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Brums [2.3K]
2 years ago
11

What are the exact solutions of x2 − 5x − 1 = 0 using x equals negative b plus or minus the square root of the quantity b square

d minus 4 times a times c all over 2 times a?
Mathematics
1 answer:
geniusboy [140]2 years ago
5 0

Answer:

x=2.5+\frac{\sqrt{29}}{2}\\\\x=2.5-\frac{\sqrt{29}}{2}

Step-by-step explanation:

So the question here is asking you to use the quadratic formula which is expressed as: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\

A quadratic can generally be expressed as: y=ax^2+bx+c

So using the equation you gave: 0=x^2-5x-1

We can identify the following values: a=1, b=-5, c=-1

Btw the equation explicitly write "1" as the coefficient of x, but since it's not provided it's implied that it's 1.

So plugging in the known values, we get the following equation:

x=\frac{5\pm\sqrt{(-5)^2-4(1)(-1)}}{2(1)}\\\\x=\frac{5\pm\sqrt{25+4}}{2}\\\\x=\frac{5\pm\sqrt{29}}{2}\\\\x=2.5+\frac{\sqrt{29}}{2}\\\\x=2.5-\frac{\sqrt{29}}{2}

The last step just consists of taking the + and - solution, and since it asks for exact solutions you leave the 29 under the radical, and you don't approximate. There is no further simplification that can be done here.

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

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}

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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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