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kobusy [5.1K]
3 years ago
6

Michael drove from Orlando to Jacksonville in 2 2/3 hours. The next day he drove from Jacksonville to Savannah in 4 4/7 hours. W

hat was Michael’s total driving time from Orlando to Savannah?
Mathematics
2 answers:
Xelga [282]3 years ago
6 0

1. 2+4=6

2. 2X7=14

3. 4X3=12

4. 7X3=21

5. 12+14=26

6. Simplify 26/21

7. 26-21= 5 so it simplifies to 1 5/21

8.  6+1 5/21=6 5/21

nataly862011 [7]3 years ago
5 0

time traveled from Orlando to Jacksonville = 2 2/3 hours

time traveled from Jacksonville to Savannah = 4 4/7 hours

So, total time traveled will be 2 2/3 + 4 4/7

convert mixed fractions to improper: 2 2/3= 8/3

                                                             4 4/7 =32/7

8/3 + 32/7

lcm = 21

96/21+ 56=+31

= 152/21

mixed fraction = 7 5/21


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Step-by-step explanation:

Data given and notation  

\bar X=136 represent the sample mean

s=11 represent the sample standard deviation

n=24 sample size  

\mu_o =140 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t 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 mean is less than 140, the system of hypothesis would be:  

Null hypothesis:\mu \geq 140  

Alternative hypothesis:\mu < 140  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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".  

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

t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=24-1=23  

Since is a one left tailed test the p value would be:  

p_v =P(t_{(23)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.  

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