Answer:
8a/b
.................
Which of the following is a step in constructing a circle circumscribed about a triangle
1 answer:
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Perimeter of a shape is the total length of its border
Figure 1, perimeter = 5 units
Figure 2, perimeter = 7 units
Figure 3, perimeter = 9 units
Figure 4, perimeter = 11 units
Notice that the perimeter is forming an Arithmetic sequence; 5, 7, 9, 11 with a common difference of 2
The general form of an Arithmetic sequence is
Where 'd' is a common difference and 'n' is the number of terms.
We have d = 2, and
zero term = the term before the first term = 3
⇒ This is the rule to find the perimeter of the next figures
Figure 1, perimeter = 5 units
Figure 2, perimeter = 7 units
Figure 3, perimeter = 9 units
Figure 4, perimeter = 11 units
Notice that the perimeter is forming an Arithmetic sequence; 5, 7, 9, 11 with a common difference of 2
The general form of an Arithmetic sequence is
Where 'd' is a common difference and 'n' is the number of terms.
We have d = 2, and
zero term = the term before the first term = 3
Answer:
(a) The probability mass function of <em>X</em> is:
(b) The most likely value for <em>X</em> is 1.32.
(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.
Step-by-step explanation:
The random variable <em>X</em> is defined as the number among the four homeowners who have earthquake insurance.
The probability that a homeowner has earthquake insurance is, <em>p</em> = 0.33.
The random sample of homeowners selected is, <em>n</em> = 4.
The event of a homeowner having an earthquake insurance is independent of the other three homeowners.
(a)
All the statements above clearly indicate that the random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 4 and <em>p</em> = 0.33.
The probability mass function of <em>X</em> is:
(b)
The most likely value of a random variable is the expected value.
The expected value of a Binomial random variable is:
Compute the expected value of <em>X</em> as follows:
Thus, the most likely value for <em>X</em> is 1.32.
(c)
Compute the probability that at least two of the four selected have earthquake insurance as follows:
P (X ≥ 2) = 1 - P (X < 2)
= 1 - P (X = 0) - P (X = 1)
Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.