A random sample of 5 paintings was studied in a museum. Please consult the data chart above to answer the following question. Wh
ich of the columns represent categorical data? A. Columns A and D O B. Columns B and C O C. Columns A and B D. Columns C and D
1 answer:
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Answer:
Hence, the surface area of cone is:
282.6 cm^2 which is approx 283 cm^2.
Step-by-step explanation:
We are given:
- The circular base of a cone has a radius of 5 centimeters.
i.e. r=5 cm.
- The height of the cone is 12 centimeters
i.e. h=12 cm.
- The slant height is 13 centimeters.
i.e. l=13 cm.
Now we know that the surface area(S.A.) of cone is given by:
Hence by putting the value of l,r and π in the formula we get:
Hence the surface area of cone is 282.6 cm^2 which is approx 283 cm^2.
Answer:
The mean for this case would increase since is defined as:
The interquartile range would not change since the definition for the IQR is and the quartiles are the same.
The standard deviation would not remain the same since by definition is:
And since we change the largest value the deviation would increase considerably.
And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.
So the best option for this case is:
Mean would increase.
Step-by-step explanation:
For this case we assume that we have a random sample given and for each observation
since the problem states that the data is approximately normal.
Let's assume that the largest value on this sample is and for this case we are going to replace this value by another one extremely higher so we satisfy this condition:
The mean for this case would increase since is defined as:
The interquartile range would not change since the definition for the IQR is and the quartiles are the same.
The standard deviation would not remain the same since by definition is:
And since we change the largest value the deviation would increase considerably.
And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.
So the best option for this case is:
Mean would increase.