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3 years ago

Convert .19 into a fraction

Mathematics

2 answers:

STatiana [176]3 years ago

8 0

Here is the decimal:

$0.19 \rightarrow .19$

To convert it into a fraction, make the denominator out of 100, and move the decimal point 2 places to the right and make it the numerator.

$0.19 \rightarrow 19$

Your fraction is:

$\frac{19}{100}$

<h3>CHECK</h3>

To see if the fraction is correct, divide the numerator by the denominator. If you get 0.9, then the fraction is correct.

$19 \div 100 = 0.19$ ✔

Your answer is $\frac{19}{100}$

oee [108]3 years ago

7 0

To write .19 as a fraction, we can multiply both the numerator and denominator to remove the two decimal places.

$\frac{0.19}{1}$ × $\frac{100}{100}$ = $\frac{19}{100}$

Therefore, 0.19 is equivalent to $\frac{19}{100}$ .

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Answer:

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.49$

$p_v =P(z>2.49)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants request financial aid is significantly higher than 0.35

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

$\hat p=\frac{67}{150}=0.447$ estimated proportion of applicants request financial aid

$p_o=0.35$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants request financial aid is higher than 0.35:

Null hypothesis: $p \leq 0.35$

Alternative hypothesis: $p > 0.35$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.49$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2.49)=0.0064$

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