Answer:
1. Yes
∆RST ~ ∆WSX
by SAS
2. Yes
∆ABC ~ ∆PQR
by SSS
3. Yes
∆STU ~ ∆JPM
by SAS
4. Yes
∆DJK ~ ∆PZR
by SAS
5. Yes
∆RTU ~ ∆STL
by SAS
5. Yes
∆JKL ~ ∆XYW
by SAS
6. No
7. Yes
∆BEF ~ ∆NML
by SAS
8. Yes
∆GHI ~ ∆QRS
by SSS
9. x=22
10. x=12
Step-by-step explanation:
1. RS/WS=ST/SX and m<RST=m<WSX
2. AB/PQ=8/6=4/3
BC/QR=AC/PR=12/9=4/3
AB/PQ=BC/QR=AC/PR
3. ST/JP=10/15=2/3
SU/JM=14/21=2/3
ST/JP=2/3=SU/JM
and m<TSU=70°=m<PJM
4. DK/PR=8/4=2
JK/ZR=18/9=2
DK/PR=2=JK/ZR
and m<DKJ=65°=m<PRZ
5. RT/ST=UT/LT
and m<RTU=m<STL
6. KL/YW=20/18=10/9
JL/XW=36/24=3/2
KL/YW=10/9≠3/2=JL/XW
7. BF/NL=24/16=3/2
BE/NM=39/26=3/2
BF/NL=3/2=BE/NM
and m<EBF=m<MNL
8. GH/QR=32/20=8/5
HI/RS=40/25=8/5
GI/QS=24/15=8/5
GH/QR=HI/RS=GI/QS=8/5
9. x/33=18/27
Simplifying the fraction on the right side of the equation:
x/33=2/3
Solving for x: Multiplying both sides of the equation by 33:
33(x/33)=33(2/3)
x=11(2)
x=22
10. x/16=9/12
Simplifying the fraction on the right side of the equation:
x/16=3/4
Solving for x: Multiplying both sides of the equation by 16:
16(x/16)=16(3/4)
x=4(3)
x=12
<u>Answer:</u>
The correct answer option is P (S∩LC) = 0.16.
<u>Step-by-step explanation:</u>
It is known that the probability if someone is a smoker is P(S)=0.29 and the probability that someone has lung cancer, given that they are also smoker is P(LC|S)=0.552.
So using the above information, we are to find the probability hat a random person is a smoker and has lung cancer P(S∩LC).
P (LC|S) = P (S∩LC) / P (S)
Substituting the given values to get:
0.552 = P(S∩LC) / 0.29
P (S∩LC) = 0.552 × 0.29 = 0.16
Answer:
T(2, -7) = (4, -10)
Step-by-step explanation:
T(x, y) = (x + 2, y - 3)
T(2, -7) = (2 + 2, -7 - 3)
T(2, -7) = (4, -10)
Answer:
34 I am pretty sure
Step-by-step explanation:
34+55 well is 89 so yeah d=34
(-8, -5)
To form a rectangle, all the lines must be perpendicular and parallel. If you draw out the points on a coordinate plane then you can find the answer pretty easily.
Also, this one is easy because all the lines are vertical or horizontal, and when that is the case then you can just see which number in the set hasn't appeared twice (1 appeared twice as an <em>x</em> value, 0 appeared twice as a <em>y </em>value, but -8 has only appeared once as an <em>x</em> and -5 has only appeared once as a <em>y </em>value)
