

Full Question

Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordless, and the other six are corded phones. Suppose that these components are randomly allocated the numbers 1, 2, . . . , 18 to establish the order in which they will be serviced.

What is the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced?

What is the probability that two phones of each type are among the first six serviced?

Answer:

a. 0.149

b. 0.182

Step-by-step explanation:

Given

Number of telephone= 18

Number of cellular= 6

Number of cordless = 6

Number of corded = 6

a.

There are 18C6 ways of choosing 6 phones

18C6 = 18564

From the Question, there are 3 types of telephone (cordless, Corded and cellular)

There are 3C2 ways of choosing 2 out of 3 types of television

3C2 = 3

There are 12C6 ways of choosing last 6 phones from just 2 types (2 types = 6 + 6 = 12)

12C6 = 924

There are 2 * 6C6 * 6C0 ways of choosing none from any of these two types of phones

2 * 6C6 * 6C0 = 2 * 1 * 1 = 2.

So, the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced is

3 * (924 - 2) / 18564

= 3 * 922/18564

= 2766/18564

= 0.149

b)

There are 6C2 * 6C2 * 6C2 ways of choosing 2 cellular, 2 cordless, 2 corded phones

= (6C2)³

= 3375

So, the probability that two phones of each type are among the first six serviced is

= 3375/18564

= 0.182

5 0
3 years ago
7 | 6.3 | =<br><br><br><br> help meeeeee
MakcuM [25]

Answer:

i would say 1.111111111111 you get it it keeps going

6 0
3 years ago
Read 2 more answers
Please help with the image below
prisoha [69]

Answer:

150

Step-by-step explanation:

7 0
2 years ago
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4[9(5-4)2] what is the answer
SVEN [57.7K]

Answer:

72

Step-by-step explanation:

5-4=1

9x1=9

9x2=18

18x4=72

4 0
3 years ago
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