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2 years ago

5

When finding the median, do we have to put the numbers in order from least to greatest

2 answers:

iris [78.8K]2 years ago

8 0

Yes you do. Place all of the numbers, even repeating ones, in order from slammers to largest and then start crossing out from both sides until you get the middle. For example, in the problem below the median would be 4 bc it’s in the middle

1,2,2,3,4,6,6,7,8

White raven [17]2 years ago

4 0

There are usually multiple ways to solve any math problem.

You don't have to put the numbers in order if there's another way to find a value such that the number of numbers greater than that value is equal to the number of numbers less than that value.*

It is usually easiest to put the numbers in order, but it is not absolutely necessary.

_____
* If the size of the set is odd, then the value found must be in the set. If the size of the set is even, the value found must be the average of the two numbers in the set that are nearest and either side of that value.

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6-1/3 divided by 1/4 equals?

bagirrra123 [75]

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2 years ago

1) 5.10-2.8.16. 10-1

const2013 [10]

Answer: -2511 (if the periods mean multiplication)

Step-by-step explanation:

I’m not sure whether you’re trying to have the periods be multiplication, but if that’s what they are, this is the answer

5x10-2x8x16x10-1

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2 years ago

How many times larger is 3*10^8 than 5*10^6

Alex777 [14]

60 times larger.
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2 years ago

labwork [276]

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2 years ago

Read 2 more answers

A leprechaun places a magic penny under a girls pillow. The next night there are 2 magic pennies under her pillow. The following

kiruha [24]

Answer:

<em>After </em><em>47</em><em> days she will have more than 90 trillion pennies.</em>

Step-by-step explanation:

At the beginning there was 1 penny. At the second day the amount of pennies under the pillow became 2.

The amount of pennies doubled each day. So the series is,

$1,2,4,8,16,32,.....$

This series is in geometric progression.

As the pennies from each of the previous days are not being stored away until more pennies magically appear so the sum of series will be,

$S_n=\dfrac{a(r^n-1)}{r-1}$

where,

a = initial term = 1,

r = common ratio = 2,

As we have find the number of days that would elapse before she has a total of more than 90 trillion, so

$\Rightarrow 90\times 10^{12}\le \dfrac{1(2^n-1)}{2-1}$

$\Rightarrow 90\times 10^{12}\le \dfrac{2^n-1}{1}$

$\Rightarrow 90\times 10^{12}\le 2^n-1$

$\Rightarrow 2^n\ge 90\times 10^{12}+1$

$\Rightarrow \log 2^n\ge \log (90\times 10^{12}+1)$

$\Rightarrow n\times \log 2\ge \log (90\times 10^{12}+1)$

$\Rightarrow n \ge \dfrac{\log (90\times 10^{12}+1)}{\log 2}$

$\Rightarrow n \ge 46.4$

$\Rightarrow n\approx 47$

8 0

2 years ago