The given above is are triangles, as per the proof the line segments on top and bottom part are parallel. Also, it is given that two pairs of the angles of the triangles are congruent.
The triangles also share one common side, CA. Since, this side is between the angles the postulate that will prove the congruence of the triangles is ASA.
The answer to this item is the third choice.
Answer:
The percentage of people should be seen by the doctor between 13 and
17 minutes is 68% ⇒ 2nd term
Step-by-step explanation:
* Lets explain how to solve the problem
- Wait times at a doctor's office are typically 15 minutes, with a standard
deviation of 2 minutes
- We want to find the percentage of people should be seen by the
doctor between 13 and 17 minutes
* To find the percentage we will find z-score
∵ The rule the z-score is z = (x - μ)/σ , where
# x is the score
# μ is the mean
# σ is the standard deviation
∵ The mean is 15 minutes and standard deviation is 2 minutes
∴ μ = 15 , σ = 2
∵ The people should be seen by the doctor between 13 and
17 minutes
∵ x = 13 and 17
∴ z = 
∴ z = 
- Lets use the standard normal distribution table
∵ P(z > -1) = 0.15866
∵ P(z < 1) = 0.84134
∴ P(-1 < z < 1) = 0.84134 - 0.15866 = 0.68268 ≅ 0.68
∵ P(13 < x < 17) = P(-1 < z < 1)
∴ P(13 < x < 17) = 0.68 × 100% = 68%
* The percentage of people should be seen by the doctor between
13 and 17 minutes is 68%
Determine the mode(s) of the data 2, 2, 2,3,5,5, 6, 7, 8, 8, 8, 9, 10.
Genrish500 [490]
To find the mode, put all the numbers in order from least to greatest, then count how many times you see a number. The number you see the most is the mode. In this problem, we have more than one mode, we have two. The number two appears three times and so does number eight. Having two modes is called bimodal, and having more than two modes is called multimodal. So we have a bimodal of two and eight from this data.
It would be 50%, because they have a 50/50 chance of getting their seat, even if the first wizard/witch got in the wrong seat.
I believe it is D but it may be B