The office of student services at a large western state university maintains information on the study habits of its full-time st
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Answer:
Part 1) Greta's right, the triangle is incorrect.
Part 2)
Part 3) The height of the tree is , Trey’s house is not at risk
Step-by-step explanation:
Part 1) we know that
In the right triangle of the figure
Compare
therefore
The triangle is not correct
Because
The side adjacent to the 30 degree angle should be and the side opposite the 30 degree angle should be
Part 2)
Let
x--------> the distance from the airplane to the SCCA (hypotenuse of the right triangle)
we know that
Part 3)
Let
x--------> the height of the tree
we know that
therefore
Trey’s house is not at risk
Answer:
(a) P (p' > 0.25) = 0.
(b) P (0.15 < p' < 0.20) = 0.7784.
(c) P (133 < X < 171) = 0.2205
Step-by-step explanation:
Let <em>p</em> = proportion of business trip that were for internal company visit.
It is provided that 19% responded that it was for an internal company visit, i.e.
<em>p</em> = 0.19.
A sample of <em>n</em> = 950 business travelers are randomly selected.
According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
The standard deviation of this sampling distribution of sample proportion is:
Thus, the distribution of is N (<em>μ</em> = 0.19, <em>σ </em>= 0.013).
(a)
Compute the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:
P (p' > 0.25) = P (Z > 4.62) = 0
*Use a <em>z</em>-table for the probability.
Thus, the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.
(b)
Compute the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:
P (0.15 < p' < 0.20) = P (-3.08 < Z < 0.77)
= P (Z < 0.77) - P (Z < -3.08)
= 0.7794 - 0.0010
= 0.7784
*Use a <em>z</em>-table for the probability.
Thus, the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.7784.
(c)
Compute the proportion of <em>X</em> = 133 and <em>X</em> = 171 as follows:
p' = 133/950 = 0.14
p' = 171/950 = 0.18
Compute the value of P (0.14 < p' < 0.20) as follows:
P (0.14 < p' < 0.18) = P (-3.85< Z < -0.77)
= P (Z < -0.77) - P (Z < -3.85)
= 0.2206- 0.0001
= 0.2205
*Use a <em>z</em>-table for the probability.
Thus, the probability that between 133 and 171 of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.2205.