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3 years ago

4 & 7/8 divided by 1 & 1/6

Mathematics

1 answer:

belka [17]3 years ago

4 0

Answer:

$4 \frac{5}{28}$

Step-by-step explanation:

$4 \frac{7}{8} \div 1 \frac{1}{6}$

Convert the mixed numbers above to improper fractions

$= \frac{39}{8} \div \frac{7}{6}$

Rule of basic operation states that when dividing fractions, the division sign changes to multiplication sign thus turn the right hand side of the fraction inversely such that 7/6 becomes 6/7.

$= \frac{39}{8} \times \frac{6}{7}$

$= \frac{39 \times 6}{8 \times 7}$

$= \frac{234}{56}$

Converting the answer above to mixed number result to:

$= 4 \frac{5}{28}$

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3 years ago

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3 years ago

The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that 77% of all fatally injured

FrozenT [24]

Answer:

The p value obtained was a low value and using the significance level given $\alpha=0.05$ 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 of driver fatalities related to alcohol is less from 0.77 or 77%.

Step-by-step explanation:

1) Data given and notation n

n=53 represent the random sample taken

X=35 represent the automobile driver fatalities in a certain county involved with an intoxicated driver

$\hat p=\frac{35}{53}=0.660$ estimated proportion of automobile driver fatalities in a certain county involved with an intoxicated driver

$p_o=0.77$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the population proportion of driver fatalities related to alcohol is less than 77% or 0.77 in Kit Carson:

Null hypothesis: $p\geq 0.77$

Alternative hypothesis: $p < 0.77$

When we conduct a proportion test we need to use the z statisitc, 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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.660 -0.77}{\sqrt{\frac{0.77(1-0.77)}{53}}}=-1.903$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This methos 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 provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is an unilateral lower test the p value would be:

$p_v =P(z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ 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 of driver fatalities related to alcohol is less from 0.77 or 77%.

8 0

3 years ago