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fredd [130]
3 years ago
11

A block of wood has a mass of 180 grams. It is 10cm long 6cm wide and 4cm thick. What is its volume and density?

Mathematics
1 answer:
Mkey [24]3 years ago
8 0

Volume = (length) x (width) x (height) =

                 (10cm) x  (6cm)  x  (4cm)  =  240 cm³

Density = (mass) / (volume) = (180 g) / (240 cm³)  =  0.75 gm/cm³ .
 
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Pleaseee helP
cupoosta [38]

\sf \: P(A|B) = P(drive \: to \: school|sophmore)

Where,

  • A is the number of sophomores who drive to school
  • B is the total number of sophomores

In our case,

  • A = 2
  • B = 2+25+3 = 30

Let's solve ! ~

\tt \: P(A|B) = P(2|30)

\tt \: P(2|30) =  \frac{2}{30}

\tt \: P =  \frac{1}{15}

\tt \: P = 0.066

\tt \: P = 1

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Read 2 more answers
What is the output value for the following function if the input value is 1?
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3 years ago
The number of motor vehicles registered in millions in the US has grown as follows:
zysi [14]

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

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