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3 years ago

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A countrys population in 1992 was 222 million. In 2001 it was 224 Million. Estimate the population growth in 2004 using the expo

nential growth formula. Round your answer to the nearest million.

1 answer:

sergejj [24]3 years ago

6 0

Find the rate of growth
224=222(1+r)^(2001-1992)
Solve for r
R=0.000997 without %
P=224(1.000997)^(2004-2001)
P=224×(1.000997)^(3)
P=224.67 million

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Find the formula for the area of the shaded region in the figure.<br> x² = 4py<br> ہے

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The formula for the area of the shaded region in the figure.

x² = 4py is mathematically given as

$A=\frac{8}{3} \sqrt{p h} \cdot h$

<h3>What is the formula for the area of the shaded region in the figure?</h3>

Parameters

$x^{2}=4 p y$

$y=h$

$\therefore x^{2}=4 p h \Rightarrow x=2 \sqrt{p h} \\$

Generally, the equation for the Area is mathematically given as

$& A=2 \int_{0}^{2 \sqrt{p h}}\left(h-\frac{x^{2}}{4 p}\right) d x \\\\& A=2\left(h x-\frac{x^{3}}{12 p}\right]_{0}^{2 \sqrt{p h}} \\\\& A=2\left[\left[h 2 \sqrt{p h}-\frac{1}{12 p} \cdot(2 \sqrt{p h})^{3}-0\right]\right. \\$

$&A=2\left[2 h^{3 / 2} \sqrt{p}-\frac{8(\sqrt{p})^{3}(\sqrt{h})^{3}}{12 p}-0\right] \\\\&A=2\left[2 h^{k / 2} \sqrt{p}-\frac{2}{3} \sqrt{p} \cdot h^{3 / 2}\right] \\\\&A=2 \times \frac{4 \sqrt{p} \cdot h^{3 / 2}}{3} \\\\&A=\frac{8}{3} \cdot \sqrt{p} \cdot \sqrt{h} \cdot h \\\\&A=\frac{8}{3} \sqrt{p h} \cdot h$

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In conclusion, the equation for the Area is

$A=\frac{8}{3} \sqrt{p h} \cdot h$

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According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones

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Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

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This is the pvalue of Z when X = 160. So

$Z = \frac{X - \mu}{\sigma}$

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It is to change different ways in order of the addends the result would be the same.

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