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Murljashka [212]
4 years ago
5

The estimated population of the world in the middle of 2010 was 6,852,472,823. Express this value in scientific notation.

Mathematics
1 answer:
Marina CMI [18]4 years ago
7 0

First we round it off to the nearest millions.

6,852,472,823 rounded off - 6,852,000,000

then you move the decimal point to the left until the number becomes between 1 and 10

-- 6.852000000

count the numbers to the right of the decimal point, remove the zeroes, then write it as

6.852 x 10^9 or 6.852e9

9 is how many numbers are in the right of the decimal point, which was 852000000

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{ \qquad\qquad\huge\underline{{\sf Answer}}}

Let's solve ~

Calculate discriminant :

\qquad \sf  \dashrightarrow \: 3 {x}^{2}  + 6x - 1

  • a = 3
  • b = 6
  • c = 1

\qquad \sf  \dashrightarrow \: discriminant =  {b}^{2}  - 4ac

\qquad \sf  \dashrightarrow \: d = (6) {}^{2}  - (4 \times 3 \times 1)

\qquad \sf  \dashrightarrow \: d = 36 - 12

\qquad \sf  \dashrightarrow \: d = 24

\qquad \sf  \dashrightarrow \:  \sqrt {d} = 2 \sqrt{6}

Now, let's calculate it's roots ( x - intercepts )

\qquad \sf  \dashrightarrow \: x =  \cfrac{ - b  \pm \sqrt{d} }{2a}

\qquad \sf  \dashrightarrow \: x =  \cfrac{ - 6\pm 2 \sqrt{6}  }{2 \times 3}

\qquad \sf  \dashrightarrow \: x =  \cfrac{ - 6\pm 2 \sqrt{6}  }{6}

So, the intercepts are :

\qquad \sf  \dashrightarrow \: x =  \cfrac{ - 6 -  2 \sqrt{6}  }{6}

and

\qquad \sf  \dashrightarrow \: x =  \cfrac{ - 6  +  2 \sqrt{6}  }{6}

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Describe how the variability of the distribution changes as the sample size increases. As the sample size increases, the variabi
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Answer:

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X bar will have std deviation as \frac{s}{\sqrt{n} }

where s is the square root of variance of sample

Thus we find the variability denoted by std deviation is inversely proportion of square root of sample size.

Hence as sample size increases, std error decreases.

As the sample size increases, the variability decreases.

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OLga [1]

Answer:

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