Kari would like to save $10,000 for a down payment on a house illustrate the difference in years it will take her to double her
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Answer:
For school A: Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17, IQR=9.5
For school B: Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19, IQR=7.5
No, the box plots are not symmetric.
Step-by-step explanation:
Part A
The given data sets are
School A : 9,14,15,17,17,7,15,6,6
School B : 12,8,13,11,19,15,16,5,8
Arrange the data in ascending order.
School A : 6,6,7,9,14,15,15,17,17
School B : 5,8,8,11,12,13,15,16,19
Divide each data set in four equal parts.
School A : (6,6),(7,9),14,(15,15),(17,17)
School B : (5,8),(8,11),12,(13,15),(16,19)
For school A:
Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17
Interquartile range of the data is
For school B:
Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19
Interquartile range of the data is
Part B:
The box plots are not symmetric because the data values are different. Five number summary and IQR of both the data set are different.
Answer:
The correct option is;
A 1
Step-by-step explanation:
The given rectangle TUVW has coordinates T(a, b), U (a + 2, b + 2), V ( a + 5, b - 1) and W (a + 3, b - 3).
Given that the coordinates of T and U are T (a, b), and U (a + 2, b + 2), we have;
The slope, m of the line TU is is found by the following relation;
Where we put (x₁, y₁) = (a, b) and (x₂, y₂) = (a + 2, b + 2), we have;
m = ((b + 2) - b)/((a + 2) - a) = 2/2 = 1
Whereby the slope of a line parallel to another line are equal, we have that the slope of the line parallel to the line that contains the side TU is also 1.
Given:
It is given that surface area must be less than 150 cm².
Solution:
The Maximum Volume With Total Surface Area Less than 150 cm² is shown in the table.
From the table, it can be concluded that for r=3.00 cm and h=4.95 cm the surface area will be less than 150 cm² and the volume will be the maximum.
Calculate the volume.
Hence, the required dimensions are r=3.00 cm and h=4.95 cm.
Answer:
The line containing the midpoints of two sides of a triangle is parallel to the third side ⇒ proved down
Step-by-step explanation:
* Lets revise the rules of the midpoint and the slope to prove the
problem
- The slope of a line whose endpoints are (x1 , y1) and (x2 , y2) is
- The mid-point of a line whose endpoints are (x1 , y1) and (x2 , y2) is
* Lets solve the problem
- PQR is a triangle of vertices P (0 , 0) , Q (2a , 0) , R (2b , 2c)
- Lets find the mid-poits of PQ called A
∵ Point P is (x1 , y1) and point Q is (x2 , y2)
∴ x1 = 0 , x2 = 2a and y1 = 0 , y2 = 0
∵ A is the mid-point of PQ
∴
- Lets find the mid-poits of PR which called B
∵ Point P is (x1 , y1) and point R is (x2 , y2)
∴ x1 = 0 , x2 = 2b and y1 = 0 , y2 = 2c
∵ B is the mid-point of PR
∴
- The parallel line have equal slopes, so lets find the slopes of AB and
QR to prove that they have same slopes then they are parallel
# Slope of AB
∵ Point A is (x1 , y1) and point B is (x2 , y2)
∵ Point A = (a , 0) and point B = (b , c)
∴ x1 = a , x2 = b and y1 = 0 and y2 = c
∴ The slope of AB is
# Slope of QR
∵ Point Q is (x1 , y1) and point R is (x2 , y2)
∵ Point Q = (2a , 0) and point R = (2b , 2c)
∴ x1 = 2a , x2 = 2b and y1 = 0 and y2 = 2c
∴ The slope of AB is
∵ The slopes of AB and QR are equal
∴ AB // QR
∵ AB is the line containing the midpoints of PQ and PR of Δ PQR
∵ QR is the third side of the triangle
∴ The line containing the midpoints of two sides of a triangle is parallel
to the third side