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3 years ago

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Mathematics

2 answers:

dangina [55]3 years ago

8 0

Both are correct compare them to geothermal

NikAS [45]3 years ago

3 0

Answer:

Both are correct

Step-by-step explanation:

All the angles are equal, therefore causing the triangles to be similar.

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If ⅔ part of a land cost Rs 45000 find the cost of ⅘ part of land

Vika [28.1K]

Answer:

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2 years ago

The population of a certain species of bird in a region after t years can be modeled by the function P(t)= 1620/1+1.15e^-0.042t

natka813 [3]

Answer:

Step-by-step explanation:

The function is $\frac{1620}{1+1.15e^{-0.042t}}$

To find the maximum population, we need to set t towards infinity to get our answer.

So, we replace time with maximum ( $\infty$ ). Let's check:

$\frac{1620}{1+1.15e^{-0.042t}}\\=\frac{1620}{1+\frac{1.15}{e^{0.042t}}}\\=\frac{1620}{1+\frac{1.15}{e^{0.042(\infty)}}}\\=\frac{1620}{1+\frac{1.15}{\infty}}\\=\frac{1620}{1+0}\\=\frac{1620}{1}\\=1620$

The population of birds approaches 1620 as t goes towards infinity. So we can say the max population of the species is 1620.

Correct answer is A

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3 years ago

Read 2 more answers

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saw5 [17]

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3 years ago

The data given below contain the state cigarette taxâ€‹ (in $) for all 20 regions of a country.

g100num [7]

Answer:

(a) $\bar x= 1.583$ -- Population Mean

(b) $s = 1.032$ --- Population standard deviation

Step-by-step explanation:

Given:

Cigarette tax for 20 regions

Solving (a): The population mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

$\sum x = 1.36 + 1.70 + 2.50 + 0.45 + 1.18 + 0.64 + 3.46 + 0.57 + 2.00 + 0.80 + 1.60 +0.98 + 0.36 + 2.24 + 4.35 + 0.62 + 2.70 + 1.78 + 1.53 + 0.84$

$\sum x = 31.66$

$n = 20$

So, we have:

$\bar x= \frac{31.66}{20}$

$\bar x= 1.583$

Solving (b): The population standard deviation

This is calculated as:

$s = \sqrt{\frac{\sum( x - \bar x)^2}{n}$

$\sum (x -\bar x)^2 = (1.36 - 1.583)^2 + (1.70 - 1.583)^2+ (2.50 - 1.583)^2+ (0.45 - 1.583)^2+ (1.18 - 1.583)^2+ (0.64 - 1.583)^2+ (3.46 - 1.583)^2+ (0.57 - 1.583)^2+ (2.00 - 1.583)^2+ (0.80 - 1.583)^2+ (1.60 - 1.583)^2+(0.98 - 1.583)^2+ (0.36 - 1.583)^2+ (2.24 - 1.583)^2+ (4.35 - 1.583)^2+ (0.62 - 1.583)^2+ (2.70 - 1.583)^2+ (1.78 - 1.583)^2+ (1.53 - 1.583)^2+ (0.84- 1.583)^2$