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Nezavi [6.7K]
3 years ago
15

The world population in 2013 was about 7.125 billion. In 2003, it was about 6.357 billion. Which number represents the increase

in the world population between 2003 and 2013?
Mathematics
2 answers:
murzikaleks [220]3 years ago
3 0
768 Im pretty sure. subtract 7.125-6.357
nexus9112 [7]3 years ago
3 0

Answer:

The number represent the increase in the world population is 0.768 billion.

Step-by-step explanation:

Given : The world population in 2013 was about 7.125 billion. In 2003, it was about 6.357 billion.

To find : Which number represents the increase in the world population between 2003 and 2013?

Solution :

The world population in 2013 was about 7.125 billion.

The world population in 2003 was about 6.357 billion.

The world population between 2003 and 2013 is

The population of 2013- The population of 2003

=7.125\text{ billion}- 6.357\text{ billion}

=0.768\text{ billion}

The number represent the increase in the world population is 0.768 billion.

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Confidence level is (380.7133, 533.2867)

Step-by-step explanation:

Responses in number of tongue flicks per 20 minutes of lizards, are: 727,217, 268, 438, 625, 319, 200, 591, 574, 727, 693, 336, 302, 761, 268, 353, 370

n = 17

Mean (μ) is:

\mu=\frac{{\Sigma}x}{n} = \frac{727+217+ 268+ 438+ 625+ 319+ 200+ 591+ 574+ 727+ 693+ 336+ 302+ 761+ 268+ 353+ 370}{17} = 439.3529

Standard deviation (σ) is:

\sigma=\sqrt{\frac{\Sigma(x-\mu^2)}{n} } =\sqrt{\frac{(727-457)^2+(217-457)^2+...+(370-457)^2}{17} } =191.2

The confidence interval (c) = 90% = 0.9

\alpha=1-0.9=0.1\\\frac{\alpha }{2} = 0.05\\Z_{\frac{\alpha }{2} }=1.64

Margin of error (e) = =Z_{\frac{\alpha}{2} }*\frac{\sigma}{\sqrt{n} }=1.64*\frac{191.2}{\sqrt{17} }=76.2867

Confidence level = μ ± e = 457 ± 76.2867 = (380.7133, 533.2867)

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<u>Option </u><u>A.</u>

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