1 Write 60% as a simplified ratio.

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

3 0

Answer:

3:10

Step-by-step explanation:

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Mark used a number line to model dividing 2 3 by 3. He divided the number line from 0 to 1 in 3 equal sections, and then divided

Oxana [17]

Answer:

No, his answer is not correct.

Step-by-step explanation:

Given expression,

$\frac{2}{3}\div 3$

$=\frac{\frac{2}{3}}{3}$

$=\frac{2}{3\times 3}$

$=\frac{2}{9}$

Thus, his answer is not correct.

By model :

Draw number line from 0 to 1 in 3 equal section,

Then divided each of those sections into 3 equal parts,

We get,

$\frac{2}{3}\div 3=\frac{2}{9}$

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3 years ago

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A company offered one half of its employees a bonus if the production of light bulbs increased by 20%. The other half of the emp

Sauron [17]

The correct order to determine if the results are significant are as follows:
1. Randomly separate the employee's individual results into two groups.
2.Calculate the mean of each group.
3. Calculate the difference of the means
4. Run the experiment many times
5. Calculate the probability of a difference of 5.
These steps are used to determine statistical significance by finding difference of means.

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3 years ago

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he American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 45 w

Sonja [21]

Answer:

The 85% confidence interval is (0.2187, 0.2613).

Step-by-step explanation:

We have a large sample of size n = 830 Americans over 45. We have observed that 631 people don't smoke. Let p be the population proportion of Americans over 45 who smoke. An point estimate of p is $\hat{p} = 199/830 = 0.24$ and an estimate of the standard deviation of $\hat{p}$ is $\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{(0.24)(0.76)/830}$ = 0.0148. Therefore, because we have a large sample, the 85% confidence interval for the population proportion is given by $\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n} = 0.24\pm z_{0.15/2}0.0148 = 0.24\pm z_{0.075}0.0148$ where $z_{0.075}$ is the 7.5th quantile of the standard normal distribution, i.e., -1.4395. Therefore the 85% confidence interval is $0.24\pm (1.4395)(0.0148)$ or equivalently (0.2187, 0.2613).