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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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NNADVOKAT [17]

Answer:

72.7

Step-by-step explanation:

First multiply 72 by 10

because 72 is 72 10s, making it a whole number.

then add 720 (72 times 10) to 7.

then all of that over 10.

then just divide.

727/10

72.7

this is what you do for all mixed numbers like that one.

7 0
2 years ago
Read 2 more answers
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
2 years ago
140 equals the sum of k and 38
Mandarinka [93]

Answer:

102

Step-by-step explanation:

140-38=102

5 0
2 years ago
Round each decimal to the nearest whole number 30.92
Serhud [2]
It would be 31. 31 is a whole number. since 0.91 is close to 1. We would round it to 31 because of that.
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3 years ago
melinda will take 15 quizzes this semester. she would like to score a b are better on at least 90% of them. so far. she gotten a
dimaraw [331]

Answer:

i hope this helps, i got  x + 5 ≤ 13 as ur equation

Step-by-step explanation:

7 0
2 years ago
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