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Andrej [43]
3 years ago
13

How many distinct permutations are there of the letters of the word BOOKKEEPER?

Mathematics
2 answers:
ASHA 777 [7]3 years ago
8 0

Answer:

The answer is 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.

Step-by-step explanation:

emmainna [20.7K]3 years ago
7 0
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!}
\\
\\=\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 
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A town’s population went from 35,000 to 42,000 in 3 years. What was the percent of change?
dolphi86 [110]

Answer:

The population changed by 20% in 3 years

Step-by-step explanation:

The percentage change, is given by the following equation;

Percentage \ change = \dfrac{Final \ value - Initial \value }{Initial \ value} \times 100

The given parameters are;

The initial population of the town = 35,000

The final population of the town = 42,000

The percentage change in the population is therefore;

Percentage \ change \ in \ population= \dfrac{42,000 - 35,000 }{35,000} \times 100 = 20 \%

The population changed (increase) by 20% in 3 years.

3 0
3 years ago
Which value of x makes this equation true? –5(– 20 ) = 35 <br>A. -11<br>B. -3 <br>C. 13 <br>D. 27​
kirza4 [7]

Answer:

c

Step-by-step explanation:

its right

3 0
2 years ago
Read 2 more answers
The radioactive isotope radium has a half-life of 1,600 years. The mass of a 100-gram sample of radium will be______ grams after
Olenka [21]
To solve this we are going to use the half life equation N(t)=N_{0} e^{( \frac{-0.693t}{t _{1/2} }) }
Where:
N_{0} is the initial sample
t is the time in years
t_{1/2} is the half life of the substance
N(t) is the remainder quantity after t years 

From the problem we know that:
N_{0} =100
t=200
t_{1/2} =1600

Lets replace those values in our equation to find N(t):
N(200) =100e^{( \frac{(-0.693)(200)}{1600}) }
N(200)=100e^{( \frac{-138.6}{1600} )}
N(200)=100e^{-0.086625}
N(200)=91.7

We can conclude that after 1600 years of radioactive decay, the mass of the 100-gram sample will be 91.7 grams.

8 0
3 years ago
What two fractions are being<br> multiplied?
tiny-mole [99]
I see 2/3 and 1/2 being multiplied
3 0
2 years ago
What is the value of 5^3i^9?
Yakvenalex [24]

Answer:

\displaystyle 125^i

Step-by-step explanation:

<u>Extended Information </u><u>on</u><u> </u><u>the</u><u> Complex Number System</u>

\displaystyle \sqrt{-1} = i \\ -1 = i^2 \\ -i = i^3 \\ 1 = i^4\:[all\:multiples\:of\:four]

I am joyous to assist you at any time.

All work is shown above. ↑

4 0
3 years ago
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