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4 years ago

70

Mathematics

2 answers:

stealth61 [152]4 years ago

5 0

Answer:

Step-by-step explanation:

Molodets [167]4 years ago

3 0

Check the picture below.

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At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 0.15 and the probability that the flight

professor190 [17]

Answer:

0.853

Step-by-step explanation:

Use conditional probability. The probability of event A, given that event B has occurred, is:

P(A|B) = P(A and B) / P(B)

The probability that it is not raining, given that the flight leaves on time, is:

P(not raining | on time) = P(not raining AND on time) / P(on time)

P(not raining | on time) = 0.81 / (1 − 0.05)

P(not raining | on time) = 0.853

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3 years ago

Are x-1 and 1-x inverse functions

Dima020 [189]

Answer:

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3 years ago

Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x

soldier1979 [14.2K]

Answer:

(a) The function is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is $x = -\frac{1}{2}$

(d) The function is concave upward on $\left(- \frac{1}{2}, \infty\right)$ and concave downward on $\left(-\infty, - \frac{1}{2}\right)$

Step-by-step explanation:

(a) To find the intervals where $f(x) = 2x^3 + 3x^2 -180x$ is increasing or decreasing you must:

1. Differentiate the function

$\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180$

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

$f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6$

These points divide the number line into three intervals:

$(-\infty,-6)$ , $(-6,5)$ , and $(5, \infty)$

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

$\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right$

Therefore f(x) is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.

f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

$f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6$