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3 years ago

PLEASE help me solve this problem

Mathematics

1 answer:

irakobra [83]3 years ago

8 0

Answer:

that looks hard?

Step-by-step explanation:

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An online company wants to establish ratings for the products they provide. To do so, they ask visitors to their website to eval

Lemur [1.5K]

Answer:

This is a group specific influence sample, it only shows a demographic sample of those that are influenced by the reviews in the first place.

Step-by-step explanation:

8 0

3 years ago

if a snail travels 200 inches in 2 hours how long will it take the snail to travel 50 inches ( i need the equation too )

Bogdan [553]

200 inches in 2 hours
200÷2= 100 inches in 1 hour
100 inches per hour divided by 50 inches = 1/2 hour

the snail will travel 50 inches in 1/2 hour

8 0

4 years ago

Read 2 more answers

What is the sum of 10000+101010

vredina [299]

Answer:

111010

Step-by-step explanation:

it is the correct answer

6 0

2 years ago

Read 2 more answers

Jamie invests $1,200 in an account that earns 3% interest. Assuming continuous compounding, how much is in his account after 7 y

goblinko [34]

Answer:

The amount in account after 7 years of investment is $1478.4

Step-by-step explanation:

Given as :

The principal invested in account by Jamie = p = $1200

The rate of interest = r = 3% compounded annually

The Time period of investment = t= 7 years

Let The Amount in Jamie account = $ A

<u>For continuous compounding</u>

Amount = Principal × $e^{r \times time}$

Or, A = p × $e^{r \times time}$

Or, A = $1200 × $e^{0.03 \times 7}$

Or, A = $1200 × $2.71^{0.03 \times 7}$

Or, A = $1200 × 1.232

∴ A = $1478.4

So, The amount after continuous compounding = A = $1478.4

Hence , The amount in account after 7 years of investment is $1478.4 Answer

5 0

4 years ago

Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con

monitta

Let $f(x)=e^{-1/x}$ . Then $f'(x)=\frac1{x^2}e^{-1/x}>0$ for all $x\ge2$ , so $f$ is strictly increasing. As $x\to\infty$ , $e^{-1/x}\to e^0=1$ , so $f$ is bounded above by 1. This is to say,

$e^{-1/x}$

and the integral of $\frac1{x^2}$ converges over the same domain, so this integral must also converge by comparison.

We have, by setting $y=-\frac1x$ ,

$\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}$

8 0

3 years ago