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DENIUS [597]
3 years ago
15

6wholes and 1/4 ÷ 5wholes

Mathematics
1 answer:
Irina-Kira [14]3 years ago
4 0
Answer: 1 1/4      (1 whole and 1/4)

6 1/4 = 25/4
5 = 5/1

(25/4)/(5/1) = (25/4)*(1/5)
(25/4)*(1/5)
25/20 
1 5/20 = 1 1/4
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An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such tha
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Answer:

t=\frac{6.3-6.0}{\frac{1}{\sqrt{26}}}=1.530  

df = n-1 = 26-1=25

Since is a right-sided test the p value would be:  

p_v =P(t_{25}>1.530)=0.0693  

If we compare the p value and the significance level given \alpha=0.025 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is significantly higher than 6.0

Step-by-step explanation:

Data given and notation  

\bar X=6.3 represent the sample mean    

s=1.0 represent the sample standard deviation

n=26 sample size  

\mu_o =6.0 represent the value that we want to test  

\alpha=0.025 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 the ture mena is higher than 6.0, the system of hypothesis would be:  

Null hypothesis:\mu \leq 6.0  

Alternative hypothesis:\mu > 6.0  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-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:  

t=\frac{6.3-6.0}{\frac{1}{\sqrt{26}}}=1.530  

P-value  

The degrees of freedom are given by:

df = n-1 = 26-1=25

Since is a right-sided test the p value would be:  

p_v =P(t_{25}>1.530)=0.0693  

Conclusion  

If we compare the p value and the significance level given \alpha=0.025 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is significantly higher than 6.0

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