Answer:
From the given, we establish this relationship:
RS + ST = RT
RS = ST since S is the midpoint of RT
Thus, let's use RS = ST to find x,
By substituting the given expressions:
RS = ST
8x + 11 = 14x - 1
12 = 6x
x = 12/6
thus,
x = 2
The equation of a circle is (x-h)²+(y-k)²=r², with h being the x value in the center, k being the y value, and r being the radius. x and y can stay variables, so we can plug numbers in to get (x-0)²+(y-3)²=5²=25=x²+(y-3)²
Answer:
a) 0.06 = 6% probability that a person has both type O blood and the Rh- factor.
b) 0.94 = 94% probability that a person does NOT have both type O blood and the Rh- factor.
Step-by-step explanation:
I am going to solve this question treating these events as Venn probabilities.
I am going to say that:
Event A: Person has type A blood.
Event B: Person has Rh- factor.
43% of people have type O blood
This means that ![P(A) = 0.43](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.43)
15% of people have Rh- factor
This means that ![P(B) = 0.15](https://tex.z-dn.net/?f=P%28B%29%20%3D%200.15)
52% of people have type O or Rh- factor.
This means that ![P(A \cup B) = 0.52](https://tex.z-dn.net/?f=P%28A%20%5Ccup%20B%29%20%3D%200.52)
a. Find the probability that a person has both type O blood and the Rh- factor.
This is
![P(A \cap B) = P(A) + P(B) - P(A \cup B)](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%20P%28A%29%20%2B%20P%28B%29%20-%20P%28A%20%5Ccup%20B%29)
With what we have
![P(A \cap B) = 0.43 + 0.15 - 0.52 = 0.06](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200.43%20%2B%200.15%20-%200.52%20%3D%200.06)
0.06 = 6% probability that a person has both type O blood and the Rh- factor.
b. Find the probability that a person does NOT have both type O blood and the Rh- factor.
1 - 0.06 = 0.94
0.94 = 94% probability that a person does NOT have both type O blood and the Rh- factor.
<h3>
Answer: 11/20</h3>
They got 33 heads out of 60 tosses, so,
33/60 = 11/20
You divide each part by 3 to reduce the fraction.