Given
P(1,-3); P'(-3,1)
Q(3,-2);Q'(-2,3)
R(3,-3);R'(-3,3)
S(2,-4);S'(-4,2)
By observing the relationship between P and P', Q and Q',.... we note that
(x,y)->(y,x) which corresponds to a single reflection about the line y=x.
Alternatively, the same result may be obtained by first reflecting about the x-axis, then a positive (clockwise) rotation of 90 degrees, as follows:
Sx(x,y)->(x,-y) [ reflection about x-axis ]
R90(x,y)->(-y,x) [ positive rotation of 90 degrees ]
combined or composite transformation
R90. Sx (x,y)-> R90(x,-y) -> (y,x)
Similarly similar composite transformation may be obtained by a reflection about the y-axis, followed by a rotation of -90 (or 270) degrees, as follows:
Sy(x,y)->(-x,y)
R270(x,y)->(y,-x)
=>
R270.Sy(x,y)->R270(-x,y)->(y,x)
So in summary, three ways have been presented to make the required transformation, two of which are composite transformations (sequence).
A) 0.8
B) 0.24
C) 0.56
D) 0.44
For A, since 1 in 5 is dry, 4 in 5 are not; 4/5 = 0.8
For B, we know that 30%, 0.3, of the ones that are not dry are contaminated:
0.3(0.8) = 0.24
For C, 100%-30%=70% of wells that are not dry are not contaminated; 0.7(0.8) = 0.56
For D, she will either have contaminated water or no water:
0.24+0.2 = 0.44
9 pounds of food at 2.00= 18.00
she has 25.00-18.00= 6.00 to buy treays
6÷1=6
Cara can buy 6 treats
Answer:
The required probability is 0.927
Step-by-step explanation:
Consider the provided information.
Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course.
That means 95% of students didn't enrolled in SAT prep course.
Let P(SAT) represents the enrolled in SAT prep course.
P(SAT)=0.05 and P(not SAT) = 0.95
30% of the SAT prep students were admitted to their first choice college, as were 20% of the other students.
P(F) represents the first choice college.
The probability he didn't take an SAT prep course is:
![P[\text{not SAT} |P(F)]=\dfrac{P(\text{not SAT})\cap P(F) }{P(F)}](https://tex.z-dn.net/?f=P%5B%5Ctext%7Bnot%20SAT%7D%20%7CP%28F%29%5D%3D%5Cdfrac%7BP%28%5Ctext%7Bnot%20SAT%7D%29%5Ccap%20P%28F%29%20%7D%7BP%28F%29%7D)
Substitute the respective values.
![P[\text{not SAT} |P(F)]=\dfrac{0.95\times0.20 }{0.05\times0.30+0.95\times0.20}](https://tex.z-dn.net/?f=P%5B%5Ctext%7Bnot%20SAT%7D%20%7CP%28F%29%5D%3D%5Cdfrac%7B0.95%5Ctimes0.20%20%7D%7B0.05%5Ctimes0.30%2B0.95%5Ctimes0.20%7D)
![P[\text{not SAT} |P(F)]\approx0.927](https://tex.z-dn.net/?f=P%5B%5Ctext%7Bnot%20SAT%7D%20%7CP%28F%29%5D%5Capprox0.927)
Hence, the required probability is 0.927
