Let First Sphere be the Original Sphere
its Radius be : r
We know that Surface Area of the Sphere is : 4π × (radius)²
⇒ Surface Area of the Original Sphere = 4πr²
Given : The Radius of Original Sphere is Doubled
Let the Sphere whose Radius is Doubled be New Sphere
⇒ Surface of the New Sphere = 4π × (2r)² = 4π × 4 × r²
But we know that : 4πr² is the Surface Area of Original Sphere
⇒ Surface of the New Sphere = 4 × Original Sphere
⇒ If the Radius the Sphere is Doubled, the Surface Area would be enlarged by factor : 4
Answer:
78.5 ft²
Step-by-step explanation:
In this situation, the dog's 5 ft leash is the radius of the circle that he forms.
So, the radius is 5.
Use the area formula, A = 
r². Plug in 5 as r, and 3.14 as
A = r²
A = (3.14)(5²)
A = (3.14)(25)
A = 78.5
So, the area of the grass he can cover is 78.5 ft²
ANSWER:
D. The answers will be different because after distributing the 4 in the first equation it will change the values.
Hope it helps u!
Question 1a - You have got this correct, the median mark is 35
Question 1b- To work out the range, you must do the largest value subtract the smallest value. For your data, this would be 57 - 13 = 44
Question 2 - You have drawn the lines in the correct places, but you have not used a ruler. In order to get full marks on a question like this you must use a ruler.
Question 3 - Your lower quartile and median is correct, to find the upper quartile, we do 3(n/4). 'n' is the number of data points - in this case 120.
120/4 = 30
30 x 3 = 90
Go across the graph at 90 to see when the line is hit.
From what I can tell, the Upper Quartile is 3.
Next just find the maximum value.
Again, by the looks of it, the Maximum value is 8.5.
Now you have all the data needed for the box plot.
Min = 0
LQ = 0.8
Med = 2.1
UQ = 3
Max = 8.5
Using this information, draw a box plot like you did on the previous question.
<em>Again, I must stress that any line that you draw (unless purposely curved) must be drawn with a ruler; even on the graph at the top, during you working out. If you do not use a ruler, marks can be lost/taken away.</em>
Question 4a - For the median on a cumulative frequency chart, you must find the halfway point in the data. For this, we do n/2. In this case, n = 40
40/2 = 20
Draw a line (using ruler) across at 20 until you reach the line.
Draw another one down to help you read the number.
The median looks like 34 seconds.
Question 4b - For this question, we already know the Min, Med and Max. Now we must work out the LQ and UQ.
Remember, LQ = n/4
In this case, n = 40
40/4 = 10
Look across at ten and you will find that the LQ = 16 seconds
To work out the UQ, we multiply the LQ place by 3 3(n/4).
3 x 10 = 30
30 on the graph takes us up to 45 seconds.
We now know that the:
Min = 9
LQ = 16
Med = 34
UQ = 45
Max = 57
With this information, draw a box plot like you have in the previous questions.
Question 5 - Good things to always compare on box plots are the Min/Max/Range and the Inter Quartile Range.
The boys had a lowest minimum and a higher maximum, this means that their range is larger, resulting in a large spread of data in comparison to the girls.
The Inter Quartile Range is the difference between the Upper Quartile and the Lower Quartile (how wide the box is).
The boy IQR = 45 - 16 = 29
The girls IQR = 34 - 23 = 11
Again, the girls have much more concise results; even though they did not get the quickest result, they were more like one another.
Question 5a - The median in a box plot is always the line down the centre of the box. In this case:
Median = 24 marks
Question 5b - The IQR is always the UQ - LQ
With this data, the IQR is the following:
36 - 17 = 19 marks
Hope this helps
