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pashok25 [27]
2 years ago
15

The height of a stack of DVD cases is proportional to the number of cases in the stack. The height of 6 DVD cases is 114 mm.

Mathematics
2 answers:
Zanzabum2 years ago
8 0
OK so my answer is from what you’ve just explained that means you multiply the six and the 114 is 684 so that means if you have to do that to 13 of them that means 684×13 is what your equation would be in the answer to 684×13 is 8892
Sholpan [36]2 years ago
5 0

88Answer:

Step-by-step explanation:

8892

You might be interested in
Which choice is equivalent to the expression below?
alekssr [168]

Answer: OPTION A

Step-by-step explanation:

We need to remember that Product of powers property, which states that:

(a^m)(a^n)=a^{(m+n)}

Let's check the options:

A. For 5^9*5^{\frac{9}{10}}*5^{\frac{6}{100}}*5^{\frac{9}{1000}} you can  apply the property mentioned before. Then:

5^{(9+\frac{9}{10}+\frac{6}{100}+\frac{9}{1000})=5^{9.969}

(It is the equivalent expression)

 B. Add the exponents:

 5^9*5^{(\frac{9}{10}+\frac{9}{10}+\frac{6}{1000})=5^{10.806}

(It is not the equivalent expression)

C. For 5^9*5^{\frac{96}{10}}*5^{\frac{9}{100}}} you can  apply the property mentioned before. Then:

 5^{(9+\frac{96}{10}+\frac{9}{100})=5^{18.69}

(It is not the equivalent expression)

D. We know that 5^{9.969}=9,290,347.808 and we maje the addition indicated in this option, we get:

 5^9+5^{\frac{9}{10}}+6^{\frac{6}{100}}=1,953,130.37

(It is not the equivalent expression)

6 0
4 years ago
Read 2 more answers
Twenty-seven is<br>% of 60<br>help mee asap​
jok3333 [9.3K]

Answer:  The answer is:  " <u> </u><u>45 </u>  % "  .    

________________________________________________

               →    " Twenty-seven is <u> 45 </u> % of 60. "

________________________________________________

Step-by-step explanation:

________________________________________________

The question asks:

 " 27 is what % {percentage] of 60 " ?  ;

________________________

So:  " 27 =  (n/100) * 60 " ;  Solve for "n" ;

________________________________________________

Method 1:

________________________________________________

  →   (n/100) * 60 = 27 ;

Divide each side by 60 :

 →   [ (n/100)  * 60 ] / 60 = 27 /60 ;

to get:

 →    (n/100) = 27/60 ;

Now:  Cross-factor multiply:

 →  60n = (27)*(100) ;

to get:

 → 60n = 2700 ;

Divide each side by "60" ;

→  60n = 2700/ 60 ;

to get:  n = 45 ;

________________________

 →  The answer is:  45 % .    

   →  " Twenty-seven is <u>45 %</u> of 60."

________________________________________________

Method 2:

________________________________________________

The question asks:

 " 27 is what % {percentage] of 60 " ?

________________________

To solve this problem:

Rephrase this question as:

________________________

" 27 is 60% of what number ? "

 →  The answer will be the same!

________________________

→  27 = (60/100)* n ;   Solve for "n" ;

Multiply each side of the equation by "100" ; to eliminate the fraction:

→  100 * 27 = 100 * [ (60/100)* n ] ;

 to get:

   →   2700 = 60n ;

↔  60n = 2700 ;

Divide Each Side of the equation by "60" ;

    →   60n/60 = 2700 / 60 ;

to get:  n = 45 ;

________________________________________________

→  The answer is:  45 % .    

       →  " Twenty-seven is <u>45 %</u> of 60."

________________________________________________

Method 2:  Variant 1 of 2:

________________________________________________

When we have:

→  27 = (60/100)* n ;   Solve for "n" ;

________________________

Note that:  "(60/100) = (60÷ 100) = (6 ÷ 10)" ;   since:  in "(60/100)" ;  the "zero" from the "<u>numerator</u>" cancels out;  <u>And</u>:  the "last zero" in "100" — from the "<u>denominator</u>" cancels out;  since we are dividing "each side" of the fraction by "10" ;

  →   "(60÷10) / (600÷10)"  =  " 6/10 " ;  

  →   " (6/10)" ; that is;  "six-tenths"} ;  

  →     can be represented by:  " 0.6 " ;

  →  {by convention;  but specifically, here is the explanation} — as follows:

________________________

  →   "(6/10)" =  " (6 ÷ 10) " ;  

<u>Note</u>:  When dividing a number by "10" ;  we take the original number; and move the decimal point to the left; & then we rewrite that number as the "answer".  

