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lys-0071 [83]
4 years ago
14

What is the percent of 1.6 hours to 0.95 hour

Mathematics
2 answers:
Stolb23 [73]4 years ago
7 0
1.6*100/0.95=168%     1.6 to 0.95
0.95*100/1.6=59%       0.95 to 1.6
Akimi4 [234]4 years ago
4 0

Answer:

168.4%  of 1.6 hours to 0.95 hour.

Step-by-step explanation:

Given : 1.6 hours to 0.95 hour.

To find : what is the percent.

Solution : We have given

1.6 hours to 0.95 hour.

Percentage  = \frac{1.6}{0.95}*100.xzs

Percentage  = 1.684 * 100

Percentage=  168.4%

Therefore, 168.4%  of 1.6 hours to 0.95 hour.

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(4/5x - 5) - (1/3x + 4)

remove the parentheses

4/5x - 5 - 1/3x - 4

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z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98    

p_v =P(Z  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=4.6 represent the sample mean

\sigma=0.9 represent the population standard deviation

n=180 sample size  

\mu_o =4.8 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if thetrue mean is below the specifications, the system of hypothesis would be:  

Null hypothesis:\mu \geq 4.8  

Alternative hypothesis:\mu < 4.8  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98    

P-value

Since is a one sided test the p value would be:  

p_v =P(Z  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.  

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Step-by-step explanation:

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