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3 years ago

A place from this table is chosen at random. Let event A = The place is a city.

Mathematics

1 answer:

Crank3 years ago

6 0

Answer:

Final answer is $P(A^c)=\frac{3}{7}$

Step-by-step explanation:

We have been given a table containing a list of few places that are either city or in North America.

Total number of places in that list = 7

That means sample space has 7 possible events.

Given that a place from this table is chosen at random. Let event A = The place is a city.

Now we need to find about what is $P(A^c)$ .

That means find find the probability that chosen place is not a city.

there are 3 places in the list which are not city.

Hence favorable number of events = 3

Then required probability is given by favorable/total events.

$P(A^c)=\frac{3}{7}$

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pshichka [43]

As shown in the figure

Height of helicopter above the ground=30 m

Distance from the base to the point where the person is standing=30 m

height of the person=2 m

Now the person cast the shadow of length x m.

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∵ ∠B= ∠D=90°

∠C is common.

So by AA similarity ΔABC and ΔEDC are similar.

As we know when triangles are similar their sides are proportional.

AB/AD =BC/DC

Let DC=x meter

⇒ $\frac{10}{x}=\frac{30+x}{x}$

⇒5=\frac{30+x}{x}[/tex]

⇒5x=30+x

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4 0

3 years ago

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information: $P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000$ , where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

$\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2$

Equating it to zero we get,

$x^2 + 36x - 6400 = 0$

We use the quadratic formula to find the values of x:

$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ , where a, b and c are coefficients of $x^2, x^1 , x^0$ respectively.

Putting these value we get x = -100, 64

Now, again differentiating

$\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x$

At x = 64, $\displaystyle\frac{d^2(P(x))}{dx^2} < 0$

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

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