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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Suppose medical records indicate that the length of newborn babies (in inches) is normally distributed with a mean of 20 and a s

andreyandreev [35.5K]

Step-by-step answer:

The normal probability curve is symmetrical about the mean.

This means that for an event that is normally distributed, there is a 50% probability that it falls below the mean, and a 50% probability that it falls above.

From the given information, the mean is 20" and we need the probability that a given infant is longer than 20", namely the mean.

Therefore by the definition of the normal probability curve, there is a 50% probability that the length falls above 20".

This can be verified by referencing a normal probability table with Z=0, meaning at the mean, the probability is equal to 0.5 for Z<0, and therefore 0.5 for Z>0.

Z=(X-mean) / Standard deviation

When Z>0, X (measurement) is greater than the mean

When Z<0, X is less than the mean.

8 0

3 years ago

Simplify. (22)-3 A) 64 B) -64 C) 1/64 D) -1/64

saveliy_v [14]

$(2^2)^{-3}=4^{-3}=\dfrac{1}{4^3}=\dfrac{1}{64}\\\\Used\ a^{-n}=\dfrac{1}{a^n}$