Which congruency theorem can be used to prove that △GHL ≅ △KHJ?
2 answers:
Answer:
B. ASA (Angle-Side-Angle).
Step-by-step explanation:
We have been given a diagram and we are asked to find the congruence theorem, which will prove that △GHL ≅ △KHJ.
Upon looking at our diagram we can see that side length GH of △GHL is equal to side length KH of △KHJ.
We are also given that .
We can see that is vertical opposite angle of
as they are formed by the intersection of line segments GK and JL. Therefore,
, by vertical angles theorem.
We can see that both triangles have two congruent angles and one congruent included side, therefore, △GHL ≅ △KHJ by ASA congruence and option B is the correct choice.
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Answer:
(√3 tan x + 1) / (√3 - tan x)
Step-by-step explanation:
Answer:
a) 0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.
b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.
Step-by-step explanation:
Binomial distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
To use the Poisson approximation for the binomial, we have that:
1 in every 2500 automobiles produced has a particular manufacturing defect.
This means that
a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.
This is when
. So
0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.
(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.
Using the approximation:
. So
0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.