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djverab [1.8K]
1 year ago
14

consider the word unusual. how many unique subsets of 5 letters (of the 7) exist? how many different strings could be made from

5 of those 7 letters?
Mathematics
1 answer:
lesantik [10]1 year ago
8 0

The total number of unique subset of 5 letter words possible are 840 and the total number of different strings that could be made 5040.

The word 'unusual' contains a total of 7 letters.

A subset of 5 letters from the seven letters of the word unusual can be found out by making the combinations of all the five letter words possible from the seven letters.

As we can see that there are total of 7 letters among which u is repeated three times. Number of the possible combinations for the subset of 5 letter word are,

= 7×6×5×4

= 840.

So, there are 840 subset possible.

The total number of Strings that are possible is,

= 7×6×5×4×3×2×1

= 5040.

So, the total number of Strings possible are 5040.

To know more about combinations, visit,

brainly.com/question/28065038

#SPJ4

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hammer [34]

Answer:

Sally has been measuring the tree for 4 months

Step-by-step explanation:

The correct question is as follows;

Sally has been measuring the tree in her yard for a science project. The tree has grown 1/2 of an inch each month and grown a total of 2 inches taller. How many months has Sally been measuring this tree?

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Solution;

The tree has grown 1/2 of an inch each month and has grown a total of 2 inch taller.

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2 years ago
F m 1=7x and m 4=3x+20 what is the m 2
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4 0
3 years ago
For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12
notsponge [240]
The binomial distribution is given by, 
P(X=x) =  (^{n}C_{x})p^{x} q^{n-x}
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) = (^{100}C_{x})(0.2)^{x} (0.8)^{100-x}
                    where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12                  
∴ P(X≤12) = (^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1} + (^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3} + (^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5} + (^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7} + (^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9} + (^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11} + (^{100}C_{12})(0.2)^{12} (0.8)^{100-12}


Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability. 
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harriet is cultivating a strain of bacteria has doubled by the end of every second hour. how many bacteria will harriet have in
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Answer:

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P is total population

n is initial population

t is time

since it doubles every second hour, and there are 6 hours

6hours/2hours= doubles 3 times, thats why t=3.

7 0
3 years ago
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