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2 years ago

ABCD is a quadrilateral Work out angle x

Mathematics

1 answer:

netineya [11]2 years ago

3 0

First of all, we can work out the length of BD: using the pythagorean theorem, we have

$BD = \sqrt{6^2+8^2}=10$

Now, we can work on triangle BCD and use the sine theorem: we have

$\dfrac{BD}{\sin(\hat{C})} = \dfrac{CD}{\sin(x)}$

Plugging the values and solving for $\sin(x)$ we have

$\dfrac{10}{\sin(29)} = \dfrac{13}{\sin(x)} \iff \sin(x)=\dfrac{13\cdot \sin(29)}{10}\approx 0.63$

We deduce

$x=\arcsin(0.63)\approx 39$

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Step-by-step explanation:

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3 years ago

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2 years ago

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Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b

Musya8 [376]

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a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$

c) cumulative distribution function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

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3 years ago

If a point is chosen inside the square, what is the probability that it will also be inside the circle?

Feliz [49]

Answer:

$79\%$

Step-by-step explanation:

The probability that the point is chosen in the circle is equal to the area of the circle divided by the area of the square.

Formulas used:

Area of a square with side length $s$ is given by $A=s^2$
Area of a circle with radius $r$ is given by $A=r^2\pi$

The segment marked as 1 represents not only the radius of the circle, but also half the side length of the square. Therefore, the side length of the square is 2, and we have:

Area of square: $A=2^2=4$