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2 years ago

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Between Ahmedabad, Bombay, and Calcutta, there are the following highways: Ahmedabad – Bombay, Ahmedabad – Calcutta, and Bombay

– Calcutta. During monsoon, when there is heavy rain, each of the road gets blocked independently with probability p. What is then the probability that Calcutta will be accessible from Ahmedabad?

1 answer:

Marizza181 [45]2 years ago

3 0

Answer: p+p^2

Step-by-step explanation: shown in the attachment

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John can type 2/3 of a manuscript in 12 hours. How much of the manuscript can John write in 1 hour?

Marina86 [1]

John can write 1/18 of the manuscript in 1 hour
Hope this helps!!

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2 years ago

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F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 x 2

34kurt

Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

$g(f(x)) = 3x^4 + 18x^2 + 18$

<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9$

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

$g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9$

Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

$g(f(x)) = 3x^4 + 18x^2 + 18$

Checking for inverse functions

$f(x) = 3x^2 + 9$

Represent f(x) with y

$y = 3x^2 + 9$

Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

$x - 9 = 3y^2 + 9 - 9$

$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

$\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}$

$y^2 = \frac{x}{3} - 3$

Take square root of both sides

$\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}$

$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

$g(x) = \sqrt{\frac{x}{3} - 3}$

Note that the resulting value of g(x) is not the same as $g(x) = \dfrac{1}{3}x^2 - 9$

<em>Hence, f(x) and g(x) and not inverse functions</em>

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2 years ago

Complete the steps for solving 7 = –2x2 + 10x. Factor out of the variable terms. inside the parentheses and on the left side of

Mamont248 [21]

we have

$7=-2x^{2} +10x$

Factor the leading coefficient

$7=-2(x^{2} -5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$7-12.50=-2(x^{2} -5x+2.5^{2})$

$-5.50=-2(x^{2} -5x+2.5^{2})$

Divide both sides by $-2$

$2.75=(x^{2} -5x+2.5^{2})$

Rewrite as perfect squares

$2.75=(x-2.5)^{2}$

Taking the square roots of both sides (square root property of equality)

$x-2.5=(+/-)\sqrt{2.75}$

Remember that

$\sqrt{2.75}=\sqrt{\frac{11}{4}}= \frac{\sqrt{11}}{2}$

$x-2.5=(+/-)\frac{\sqrt{11}}{2}$

$x=2.5(+/-)\frac{\sqrt{11}}{2}$

$x=2.5+\frac{\sqrt{11}}{2}=\frac{5+\sqrt{11}}{2}$

$x=2.5-\frac{\sqrt{11}}{2}=\frac{5-\sqrt{11}}{2}$

<u>the answer is</u>

The solutions are

$x=\frac{5+\sqrt{11}}{2}$

$x=\frac{5-\sqrt{11}}{2}$

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2 years ago

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Determine if this conjecture is true. If not, give a counterexample. The difference of two negative numbers is a negative number

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That would be a false statement

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2 years ago

To avoid collisions with invasive species of aliens, new imperial regulations allow only positive integer space jumps parallel t

Sunny_sXe [5.5K]

Answer:24,942,060

Step-by-step explanation:

(10,10,4) - (3,2,0) = 7,8,4

7+8+4=19

19!/(7!*8!*4!)= 24'942'060

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1 year ago