How many square units are in the area of the triangle whose vertices are the x and y intercepts of the curve y = (x-3)^2 (x+2)?

Mathematics

1 answer:

pantera1 [17]4 years ago

6 0

The intercepts can be read from the equation as (-2, 0), (0, 18), (3, 0). These define a triangle with a base of 5 and an altitude of 18, so the area is

... A = (1/2)·5·18 = 45 . . . . square units

_____

The x-intercepts are those values of x that make one or another of the factors be zero. (x-3) is zero when x=3; (x+2) is zero when x=-2.

The y-intercept is the value when x=0. That is (-3)²·(2) = 9·2 = 18.

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The two intervals (114.9, 115.7) and (114.6, 116.0) are confidence intervals (computed using the same sample data) for μ = true

kenny6666 [7]

Answer:

115.3 hertz

Step-by-step explanation:

For either interval, the value of the sample mean is given by the average of the lower and upper bound of the confidence interval. Since both intervals were constructed by using the same sample, both values should be equal.

For the first interval:

$M=\frac{114.9 + 115.7}{2} \\M=115.3\ hertz$