Answer:
Hence, option (3) is correct.
Step-by-step explanation:
The probability that Roger wins a tennis tournament (event A) is 0.45 i.e. P(A)=0.45
The probability that Stephan wins the tournament (event B) is 0.40 i.e. P(B)=0.40
The probability of Roger winning the tournament, given that Stephan wins, is 0 i.e. 
The probability of Stephan winning the tournament, given that Roger wins, is 0 i.e. 
We know that 
⇒ P(A∩B)=0
But if A and B are independent events then P(A∩B)=P(A)×P(B)
Hence, P(A|B)=P(A)
Similarly 
Hence, if A and B are independent events then,
P(B|A)=P(B)
But in this situation such a thing is not possible.
Hence, option (3) is correct.
Answer:
She will cycle 85.169 miles in the third week.
(round if needed)
Step-by-step explanation:
56÷100=0.56 (this is 1%)
0.56 x 15 + 56 = 64.4 (you find 15% then add 56 (the original amount) because it's an increase of 15%)
repeat for the other 3 weeks
64.4÷100=0.644
0.644 x 15 + 64.4 = 74.06
74.06÷100=0.7406
0.7406 x 15 + 74.06 = 85.169
Answer:
Deb will have $93.17 in 3 years
Step-by-step explanation:
FV = P(1 + r)^n
FV = 70(1+0.1)^3
FV = 70(1.331)
FV = 93.17
A couple of shapes that have perpendicular lines are right triangles, and rectangles.
Answer:
The solution to the system of equations is y = -5 and x = -2.
Step-by-step explanation:
The question tells us to use substitution to solve the system. This means that the given value for x (in terms of y) should be substituted into the other equation. This is modeled below:
-4y - 5x = 30
-4y - 5(y+3) = 30
Next, we should use the distributive property to simplify the left side of the equation.
-4y -5y - 15 = 30
The next step is to combine like terms on the left side of the equation.
-9y - 15 = 30
Then, we can add 15 to both sides of the equation.
-9y = 45
Finally, we can divide both sides of the equation by -9.
y = -5
To find the value for x, we substitute in the value we just found for y into either of our original equations.
x = y + 3
x = -5 + 3
x = -2
Therefore, the correct answer is y = -5 and x = -2.
Hope this helps!
