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3 years ago

Please help me solve this

Mathematics

1 answer:

OlgaM077 [116]3 years ago

4 0

Answer:

645320

Step-by-step explanation:

Given: 6_5_2_

If a multiplication sign is placed between the second and third digit, and an equal sign is placed between the third and fourth digit, then

6_ × 5 = _2_

The passcode contains 6 DIFFERENT digits, so we know the missing numbers cannot be 6, 5, or 2.

So, do a trial and error session of multiplying 5 by numbers it could possibly be (60, 61, 63, 64, 67, 68, 69).

64 × 5 = 320. It is the only equation that is both correct and matches the digits given, so the missing digits are 4, 3, and 0. Barry's passcode is 645320.

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Find the Product<br>2/15*4

Fynjy0 [20]

Answer:

0.5333333333

3 is infinite

6 0

4 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:

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

$P(T\leq 0.1667| T>0.0833)$

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

At one farmer's market, bananas cost $0.80 per pound. At another farmer's market, bananas are sold in 5-pound bags for $4.50 per

kozerog [31]

Divide the second amount by five to find out how much it costs at the second farmers market to get 1 pound of bananas.
5/5=1
4.50/5=.9
.9>.8 so the first farmers market had a better deal.

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lesya [120]

Answer:

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Step-by-step explanation:

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