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3 years ago

How do you turn a fraction into a percentage

Mathematics

2 answers:

alexira [117]3 years ago

7 0

Answer:

you turn a fraction into a percentage by dividing the numerator by the denominator. then multiply the decimal by 100.

Step-by-step explanation:

RoseWind [281]3 years ago

6 0

Convert a fraction to a percent, first divide the numerator by the denominator. Then multiply the decimal by 100 .

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Lisa walked 8 km more than Tim. Lisa walked twice as far as Tim. How far did each walk?

valentinak56 [21]

$x-\ Tim's\ distance\\\\
x+8-\ Lisa's\ distance\\\\
2x=x+8\ \ |Subtract\ x\\
2x-x=8\\
x=8\ \ -Tim\\
x+8=8+8=16- \ Lisa\\\\Lisa\ walked\ 16\ km\ and\ Tim\ 8km.
$

6 0

3 years ago

mustafa is putting a sidewalk using two diffrent style bricks, one style brick is 8 inches long, and the other style brick is 6

Svetlanka [38]

Answer:

8y+6y=288

Step-by-step explanation:

8y+6y= 12y

12y=288

12y divide by 12 = 288 divided by 12 =

y=24

3 0

2 years ago

In Spring 2017, data was collected from a random selection of STA 2023 students. One of the questions asked how many hours they

Arte-miy333 [17]

Answer: 0.22

Step-by-step explanation:

We know that the best point estimate for the difference between two population mean is the difference between their sample means.

Given : For the 39 randomly selected upperclassmen, the sample mean was 0.12 and sample standard deviation was 0.42. For the 35 randomly selected underclassmen, the sample mean was 0.34 and the sample standard deviation was 0.87.

Let A denotes the population of upperclassmen and B denotes the population of underclassmen .

$\overline{x}_A=0.12$

$\overline{x}_B=0.34$

Then, the point estimate of the difference in the population mean volunteered between underclassmen and upperclassmen will be :-

$\overline{x}_B-\overline{x}_A=0.34-0.12=0.22$

Hence, the point estimate of the difference in the population mean volunteered between underclassmen and upperclassmen =0.22

5 0

3 years ago

Help!!!!<br>Given log7⁡3≈0.5646 and log7⁡16≈1.4248, evaluate the expressions.

Alinara [238K]

Answer:

a) 0.356

b) 1.1397

Step-by-step explanation:

a) log₇2

log(2) / log(7)
0.356

b) log₇ (¹⁴⁷/₁₆)

<u>log (¹⁴⁷/₁₆)</u>

log (7)

1.1397

6 0

2 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0

3 years ago