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alina1380 [7]
3 years ago
11

The distance between Jamie's house and her friend's house is 75 meters. If Jamie walks to her friend's house and back,how many c

enmeters does she walk? Explain. (ONLY IF YOU KNOW)
Mathematics
2 answers:
Ratling [72]3 years ago
8 0
75meters*2(way there and way back)=150mt
150meters= 15'000centimeters
Grace [21]3 years ago
5 0
So when we look at the question, it wants the answer to be in cm. So we should firstly convert m to cm

In 1m there are 100cm
So our conversion is 1m=100cm

So if we have 75m then we multiply it by 100
75×100= 7500cm

So from her house to her friends house is 7,500cm
She travels to her friend's house ANS back, So the distance will be times 2

So 7,500×2= 15,000cm

She walked 15,000cm total
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The number of motor vehicles registered in millions in the US has grown as follows:
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Answer:

a) Figure attached

b) r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}  

For our case we have this:

n=10 \sum x = 75.819, \sum y = 43.5231, \sum xy = 330.0321, \sum x^2 =574.8598, \sum y^2 =192.8274  

r=\frac{10(330.0321)-(75.81948)(43.5231)}{\sqrt{[10(574.8598) -(75.819)^2][10(192.8274) -(43.5231)^2]}}=0.989  

So then the correlation coefficient would be r =0.989

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

Solution to the problem

Part a

Year (x): 1940, 1945 1950, 1955, 1960, 1965, 1970, 1975, 1980, 1985

Vehicles (Y): 32.4, 31.0, 49.2, 62.7, 73.9, 90.4, 108.4, 132.9, 155.8, 171.7

After apply natural log for the two variables and create the scatterplot in excel we got the following result on the figure attached.

Part b

And in order to calculate the correlation coefficient we can use this formula:  

r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}  

For our case we have this:

n=10 \sum x = 75.819, \sum y = 43.5231, \sum xy = 330.0321, \sum x^2 =574.8598, \sum y^2 =192.8274  

r=\frac{10(330.0321)-(75.81948)(43.5231)}{\sqrt{[10(574.8598) -(75.81948)^2][10(192.8274) -(43.5231)^2]}}=0.989  

So then the correlation coefficient would be r =0.989

4 0
3 years ago
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