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3 years ago

HELP WHAT NUMBER DOES IT SHOW??

Mathematics

2 answers:

ollegr [7]3 years ago

6 0

Step-by-step explanation:

28,4 most likely I think

amid [387]3 years ago

4 0

Answer:

$27.6$

Step-by-step explanation:

Hope it helps! :)

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Who was the first United States president

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dalvyx [7]

The first digit may be 1 or 8.

The second digit may be 1, 2, 4, 8.

The third digit may be 1, 3, 5, 7, 9.

There are $2\times4\times5=40\:variants$

The probability that this is the correct code is $\frac{1}{40}$

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An ice cream cone is filled with a vanilla and chocolate ice cream at a ratio of 2:

nirvana33 [79]

$\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\quad 
\begin{cases}
r=radius\\
h=height\\
----------\\
diameter=2\\
r=\frac{diameter}{2}\\
\quad = 1\\
h=6
\end{cases}\implies V=\cfrac{\pi\cdot 1^2\cdot 6}{3}
\\\\\\
V=2\pi \\\\
-------------------------------$

$\bf \cfrac{\stackrel{vanilla}{volume}}{\stackrel{chocolate}{volume}}\qquad 2:1\qquad \cfrac{2}{1}\implies \cfrac{2\cdot \frac{V}{2+1}}{1\cdot \frac{V}{2+1}}\implies \cfrac{2\cdot \frac{2\pi }{2+1}}{1\cdot \frac{2\pi }{2+1}}
\\\\\\
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3 years ago

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The managers of 21 supermarkets counted the number of cars in their parking lots on the same day. The results are shown in the l

mixer [17]

The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: $(\frac{n+1}{2}) th\ term$

Putting n=21

$(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term$

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

For First Half:

98, 100, 101, 102, 108, 109, 111,118, 129, 132

$Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5$

For Second Half:

135, 135, 145, 146, 146, 156, 170 176, 180, 180

$Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151$

Now

Interquartile Range:

$IQR = Q_3-Q_1\\= 151-108.5\\=42.5$

Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

#LearnwithBrianly

7 0

3 years ago