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devlian [24]
3 years ago
15

Can someone plz help me!!!

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
8 0

Answer:

The answer is no because 5 is greater than 3

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Birds arrive at a birdfeeder according to a Poisson process at a rate of six per hour.
m_a_m_a [10]

Answer:

a) time=10 \frac{1}{6}=\frac{10}{6}=1.67 hours

b) P(T\geq 0.25h)=e^{-(6)0.25}=0.22313

c) P(T\leq 0.0833)=1-e^{-(6)0.0833}=0.39347

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

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

And the parameter \lambda represent the average ocurrence rate per unit of time.

The exponential distribution is useful when we want to describ the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time btween two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:

P(T>t)= e^{-\lambda t}

a. What is the expected time you would have to wait to see ten birds arrive?

The original rate for the Poisson process is given by the problem "rate of six per hour" and on this case since we want the expected waiting time for 10 birds we have this:

time=10 \frac{1}{6}=\frac{10}{6}=1.67 hours

b. What is the probability that the elapsed time between the second and third birds exceeds fifteen minutes?

Assuming that the time between the arrival of two birds consecutive follows th exponential distribution and we need that this time exceeds fifteen minutes. If we convert the 15 minutes to hours we have 15(1/60)=0.25 hours. And we want to find this probability:

P(T\geq 0.25h)

And we can use the result obtained from the definitions and we have this:

P(T\geq 0.25h)=e^{-(6)0.25}=0.22313

c. If you have already waited five minutes for the first bird to arrive, what is the probability that the bird will arrive within the next five minutes?

First we need to convert the 5 minutes to hours and we got 5(1/60)=0.0833h. And on this case we want a conditional probability. And for this case is good to remember the "Markovian property of the Exponential distribution", given by :

P(T \leq a +t |T>t)=P(T\leq a)

Since we have a waiting time for the first bird of 5 min = 0.0833h and we want that the next bird will arrive within 5 minutes=0.0833h, we can express on this way the probability of interest:

P(T\leq 0.0833+0.0833| T>0.0833)

P(T\leq 0.1667| T>0.0833)

And using the Markovian property we have this:

P(T\leq 0.0833)=1-e^{-(6)0.0833}=0.39347

3 0
3 years ago
Mohit pays rupees 9000 as an amount on the sum of rupees 7000 that he had borrowed for 2 years.
Sphinxa [80]

Answer:

The required rate of interest   =   14.285%

Step-by-step explanation:

Here, The principal amount = 7,000

The final paid amount = 9,000

Time = 2 years

Let us assume the rate of interest = R

Now, Simple Interest  =  Amount - Principal

                                      =  9,000 - 7,000  = 2,000

So, the SI on the given principal for 2 years  = 2,000

\textrm{SIMPLE INTEREST } = \frac{P \times  R \times T}{100}\\\implies 2000 = \frac{7,000 \times  R \times 2}{100}\\\implies R = \frac{2,000 \times 100}{7,000 \times 2}  = 14.285

or, R =  14.285%

Hence, the required rate of interest   =   14.285%

8 0
3 years ago
solve the equation below for y. 8x − 2y = 24 a. y = 4x − 12 b. y = 12 − 4x c. y = 8x − 24 d. y = 4x − 8
Feliz [49]
 the answer is a. 4x-12
7 0
3 years ago
11000 is compounded semiannually at a rate of 12% for 21 years. Find the total amount in the compound interest account
marishachu [46]

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\dotfill &\$11000\\
r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semiannually, thus twice}
\end{array}\dotfill &2\\
t=years\dotfill &21
\end{cases}


\bf A=11000\left(1+\frac{0.12}{2}\right)^{2\cdot 21}\implies A=11000(1.06)^{42}\implies A\approx 127127.359416

5 0
4 years ago
Is this directly proportional?<br> Speed and time, if the distance traveled is 120 km.
Andru [333]

Answer:

No

Step-by-step explanation:

If the speed rises then it will take you less time

If the speed decreases then it will take you more time

8 0
3 years ago
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