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snow_tiger [21]
2 years ago
5

Anjana has 4 pencil in her box. Their average length is 10cm. Which of the following is NOT POSSIBLE? A) The shortest pencil in

the box is 2cm long B) The longest pencil in the box is 8cm long C) Two of the pencil's are 10cm each D) all the pencils are of the same length Guys can you quickly give an answer to the Question WITH AN EXPLANATION
Mathematics
1 answer:
Troyanec [42]2 years ago
5 0

The situation which is not possible about the pencils in her box is; Choice B: The longest pencil in the box is 8cm long.

<h3>Mean and Measure of central tendency</h3>

By convention, the mean is a statistical measure of central tendency.

On this note, if the mean of a set of data values is 10, this means the data values have data points greater, equal and less than the mean value.

  • On this note, it is impossible to have the longest pencil in the box to have a length, 8cm.

Read more on Mean;

brainly.com/question/281308

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uranmaximum [27]
20-17.92 is 2.08. .13x=2.08.


divide by 2.08 to get your answer which is .0625 or 06 minutes.

8 0
4 years ago
Please help thanks! Brainliest <br> Sorry for spamming.
Bad White [126]

Answer: −10

Step-by-step explanation: 9^2−21/−6

81−21/−6

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6 0
3 years ago
Troy signed up for a new cell phone plan which charges him a fee of $40 per month, plus $0.05 for each text message that he send
nataly862011 [7]
He can send 600 messages.
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3 years ago
X+2/x–2 –x–2/x+2 = 5/6<br>​
Anna71 [15]
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4 0
3 years ago
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\qquad \qquad \textit{direct proportional variation}\\\\\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill }\\\\\textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill }\\\\[-0.35em]~\dotfill

\stackrel{\textit{"th" varies directly with "p"}}{th=kp}\qquad \textit{we also know that} \begin{cases} th=1.8\\ p=350 \end{cases} \\\\\\ 1.8=k(350) \implies \cfrac{9}{5}=350k\implies \cfrac{9}{1750}=k~\hfill \boxed{th=\cfrac{9}{1750}k} \\\\\\ \textit{when p = 84, what is "th"?}\qquad th=\cfrac{9}{1750}(84)\implies th=\cfrac{54}{135}\implies th=0.432

8 0
2 years ago
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