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KATRIN_1 [288]
3 years ago
6

Sarah rides her horse for 2 1/6 hrs with a constant speed of 6 km/hr and then for another 2 hours and 5 minutes with a constant

speed of 12 km/hr...
Mathematics
1 answer:
NemiM [27]3 years ago
7 0
The complete question is
<span>Sarah rides her horse for 2 1/6 hrs with a constant speed of 6 km/hr and then for another 2 hours and 5 minutes with a constant speed of 12 km/hr what is her average speed for the total trip?</span>

we know that
speed=distance/time

1) Sarah rides her horse for 2 1/6 hrs with a constant speed of 6 km/hr
time=2 1/6 hrs----> 13/6 hrs
speed=6 km/h
distance=speed*time----> 6*(13/6)---> 13 km

2)for another 2 hours and 5 minutes with a constant speed of 12 km/hr
time=2hrs+5 minutes----> 2+(5/60)--> 125/60 hrs
speed=12 km/hrs
distance=speed*time----> 12*(125/60)---> 25 km

3) find the average speed for the total trip
total distance=13 km+25 km----> 38 km
total time=(13/6)+(125/60)----> (10*13+125)/60----> 4.25 hours
speed=38/4.25----> 8.94 km/hrs

the answer is
8.94 km/hrs
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The surface area of a cube is 1,014^2. Find the volume of the cube.
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HELP ME <br> What is the expression in simplest form? 21 - 6n + 2(2n - 5 ) - 3n
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The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
Kruka [31]

Answer:

6.76-2.01\frac{2.55}{\sqrt{50}}=6.03    

6.76+2.01\frac{2.55}{\sqrt{50}}=7.49    

The 95% confidence interval would be given by (6.03;7.49)    

Step-by-step explanation:

Notation

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n=50 represent the sample size  

Solution

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

We can calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=6.76

The sample deviation calculated s=2.55

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=50-1=49

We assume a standard confidence level of 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that t_{\alpha/2}=2.01

Now we have everything in order to replace into formula (1):

6.76-2.01\frac{2.55}{\sqrt{50}}=6.03    

6.76+2.01\frac{2.55}{\sqrt{50}}=7.49    

The 95% confidence interval would be given by (6.03;7.49)    

3 0
3 years ago
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