A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3
miles.
1 answer:
Answer:
The answer is below
Step-by-step explanation:
A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3 miles.
a) Start by drawing a diagram of this situation. Your diagram might include a coordinate plane, the cell tower, and a circle representing the cell tower's coverage boundary.
a) A house is located 5 miles north of the center of the town and is to the east of the cell tower. If the house lies on the boundary of the cell tower's coverage, how far east of the center of the town is the house?
Answer:
Let the center of the town represent the origin (0,0). Also 1 unit = 1 mile. Since the cell tower is located 3 miles east and 4 miles north of the center of a small town, it is represented by A(3, 4)
The cell tower has a coverage of radius 3 miles. This can be represented by a circle with equation:
(x - a)² + (y - b)² = r². where (a,b) is the center of the circle and r is the radius. Hence:
(x - 3)² + (y - 4)² = 3²
(x - 3)² + (y - 4)² = 9
The diagram is drawn using geogebra.
b) The house is 5 miles north. It can be represented by y = 5 line.
To find the distance east of the house we have to substitute y = 5 and solve for x, hence:
(x - 3)² + (y - 4)² = 9
(x - 3)² + (5 - 4)² = 9
(x - 3)² + 1 = 9
(x - 3)² = 8
(x - 3) = √8
x - 3 = ±2.83
x = 3 ± 2.83
x = 5.83 or 1.83
Since it is to the east to the cell tower, hence x = 5.83.
Therefore the house is located 5.83 miles to the east of the cell tower
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Answer:
Y will arrive earlier than X one fourth of times.
Step-by-step explanation:
To solve this, we might notice that given that both events are independent of each other, the joint probability density function is the product of X and Y's probability density functions. For an uniformly distributed density function, we have that:
Where L stands for the length of the interval over which the variable is distributed.
Now, as X is distributed over a 1 hour interval, and Y is distributed over a 0.5 hour interval, we have:
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Now, the probability of an event is equal to the integral of the density probability function:
Where A is the in which the event happens, in this case, the region in which Y<X (Y arrives before X)
It's useful to draw a diagram here, I have attached one in which you can see the integration region.
You can see there a box, that represents all possible outcomes for Y and X. There's a diagonal coming from the box's upper right corner, that diagonal represents the cases in which both X and Y arrive at the same time, under that line we have that Y arrives before X, that is our integration region.
Let's set up the integration:
We have used here both the independence of the events and the uniformity of distributions, we take the 2 out because it's just a constant and now we just need to integrate. But the function we are integrating is just a 1! So we can take the integral as just the area of the integration region. From the diagram we can see that the region is a triangle of height 0.5 and base 0.5. thus the integral becomes:
That means that one in four times Y will arrive earlier than X. This result can also be seen clearly on the diagram, where we can see that the triangle is a fourth of the rectangle.
