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tiny-mole [99]
2 years ago
15

Multiply. Write the answer in simplest form. 3/7 x 2 5/6

Mathematics
1 answer:
aev [14]2 years ago
8 0

193/42

  • 3×6+25×7÷42
  • 18+175÷42
  • 193/42
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Step-by-step explanation:

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Please help me solve this 200(20)^x>500x+400
s2008m [1.1K]
Hi there!

200(20)^x > 500x + 400

Let's find the critical points of the inequality

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6 0
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

6 0
3 years ago
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