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Varvara68 [4.7K]
3 years ago
9

Work out the percentage change when a price of £50 is increased to £80.

Mathematics
2 answers:
professor190 [17]3 years ago
7 0

Answer:

Increase =increase /the original price ×100%

Step-by-step explanation:

80-50=30

30/50×100

=60%

sattari [20]3 years ago
3 0

Answer:

50 to 80 = 60%

Step-by-step explanation:

(80 -50) / 50 x 100 = 60%

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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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a) 7.79%

b) 67.03%

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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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