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3 years ago

8k-1=15 ? Help I have a question were we have to work this out what do I do?

2 answers:

7 0

You have to add 1 to the -1 and you will get 0.What you have to do next is also add 1 to 15 because you want a variable on one side and a number on the other.So now you have 8k=16. So you divide 8 to 8k and you get nothing divide 8 and 16 and your answer for k =2 hope you get it right.trust mw it is very easy when you get use to it

adell [148]3 years ago

5 0

$8k-1=15\ \ \ \ /+1\\\\8k-1+1=15+1\\\\8k=16\ \ \ \ /:8\\\\8k:8=16:8\\\\k=2$

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Which of the ordered pairs below satisfy the following equation? Check all that apply. y = 7x A. (84, 12) B. (42, 7) C. (12, 84)

m_a_m_a [10]

So you have many ordered pairs here that look like (x , y) So you take each one, for example (84,12) and plug it into that equation you have y=7x (84 , 12) means (x = 84, y = 12 ) so we plug those in 12=7(84) Does that come out true? if it doe...we have a solution..if not...obviously not :)

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3 years ago

Flight 202's arrival time is normally distributed with a mean arrival time of 6:30 p.m. and a standard deviation of 15 minutes.

natta225 [31]

Answer:

The probability is 0.0015

Step-by-step explanation:

We know that the mean $\mu$ is:

$\mu=6:30\ p.m$

The standard deviation $\sigma$ is:

$\sigma=0:15\ minutes$

The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

We seek to find

$P(x>7:15\ p.m.)$

The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{7:15-6:30}{0:15}$

$Z=\frac{0:45}{0:15}$

$Z=3$

The score of Z = 3 means that 7:15 p.m. is 3 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the conficion of 3 deviations from the mean has percentage of 0.15%

$P(x>7:15\ p.m.)=P(Z>3)=0.0015$

3 0

3 years ago

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:

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