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DiKsa [7]
3 years ago
12

Paul Colin and Brian are waiters one night the restaurant earns tips totalling £78.40 they share the tips in the ratio 2 to 5 an

d 1 how much does Colin get over Paul ​
Mathematics
1 answer:
alexdok [17]3 years ago
3 0

Answer: £29.40

Step-by-step explanation:

The fraction for each one one of them based on the ratio will be:

Paul = 2/(2+5+1) = 2/8

Colin = 5/(2+5+1) = 5/8

Brian = 1/(2+5+1) = 1/8

Amount gotten by each person will be:

Paul = 2/8 × £78.40 = £19.60

Colin = 5/8 × £78.40 = £49.00

Brian = 1/8 × £78.40 = £9.80

The amount that Colin get over Paul ​ will be:

= £49.00 - £19.60

= £29.40

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10. An arithmetic sequence has this recursive formula. What is the explicit formula?​
Mrrafil [7]

Answer:

a_{n}=45-3n

Step-by-step explanation:

Method 1:

Arithmetic sequence is in the form

a_{n} =a_{1} +(n-1)d\\

d is the common difference, can be found by:

d=a_{n}-a_{n-1}=-3

Subtituting the a_{1} and d

You get:

a_{n}=42+(-3)(n-1)=45-3n

Method 2 (Mathematical induction):

Assume it is in form a_{n}=45-3n

Base step: a_{1} =45-3(1)=42

Inducive hypophesis: a_{n}=45-3n

GIven: a_{n+1} =a_{n}-3

a_{n+1}=45-3n-3=45-3(n+1)

Proved by mathematical induction

a_{n}=45-3n

5 0
3 years ago
Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

P(X

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}

The cumulative distribution for this function is given by:

F(X) = 1- e^{-\lambda x}, x\ geq 0

We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

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Answer:0.16

Step-by-step explanation:

7 0
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