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2 years ago

Help help help help help help help

Mathematics

1 answer:

earnstyle [38]2 years ago

8 0

Answer:

The first one is 90, and the second one is 150

Explanation:

40% of 90 is 36, and

40% of 150 is 60.

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2+cotA=1 homework help

Alex Ar [27]

$2+\cot A=1\implies \cot A=-1\implies \tan A=-1$

This happens whenever $A=\dfrac{3\pi}4$ or $A=\dfrac{7\pi}4$ . More generally, $\tan A=-1$ whenever you start with one of these angles and add any multiple of $\pi$ , so the general solution would be $A=\dfrac{3\pi}4+n\pi$ , where $n$ is any integer. (Notice that when $n=1$ , you end up with $\dfrac{7\pi}4$ .)

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Effectus [21]

45: 100% - 17% = 83%

47: 44/2 = 22
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See attached file

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2 years ago

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2 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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A density graph for all of the possible temperatures from 60 degrees to 260

musickatia [10]

C because it’s the only one in between 60 and 260

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