Suppose that receiving stations X, Y, and Z are located on a coordinate plane at the points (4,5), (-6,-6), and (-14,2
), respectively. The epicenter of an earthquake is determined to be 10 units from X, 5 units from Y, and 13 units from Z. Where on the coordinate plane is the epicenter located?
Find the coordinates of the epicenter.
Find the coordinates of the epicenter.
1 answer:
Answer:
(-2, -3)
Step-by-step explanation:
A careful graph shows the point (-2, -3) is at the intersection of the circles whose radii are the given distances from the receiving stations.
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The simultaneous equations for the circles can be solved algebraically.
The epicenter is 10 units from X, so lies on the circle ...
(x -4)^2 +(y -5)^2 = 10^2
x^2 -8x +16 +y^2 -10y +25 = 100
x^2 +y^2 -8x -10y = 59
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The epicenter is 5 units from Y, so lies on the circle ...
(x +6)^2 +(y +6)^2 = 5^2
x^2 +12x +36 +y^2 +12y +36 = 25
x^2 +y^2 +12x +12y = -47
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The epicenter is 13 units from Z, so lies on the circle ...
(x +14)^2 +(y -2)^2 = 13^2
x^2 +28x +196 +y^2 -4y +4 = 169
x^2 +y^2 +28x -4y = -31
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Subtracting the second equation from each of the other two, we get ...
(x^2 +y^2 -8x -10y) -(x^2 +y^2 +12x +12y) = (59) -(-47)
-20x -22y = 106 . . . . eq1 -eq2
(x^2 +y^2 +28x -4y) -(x^2 +y^2 +12x +12y) = (-31) -(-47)
16x -16y = 16 . . . . . . . .eq3 -eq2
These simultaneous linear equations can be solved a variety of ways. We might use substitution:
x = y+1 . . . . . from eq3 -eq2 divided by 16
10(y +1) +11y = -53 . . . . . from eq1 -eq2 divided by -2
21y = -63 . . . . . . . . . . . . simplify, subtract 10
y = -3
x = y+1 = -2
The epicenter is located at (x, y) = (-2, -3).
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Answer:
Step-by-step explanation:
Hello!
a)
To compare both datasets I've made a box plot for them.
The yellow one belongs to the points scored by the men and the green one is for the points scored by the women.
Women: The box seems symmetrical, the median and the mean (black square) are almost the same. The whiskers are almost the same length, you could say that in general the points scored by the women have a symmetrical distribution.
Men: The box looks a little skewed to the right, the median is closer to the 1st quantile and the mean is greater than the median. The whiskers are the same length, in general the distribution seems almost symmetrical.
b)
As you can see in the box plot the median and mean of the points scored by the women are almost the same. Using the data I've calculated both values:
Me= 24.50
X[bar]= 23.80
The median divides the sample in halves (50-50), it shows you where the middle of the distribution is.
The average shows you the value that centers the distribution, meaning, it is the value around which you'll find most of the data set. It summarizes best the sample information and therefore is a better measure of center.
c.
As you can see in the box plots, there are no outliers in both distributions.
An outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on a statistical analysis.
Considering the 1st quartile (Q₁), the 3rd quartile (Q₃) and the interquartile range IQR, any value X is considered an outlier if:
X < Q₁ - 1.5 IQR
X > Q₃ + 1.5 IQR
Or extreme outliers if:
X < Q₁ - 3 IQR
X > Q₃ + 3 IQR
For the women data set:
Q₁= 17; Q₃= 30; IQR= 30 - 17= 13
Lower outliers: X < Q₁ - 1.5 IQR= 17-1.5*13= -2.5 ⇒ The minimum value recorded is 03, so there are no outliers.
Upper outliers: X > Q₃ + 3 IQR= 30 + 1.5*13= 49.5 ⇒ The maximum value registered is 41, so there are no outliers.
I hope this helps!
