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koban [17]
3 years ago
7

How do you find the median

Mathematics
2 answers:
Dafna1 [17]3 years ago
4 0
In a list of numbers, such as 1,2,3,4,5,6,7, start from the right and cross off the last number. In this case, it's the 7. Then, go to the left and cross of the first number, the one. Keep going from right to left, and when you get down to one number in the middle, that is the median. However if there are two numbers in the middle, I think that you have to find the mean of those two numbers, and that should be the median. That's only if there are two numbers left in the middle.  Hope this helps!
damaskus [11]3 years ago
4 0
The median is the middle score for a set of data that has been arranged in order of magnitude.

So order your vales, and then pick the middle record.

This works fine when you have an odd number of scores.



BUT

What happens when you have an even number of scores?

It is easy...

You would calculate the mean of the two middle records ..

in other words:
you simply have to add up the middle two scores and average the result.


I hope that helps!



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B

Step-by-step explanation:

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Evaluate the expression (2.5)^3
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Answer:

15.625

Step-by-step explanation:

  • (2.5)^3 = 2.5 · 2.5 · 2.5
  • 2.5 · 2.5 · 2.5 = 6.25 · 2.5
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Therefore, (2.5)^3 = 15.625.

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How do I solve for B in the formula I=M/B+a<br><br>Is this correct: B=M/I -a
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Mariulka [41]

Given:

The equation is,

2\log _3x-\log _3(x-2)=2

Explanation:

Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

Simplify further.

\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

Solve the quadratic equation for x.

\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

\begin{gathered} x-6=0 \\ x=6 \end{gathered}

For (x - 3) = 0,

\begin{gathered} x-3=0 \\ x=3 \end{gathered}

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0
1 year ago
What is a/5+3=8 and what does a=?
jonny [76]

Answer:

a=25

Step-by-step explanation:

Solve by subtracting 3 from both sides.

a/5=5

Multiply both sides by 5.

a=25

8 0
2 years ago
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