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AVprozaik [17]
2 years ago
6

Suppose that a company's sales were $1,000,000 6 years ago and are $9,000,000 at the end of the 6 years. Find the geometric mean

growth rate of sales. (Round your answer to 4 decimal places.)

Mathematics
1 answer:
Dmitriy789 [7]2 years ago
6 0

Answer:

The geometric mean growth rate of sales is 1.4422.

Step-by-step explanation:

We have two sales values, one from 6 years ago and the other from now.

We have to calculate the geometric growth rate of sales.

We have:

y(-6)=1,000,000\\\\y(0)=9,000,000

We can write the relation between these two values as:

y(0)=y(-6)k^{0-(-6)}\\\\9,000,000=1,000,000k^6\\\\k^6=9\\\\k=9^{1/6}= 1.4422

The geometric mean growth rate of sales is 1.4422.

You might be interested in
Find the sum of the first 30 terms of the sequence below an=3n+2
Solnce55 [7]
First off, let's find the 1st term's value, and the 30th term's value,

\bf a1=3(1)+2\implies a1=5\qquad \qquad \quad   a30=3(30)+2\implies a30=92\\\\
-------------------------------\\\\
~~~~~~~\textit{ Sum of an arithmetic sequence}\\\\
S_n=\cfrac{n(a1+an)}{2}~ 
\begin{cases}
n=n^{th}\ term\\
a1=\textit{first term's value}\\
----------\\
a1=5\\
a30=92\\
n=30
\end{cases} \implies S_{30}=\cfrac{30(5+92)}{2}
\\\\\\
S_{30}=15(97)
6 0
2 years ago
Read 2 more answers
) there are exactly 20 students currently enrolled in a class. how many different ways are there to pair up the 20 students, so
Brilliant_brown [7]
<span>Assuming that "pair up students" means "divide up all 20 of the students into groups of two," and we regard two pairings as the same if and only if, in each pairing, each student has the same buddy, then I believe that your answer of 20! / [(2!)^10 * (10!)] is correct. (And I also believe that this is the best interpretation of the problem as you've stated it.) 

There are at least two ways to see this (possibly more). 


One way is to note that, first, we have to select 2 students for the first pair; that's C(20, 2) (where by C(20, 2) I mean "20 choose 2"; that is, 20! / (18! * 2!). ) 

Then, for each way of selecting 2 students for the first pair, I have to select 2 of the remaining 18 students for the second pair, so I multiply by C(18, 2). 

Continuing in this manner, I get C(20, 2) * C(18, 2) * ... * C(2, 2). 

But it doesn't matter in this situation the order in which I pick the pairs of students. Since there are 10! different orders in which I could pick the individual pairs, then I want to divide the above by 10!, giving me the answer 

[C(20, 2) * C(18, 2) * ... * C(2, 2)] / 10!. 

This is the same as your answer, because C(n , 2) = n(n - 1) / 2, so we can simplify the above as 

[(20 * 19) / 2 * (18 * 17) / 2 * ... * (2 * 1) / 2] / 10! 
= 20! / [2^10 * 10!] 
= 20! / [(2!)^10 * (10!)]. 



Another way is to reason as follows: 

1. First, arrange the 20 students in a line; there are 20! ways to do this 
2. We can get a pairing by pairing the 1st and 2nd students in line together, the 3rd and 4th students together, etc. 
3. But if I switch the order of the 1st and 2nd student, then this doesn't give a different pairing. I don't want to count both orderings separately, so I divide by 2! 
4. The same argument from step 3 holds for the 3rd and 4th student, the 5th and 6th student, etc., so I need to divide by 2! nine more times 
5. Finally, the particular order in which I selected the ten pairings are unimportant--for example, the following orderings don't produce different pairings: 

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 
3, 4, 1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 

So I need to further divide by the number of ways I can arrange the ten pairs, which is 10!. 


</span>
8 0
2 years ago
Use digits to write the value of the 4 in this number.<br><br> 842,963
Sidana [21]

Answer:

40,000

Step by step explanation:

8<u>4</u>2,963

8<u>40,000</u>

<u>40,000</u>

6 0
2 years ago
Read 2 more answers
RANDOMLY CHOOSING TWO POTENTIAL CLIENTS
gregori [183]
Let <span>Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be represented by the letters J, C, G, M, E, D, A, and S respectively. </span>

<span>In part IV we are asked:

</span><span>What is the sample space of the pairs of potential clients that could be chosen?
</span><span>
Since the Sample Space is the set of all possible outcomes, we need to make a set (a list) of all the possible pairs, which are as follows:

{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)

          , </span>(C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
<span>                   
</span>                      , (G, M), (G, E), (G, D), (G, A), (G, S)
<span>             
                                    ,</span>(M, E), (M, D), (M, A), (M, S)    
<span> 
                                               , </span>(E, D),  (E, A),  (E, S) <span>   
                                                          
                                                           , </span>(D, A), (D, S)
              
                                                                       , (A, S).}

We can check that the number of the elements of the sample space, n(S) is 

1+2+3+4+5+6+7=28.


This gives us the answer to the first question: <span>How many pairs of potential clients can be randomly chosen from the pool of eight candidates? 

(Answer: 28.)


II) </span><span>What is the probability of any particular pair being chosen?
</span>
The probability of a particular pair to be picked is 1/28, as there is only one way of choosing a particular pair, out of 28 possible pairs.

III) <span>What is the probability that the pair chosen is Jacob and Meg or Geraldo and Sally? 

The probability of choosing (J, M) or (G, S) is 2 out of 28, that is 1/14.


Answers:

I) 28
II) 1/28</span>≈0.0357
III) 1/14≈0.0714
IV)


{(J, C),  (J, G), (J, M),  (J, E),   (J, D),  (J, A),   (J, S)
          , (C, G), (C, M), (C, E),   (C, D), (C, A),  (C, S)
                      , (G, M), (G, E),  (G, D), (G, A),  (G, S)
                                   ,(M, E),  (M, D), (M, A),  (M, S)    
                                               , (E, D),  (E, A),  (E, S)    
                                                            , (D, A), (D, S)
                                                                        , (A, S).}
6 0
2 years ago
The figure shows a map of five streets that meet at Concord Circle. The measure of the angle formed by Melville Road and Emerson
DiKsa [7]
Do you have a picture?
4 0
2 years ago
Read 2 more answers
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