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3 years ago

Put the equation in standard form with a positive x coefficient.

Mathematics

1 answer:

serg [7]3 years ago

6 0

Answer:

y+2=5/3(x+6)

3(y+2)=5(x+6) [multiply both sides by 3 to get rid of fraction]

3y+6=5x+30 [distribute]

3y=5x+24 [subtract 6 from both sides]

3y-5x=24 [subtract 5x from both sides]

-3y+5x=-24 [multiply everything by -1]

5x-3y=-24 [rearrange]

Step-by-step explanation:

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3 years ago

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

Musya8 [376]

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:

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

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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4 years ago

I need help ASAP people

nirvana33 [79]

Answer:

point D: (2, -4)

Step-by-step explanation:

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4 years ago