Suppose the lengths, in seconds, of the songs in an online database are normally distributed. For a random sample of songs, the
1 answer:
Answer:
186
Step-by-step explanation:
The sample mean, x, is the central value in the confidence interval, that is, the average between the upper and lower bounds of the interval.
In this case, the Lower bound is 184.00 and the upper bound is 188.00. Therefore, the sample mean is given by:
The mean length, in seconds, of the sampled songs is 186.00.
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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).