<u>Note</u>:  When multiplying or dividing by a positive, non-zero integer factor of "10" that has at least 1 (one) "zero" after that particular factor of "10".  We can get the answer by taking the original number & moving the decimal point the number of spaces as designated by the number of zeros following the particular [aforementioned factor of "10".].

We move the decimal point to the right if we are multiplying;  and to the left  if we are dividing.  In this case, <u>we are dividing</u> "6" by "10 " :

 →  " 6   ÷  10  =  ? " ;  

 →  " 6.  ÷ 10 =  ? " ;

   We take the: " 6. " ;  and move the decimal point "<u>one space backward [i.e. "to the left</u>"];  since we are <u>dividing by "</u><u>10</u><u>"</u> ;

 →  to get:  " .6 " ;  & we rewrite this value as "0.6" in a rewritten equation:

________________________

So; we take our equation:

→  27 = (60/100)*n ;  And rewrite—substituting "0.6" for

"(60/100)"— as follows:

________________________

→  27 = (0.6)n ;  ↔ (0.6) n = 27 ;

Multiply each side of the equation by "10" ; to eliminate the decimal:

   →  10 * [ (0.6)n ]  = 27 * 10 ;

to get:

  →  6n = 270 ;

Divide each side of the equation by "6" ; to isolate "n" on one side of the equation; & to solve for "n" ;

 →  6n / 6  =  270 / 6 ;

to get:   n = 45 ;

________________________________________________

→  The answer is:  45 % .    

      →  " Twenty-seven is <u>45 %</u> of 60."

________________________________________________

Method 2 (variant 2 of 2):

________________________________________________

We have the equation:  27 = (60/100)* n ;   Solve for "n" ;

________________________

<u>Note</u>:  From Method 2 (variant) 1 of 2— see above):

________________________

<u>Note</u>:  Refer to the point at which we have:

________________________

→   " {  (60÷10) / (600÷10)"  =  " (6/10) " ;  that is;  "six-tenths"} ;

________________________

Note that the fraction— "(6/10)" ;  can be further simplified:

→  "(6/10)" =  "(6÷2) / (10÷2)" = "(3/5)" ;

Now, we can rewrite the equation;

→ We replace "(60/100)" ;  with:  "(3/5)" :

    →  27 = (3/5)* n ;   Solve for "n" ;

↔ (3/5)* n = 27 ;  

↔    (3n/5) = 27 ;

Multiply Each Side of the equation by "5" ;

→  5* (3n/5) = 27 * 5 ;  

to get:

→   3n = 135 ;

Divide Each side of the equation by "3" ;  to isolate "n" on one side of the equation;  & to solve for "n" ;

→  3n / 3 = 135 / 3  ;

to get:   n = 45 ;

________________________________________________

 →  The answer is:  45 % .    

       →  " Twenty-seven is <u>45 %</u> of 60."

________________________________________________

Hope this answer is helpful!

        Wishing you the best in your academic endeavors

           — and within the "Brainly" community!

________________________________________________

7 0
3 years ago
Read 2 more answers
5 burgers and 3 fries cost $23.
Mashutka [201]
...........................
5 0
3 years ago
What 1 thruogh 100 added up them divided
Marina CMI [18]
We can use the sum of an aritmetic sequence
the sum from n=1 to n=r when the first term is a1 and the nth term is an is
S=\frac{n(a_1+a_n)}{2}
first term is 1
last term is 100
there are 100 terms so n=100

so the sum is S=\frac{100(1+100)}{2}
S=(50)(101)
S=5050
now you want us to divide by 10
5050/10=505


fun fact, gauss (famous math guy) did this when he was younger, legend has it that he was assigned this as an in class assigment to kill time but gauss found a neat pattern, he noticed that adding the end terms wer giving the same sum, example, 100+1=101, 2+99=101, etc, so he just needed to find al the pairs and add them all up




answer is 505
the result is 505
5 0
3 years ago
The probability of winning a certain lottery is 1/77076 for people who play 908 times find the mean number of wins
Alex73 [517]

The mean is 0.0118 approximately. So option C is correct

<h3><u>Solution:</u></h3>

Given that , The probability of winning a certain lottery is \frac{1}{77076} for people who play 908 times

We have to find the mean number of wins

\text { The probability of winning a lottery }=\frac{1}{77076}

Assume that a procedure yields a binomial distribution with a trial repeated n times.

Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial.

n=908, \text { probability } \mathrm{p}=\frac{1}{77076}

\text { Then, binomial mean }=n \times p

\begin{array}{l}{\mu=908 \times \frac{1}{77076}} \\\\ {\mu=\frac{908}{77076}} \\\\ {\mu=0.01178}\end{array}

Hence, the mean is 0.0118 approximately. So option C is correct.

4 0
3 years ago
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