Answer:
745.44 mm² (nearest hundredth)
Step-by-step explanation:
Section the shape into a semicircle, a rectangle and a triangle by drawing a line vertically through the point, and a vertical line upwards from the corner of the 12.5 mm and 32.5 mm corner.
area of semicircle = (1/2) r² = 78.125
mm²
area of the rectangle = 12.5 x 25 = 312.5 mm²
area of triangle = 1/2 x (55 - 25) x 12.5 = 187.5 mm²
total area = 312.5 + 78.125 + 187.5 = 745.44 mm² (nearest hundredth)
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.