The vertices of a right- angled triangle are A(1,0), B(7.0) and C(1,6). Plot the points A, B and C on the cartesian plane below.
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Answer:
With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.
Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.
Step-by-step explanation:
For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric distribution.
Binomial probability 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.
Hypergeometric distribution:
The probability of x sucesses is given by the following formula:
In which:
x is the number of sucesses.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
is the number of different combinations of x objects from a set of n elements, given by the following formula.
Sampling with replacement:
I consider a success choosing a black ball, so
We want 2 black balls and 2 white, 2 + 2 = 4, so , and we want P(X = 2).
With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.
Sampling without replacement:
150 + 50 = 200 total balls, so
Sample of 4, so
50 are black, so
We want P(X = 2).
Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.
Answer:
Step-by-step explanation:
a). Given a parametric equation, we are describing a set of coordinates based on the value of t. The variable t is called the parameter.
b) we have the following equations. x=t y=t^2, so in order for us to know where the object is at t=t' we must replace t with the specific value t'. Hence, when t=0 the object is at (0,0^2) = (0,0) (the origin). When t=6, the object is at (6,6^2) = (6,36).
c). To eliminate the parameter, we replace the parameter in one equation by using the second equation. Recall that we have that x=t. Then, by replacing in the second equation, we have the following
where