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Nataly [62]
3 years ago
8

Solve for x and show your work 2x=36 x= ?

Mathematics
2 answers:
Vanyuwa [196]3 years ago
6 0

Answer:

x = 18

Step-by-step explanation:

2x = 36

We want to get x alone, so we need to divide by 2

2x/2 = 36/2

x = 18

Harlamova29_29 [7]3 years ago
3 0

Answer:

x=18

36/2= 18

therefor 2(18)=36

Step-by-step explanation:


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how many 4 letter words can you make from a list of 12 letters, if you use each letter only once in each word?​
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Which function has the same domain as y=2 sqrt x
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The domain of a function is the set of values for which the function is defined; i.e. the function outputs a value for an input from the domain. That means that we must be able to plug in numbers for x and get numbers from y.
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4 years ago
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3 years ago
Of the 9-letter passwords formed by rearranging the letters AAAABBCCC (4 A's, 2 B's, and 3 C's), I select one at random. Determi
Tanya [424]

Answer:

a) 3

b) (8!/9!)-(7!/9!)

c) (1-(8!/9!))*(7!/9!)

Step-by-step explanation:

a)With 4 As ;  2Bs and 3Cs it is possible to get a palindrome if you fixed the  letters C according to: (2) in the extremes of the word and the other one at the center therefore you only have palindrome in the following cases

<u>C</u> (       ) <u>C</u> (       ) <u>C</u>

To fill in the gaps we have  4 letters  A and 2 letters B, wich we have two divide in two palindrome gaps,  

AAB         and    BAA the palindrome is  C  AAB C BAA C

BAA         and    AAB    "           "           is  C  BAA C AAB C  

ABA         and    ABA    "           "           is  C  ABA C ABA C

b) 4 A  ;   2B  ; 3C

We have the total number of elements  9, so the total number of possible outcomes is : 9!

Total events: 9!

if we fixed 3 C we have (the group of 3 Cs becoming one element) so the total amount of events with 3 adjacent Cs is: 7!

Therefore the probability of having 3 adjacent Cs is: 7!/9!

If we fixed only 2 Cs we have:

4 A  ; 2 B  ; 2C  : 1C

Total number of words (events) in this case is 8! (2C becomes 1 element)

so the total numbers of events is 8! the probability in this case is 8!/9!(this value includes cases of adjacent 3 Cs previous calculated ) so this value minus the case of 3 adjacent Cs ) give us 2 adjacent C and the other no next to them

Probability (of words with 2 adjacent Cs and the other no next to them is); 8!/9! - 7!/9!

c) Probability of B apart from each other is the whole set of events minus those where 2 B are adjacent or (become 1 element)

4 A ; 2B ; 3C

Total of events 9! and events with adjacent B is 8!/9!

Therefore the probability of words with 3 adjacent Cs and 2 B separeted is

the probability of 3 adjacent Cs (7!/9!) times probability of words with no adjacent Bs wich is (1-(8!/9!))*(7!/9!)

5 0
3 years ago
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