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Sholpan [36]
3 years ago
11

In a race, 47 out of the 50 swimmers finished in less than 35 minutes. What percentage of the swimmers finished the race less th

an 35 minutes?
47*_____=_______
50*_____=_______=______%
Mathematics
2 answers:
Dmitry [639]3 years ago
6 0

Answer:

94%

47*2=94

50*2=100

Step-by-step explanation:

Yuri [45]3 years ago
3 0

Answer: 94% of the swimmers finished the race in less than 35 minutes.

Step-by-step explanation:

There are several ways to go about finding the answer. The best way to solve this, which can be easily applied to other percentage problems is dividing the number of swimmers who finished in this time (47) by the total number of swimmers (50) (Using a calculator).

47/50 = 0.94

To change a decimal to a fraction, move the decimal place over two spaces to the right. Therefore, 0.94 becomes 94%.

Way 2:

This only works because there the number of swimmers are a multiple of 100.

50 is 1/2 of 100. To get the full 100, you can multiply both the numerator and denominator by 2.

Multiply 47 by 2 to get 94 and 50 by 2 to get 100. Thus, 94/100 swimmers would have finished in under 35 minutes. 94/100 = 94%

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Step-by-step explanation:

When you start raising i to certain powers, you begin to notice a pattern.

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Integrate this two questions ​
tankabanditka [31]

Simplify the integrands by polynomial division.

\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)

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Now computing the integrals is trivial.

5.

\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}

where we use the power rule,

\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)

and a substitution to integrate the last term,

\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)

8.

\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}

using the same approach as above.

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2 years ago
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