Answer: False
Reason: It should be x-a instead of x+a
Answer:
- area of rectangle: 48 cm²
- area of triangle: 18 cm²
- shaded area: 30 cm²
Step-by-step explanation:
The area of a rectangle is given by the product of its length and width.
... A = LW
For your rectangle, the overall area is ...
... A = (12 cm)(4 cm) = 48 cm² . . . . rectangle area
___
The area of a triangle is half the product of its base and height. Your triangle is shown with a base of 9 cm, and its height is the height of the rectangle, 4 cm. The triangle area is ...
... A = (1/2)bh
... A = (1/2)(9 cm)(4 cm) = 18 cm² . . . . triangle area
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The shaded area of the figure is the difference between the rectangle area and the triangle area. (The triangle is fully included in the rectangle, but is not shaded.) That difference is ...
... shaded area = rectangle area - triangle area
... shaded area = 48 cm² - 18 cm² = 30 cm² . . . . shaded area
The question is incomplete. The complete question is :
Victor is selling lemonade to raise money for his school. The amount of money he collected from his sales each day is summarized on the box plot below.
What is the interquartile range of the data ?
Answer:
$ 7
Step-by-step explanation:
It is given that Victor sold lemonade in order to raise money for his school.
The money that Victor collected is shown on the diagram below.
Therefore, we find the interquartile range. It is the difference between the 3rd quartile and the 1st quartile, i.e.
IQR = 
Now from the figure, the 3rd quartile is the value that is at the end of the rectangular box towards the right. Therefore, the 3rd quartile is = $ 29
Similarly from the figure the 1st quartile at the beginning of the rectangular box to the left is = $ 22
So the interquartile range is = 3rd quartile - 1st quartile
= 29 - 22
= $ 7
Thus the amount of money Victor collected is $ 7.
Answer:
C. The typical value is greater in set . The spread is greater in set A.
Step-by-step explanation:
Set A data is all over the place meanwhile set b data is clumped in one spot. This means Set A data has its data more spread. Then the value is greater in set B since the arrow is on 8.5 while in set A it is on 4.