The simulation of the medicine and the bowler hat are illustrations of probability
- The probability that the medicine is effective on at least two is 0.767
- The probability that the medicine is effective on none is 0
- The probability that the bowler hits a headpin 4 out of 5 times is 0.3281
<h3>The probability that the medicine is effective on at least two</h3>
From the question,
- Numbers 1 to 7 represents the medicine being effective
- 0, 8 and 9 represents the medicine not being effective
From the simulation, 23 of the 30 randomly generated numbers show that the medicine is effective on at least two
So, the probability is:
p = 23/30
p = 0.767
Hence, the probability that the medicine is effective on at least two is 0.767
<h3>The probability that the medicine is effective on none</h3>
From the simulation, 0 of the 30 randomly generated numbers show that the medicine is effective on none
So, the probability is:
p = 0/30
p = 0
Hence, the probability that the medicine is effective on none is 0
<h3>The probability a bowler hits a headpin</h3>
The probability of hitting a headpin is:
p = 90%
The probability a bowler hits a headpin 4 out of 5 times is:
P(x) = nCx * p^x * (1 - p)^(n - x)
So, we have:
P(4) = 5C4 * (90%)^4 * (1 - 90%)^1
P(4) = 0.3281
Hence, the probability that the bowler hits a headpin 4 out of 5 times is 0.3281
Read more about probabilities at:
brainly.com/question/25870256
Answer:
Step-by-step explanation:
If a point (x, y) is translated by 4 units horizontally right and 1 unit upwards, coordinates of the image point will be,
(x, y) → (x + 4. y + 1)
Therefore, vertices of the image triangle ABC will be,
A(2, 2) → A'(2+4, 2+1)
→ A'(6, 3)
B(5, -1) → B'(5+4, -1+1)
→ B'(9, 0)
C(1, -2) → C'(1+4, -2+1)
→ C'(5, -1)
Then reflected across y-axis.
Rule for the reflection across y-axis will be,
(x, y) → (-x, y)
A'(6, 3) → A"(-6, 3)
B'(9, 0) → B"(-9, 0)
C'(5, -1) → C"(-5, -1)
Answer:
3/2
Step-by-step explanation:
(3/4)/(1/2)=(3/4)(2/1)=6/4=3/2