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tatiyna
4 years ago
9

How does a percent relate to a proportion

Mathematics
1 answer:
MaRussiya [10]4 years ago
7 0
Percent means per hundred, which means, when you are using a percent, you are actually using a ratio to compare  the number of your percent to one hundred.
For example, suppose we have 5%. The percent symbol (%) indicates that we are using a ratio to compare 5 to 100, and remember that we use fractions to express ratios, so we can express 5% as the ratio \frac{5}{100}. Notice that we can simplify our ratio, both numerator and denominator are divisible by 5, so lets simplify it: \frac{5}{100} = \frac{1}{20}. We can conclude that we can express 5% as the ratio \frac{1}{20}.

Now that we know we can express a percent as ratio, lets show how the percent is related to a proportion:
Suppose that in a class of 30 students are 12 girls, and you want to find <span>how many percent of girls are in your class using a proportion:
Let </span>x% be the percent girls. Since we know that percents can be expressed as ratios, x%=\frac{x}{100}: Now we can use the fact that the students are in a ratio 12 to 30 to establish our proportion:
\frac{12}{30} = \frac{x}{100}
x= \frac{(12)(100)}{30}
x=40
We just prove that 40% of the students are girls.

The proportion that we used in this problem is called a percent proportion, and its formula: \frac{a}{b} = \frac{x}{100} can be use to relate a percent with a proportion.
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The resting heart rate for an adult horse should average about µ = 47 beats per minute with a (95% of data) range from 19 to 75
KatRina [158]

Answer:

a. 0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. 0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. 0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean:

\mu = 47

(95% of data) range from 19 to 75 beats per minute.

This means that between 19 and 75, by the Empirical Rule, there are 4 standard deviations. So

4\sigma = 75 - 19

4\sigma = 56

\sigma = \frac{56}{4} = 14

a. What is the probability that the heart rate is less than 25 beats per minute?

This is the p-value of Z when X = 25. So

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

Z = \frac{25 - 47}{14}

Z = -1.57

Z = -1.57 has a p-value of 0.0582.

0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. What is the probability that the heart rate is greater than 60 beats per minute?

This is 1 subtracted by the p-value of Z when X = 60. So

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

Z = \frac{60 - 47}{14}

Z = 0.93

Z = 0.93 has a p-value of 0.8238.

1 - 0.8238 = 0.1762

0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. What is the probability that the heart rate is between 25 and 60 beats per minute?

This is the p-value of Z when X = 60 subtracted by the p-value of Z when X = 25. From the previous two items, we have these two p-values. So

0.8238 - 0.0582 = 0.7656

0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

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A Party Store sells large plates in packs of 12 and small plates in packs of 8. In order to have an equal number of both, what i
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Answer:

8 12 packs and 12 8 packs.

Step-by-step explanation:

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At time t ≥ 0, the velocity of a body moving along the s-axis is v = t² -5t +4. When is the body moving backwards
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Answer:

A

Step-by-step explanation:

The velocity of a moving body is given by the equation:

v=t^2-5t+4 ,\, t\geq0

Is the velocity is <em>positive </em>(v>0), then our object will be moving <em>forwards</em>.

And if the velocity is negative (v<0), then our object will be moving <em>backwards</em>.

We want to find between which interval(s) is the object moving backwards. Hence, the second condition. Therefore:

v

By substitution:

t^2-5t+4

Solve. To do so, we can first solve for <em>t</em> and then test values. By factoring:

(t-4)(t-1)=0

Zero Product Property:

t=1, \text{ and } t=4

Now, by testing values for t<1, 1<t<4, and t>4, we see that:

v(0)=4>0,\, v(2)=-20

So, the (only) interval for which <em>v</em> is <0 is the second interval: 1<t<4.

Hence, our answer is A.

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