F= -3/5 x+2. Hope this helps
Answer:
x = 12
m(QS) = 52°
m(PD) = 152°
Step-by-step explanation:
Recall: Angle formed by two secants outside a circle = ½(the difference of the intercepted arcs)
Thus:
m<R = ½[m(PD) - m(QS)]
50° = ½[(12x + 8) - (4x + 4)] => substitution
Solve for x
Multiply both sides by 2
2*50 = (12x + 8) - (4x + 4)
100 = (12x + 8) - (4x + 4)
100 = 12x + 8 - 4x - 4 (distributive property)
Add like terms
100 = 8x + 4
100 - 4 = 8x
96 = 8x
96/8 = x
12 = x
x = 12
✔️m(QS) = 4x + 4 = 4(12) + 4 = 52°
✔️m(PD) = 12x + 8 = 12(12) + 8 = 152°
Answer:
y = 18.25
Step-by-step explanation:
The 2-point form of the equation of a line can be used to write the equation of the line through the given points.
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
y = (-2 -7)/(3 -(-1))(x -(-1)) +7
y = -9/4(x +1) +7
For x = -6 this evaluates to ...
y = -9/4(-6 +1) +7 = (9/4)(5) +7 = 45/4 +7
y = 18 1/4
Answer:
probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3
Step-by-step explanation:
First of all;
Let B1 be the event that the card with two red sides is selected
Let B2 be the event that the
card with two black sides is selected
Let B3 be the event that the card with one red side and one black side is
selected
Let A be the event that the upper side of the selected card (when put down on the ground)
is red.
Now, from the question;
P(B3) = ⅓
P(A|B3) = ½
P(B1) = ⅓
P(A|B1) = 1
P(B2) = ⅓
P(A|B2)) = 0
(P(B3) = ⅓
P(A|B3) = ½
Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;
P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]
Thus;
P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]
P(B3|A) = (1/6)/(⅓ + 0 + 1/6)
P(B3|A) = (1/6)/(1/2)
P(B3|A) = 1/3
