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Mariana [72]
3 years ago
13

Mary bought a handbag for 15.00naira. if she sold it for 18.00,what was her percentage gain?​

Mathematics
1 answer:
melomori [17]3 years ago
8 0

Answer:

total profit gained = 20%

Step-by-step explanation:

profit = SP - CP = 18 - 15 = 3 naira

total profit percentage =

  • ( profit / CP ) × 100
  • 3 / 15 × 100
  • 20%

hence her total percentage gain was 20%

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Write the expression in terms of a single trigonometric identity or a constant.
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2 years ago
Jessica works at a store. She sold a computer for $950. She receives a 5% commission for each sale. How much was her commission
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What is the midpoint of a line segment whose endpoints are (-4,3) and (5,-6)?
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4 0
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Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t
worty [1.4K]

Answer:

a) P(X \leq 100) = 1- e^{-0.01342*100} =0.7387

P(X \leq 200) = 1- e^{-0.01342*200} =0.9317

P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930

b) P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498

c) m = \frac{ln(0.5)}{-0.01342}=51.65

d) a = \frac{ln(0.05)}{-0.01342}=223.23

Step-by-step explanation:

Previous  concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

Solution to the problem

For this case we have that X is represented by the following distribution:

X\sim Exp (\lambda=0.01342)

Is important to remember that th cumulative distribution for X is given by:

F(X) =P(X \leq x) = 1-e^{-\lambda x}

Part a

For this case we want this probability:

P(X \leq 100)

And using the cumulative distribution function we have this:

P(X \leq 100) = 1- e^{-0.01342*100} =0.7387

P(X \leq 200) = 1- e^{-0.01342*200} =0.9317

P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930

Part b

Since we want the probability that the man exceeds the mean by more than 2 deviations

For this case the mean is given by:

\mu = \frac{1}{\lambda}=\frac{1}{0.01342}= 74.516

And by properties the deviation is the same value \sigma = 74.516

So then 2 deviations correspond to 2*74.516=149.03

And the want this probability:

P(X > 74.516+149.03) = P(X>223.547)

And we can find this probability using the complement rule:

P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498

Part c

For the median we need to find a value of m such that:

P(X \leq m) = 0.5

If we use the cumulative distribution function we got:

1-e^{-0.01342 m} =0.5

And if we solve for m we got this:

0.5 = e^{-0.01342 m}

If we apply natural log on both sides we got:

ln(0.5) = -0.01342 m

m = \frac{ln(0.5)}{-0.01342}=51.65

Part d

For this case we have this equation:

P(X\leq a) = 0.95

If we apply the cumulative distribution function we got:

1-e^{-0.01342*a} =0.95

If w solve for a we can do this:

0.05= e^{-0.01342 a}

Using natural log on btoh sides we got:

ln(0.05) = -0.01342 a

a = \frac{ln(0.05)}{-0.01342}=223.23

5 0
3 years ago
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