HELP ASAP WILL MARK BRAINLIEST!!!!! The coordinates of point T are (0,3). The midpoint ST of is (1,-5). Find the coordinat
es of point S. The other endpoint is
1 answer:
<em>Greetings from Brasil...</em>
See the attached chart. The midpoint, M, has 8 units down on the Y axis. So the other half will also have 8 units after that point M. This is also true for the X axis.
Just make the difference between points T and M. For M and S they are the same quantities.
(0; 3) and (1; -5)
X: 1 - 0 = 1 (one unit until mid point)
Y: - 5 - 3 = - 8 (8 units until mid point)
To S:
S(W; Z) <em>M = center point coordinate value</em>
W = M + 1 ⇒ W = 1 + 1 = 2
Z = M - 8 ⇒ Z = - 5 - 8 = - 13
<h3>S(2; -13)</h3>
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Answer:
(a) The probability distribution of the random variable <em>X</em> is Binomial.
(b) The mean and standard deviation of the 25-question test are 5 and 2 respectively.
(c) The value of <em>x</em> is 0.34.
Step-by-step explanation:
The random variable <em>X </em>is defined as the number of correct answers given by a person who is guessing each answer on a 25-question exam.
There are five possible answer for every question.
This implies that the probability of getting a correct answer is:
<em>P</em> (X) = 0.20.
There are a total of <em>n</em> = 25 questions.
Every answer is independent of the others.
(a)
The random variable <em>X</em> has finite number of independent trials (i.e. 25 questions). There are only two outcomes for each trial, i.e. Success = correct answer and Failure = wrong answer. Each trial has the same probability of success (, i.e. P (X) = 0.25).
Thus, the probability distribution of the random variable <em>X</em> is Binomial with parameters <em>n</em> = 25 and <em>p</em> = 0.20.
(b)
Compute the mean of the random variable <em>X</em> as follows:
Compute the standard deviation of the random variable <em>X</em> as follows:
Thus, the mean and standard deviation of the 25-question test are 5 and 2 respectively.
(c)
The sample is large and the probability of success is close to 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
1. np ≥ 5
2. n(1 - p) ≥ 5
Check the conditions as follows:
Thus, a Normal approximation to binomial can be applied.
So,
It is provided that the minimum passing score foe the test is such that only 1% of students who are guessing will pass the test.
That is P (X < x) = 0.01.
⇒ P (Z < z) = 0.01
The value of <em>z</em> is -2.33.
Compute the value of <em>x</em> as follows:
Thus, the value of <em>x</em> is 0.34.