Answer:
Proved
Step-by-step explanation:
Given: EC || AC, DB || AC, ∠A = ∠F
Prove: ΔMDF ∼ ΔNCA
Solution
See diagram attached to the solution to better understand the following workings.
Redrawing ΔMDF or rotating to be facing the same direction.
EC is parallel to AC
DB parallel to AC
Using similar triangle theorem:
If ΔMDF ∼ ΔNCA
Ratio of Corresponding sides would be equal
(adjacent of ΔMDF)/(adjacent of ΔNCA) = (Opposite of ΔMDF)/(opposite of ΔNCA) = (hypotenuse of ΔMDF)/(hypotenuse of ΔNCA)
DF/ CA = MD/NC = FM/AN
∠A = ∠F
∠M = ∠N
∠D = ∠C
Since the ratio of Corresponding sides and angle are equal, ΔMDF is similar to ΔNCA.
ΔMDF ∼ ΔNCA
Answer:When n is an odd number, ![\sqrt[n]{a}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D)
is a real number for all values of a. Then, the domain is the real domain.
The probability that a randomly selected score is greater than 334 will be 0.02275.
<h3>What is a normal distribution?</h3>
The Gaussian Distribution is another name for it. The most significant continuous probability distribution is this one. Because the curve resembles a bell, it is also known as a bell curve.
The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.
The GRE is an entrance exam that many students are required to take in order to apply to graduate school. In 2014, the combined scores for the Verbal and Quantitative sections were approximately normally distributed with a mean of 310 and a standard deviation of 12.
Then the probability that a randomly selected score is greater than 334 will be
The z-score is given as
z = (x - μ)/σ
z = (334 - 310)/12
z = 24/12
z = 2
Then the probability will be
P(x > 334) = P(z > 2)
P(x > 334) = 1 - P(x<334)
P(x > 334) = 1 - 0.97725
P(x > 334) = 0.02275
More about the normal distribution link is given below.
brainly.com/question/12421652
#SPJ1
If you are asking for the probability of getting a certain card you have a 1/ 20th chance
If you are asking for a percentage there is a 5% chance of getting any card
