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Studentka2010 [4]
3 years ago
10

What is the image point of (0,-9)after a translation right 5 units and down 1 unit?

Mathematics
1 answer:
nekit [7.7K]3 years ago
6 0

Answer:

I think 0 -10 or mabe 0,-11

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Over 500 million tweets are sent per day (digital Marketing ramblings website, December 15, 2014). Assume that the number of twe
Aleksandr [31]

Answer:

(a) The probability that Bob receives no tweets during his lunch hour is 0.0002.

(b) The probability that Bob receives at least 4 tweets during his lunch hour is 0.9190.

(c) The expected number of tweets Bob receives during the first 30 minutes of his lunch hour is 3.5.

(d) The probability that Bob receives no tweets during the first 30 minutes of his lunch hour is 0.0302.

Step-by-step explanation:

Let <em>X</em> = number of tweets.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 7.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-7}(7)^{x}}{x!};\ x=0, 1, 2, 3...

(a)

Compute the probability that Bob receives no tweets during his lunch hour as follows:

P(X=0)=\frac{e^{-7}(7)^{0}}{0!}\\=\frac{0.000192\times1}{1} \\=0.000192\\\approx0.0002

Thus, the probability that Bob receives no tweets during his lunch hour is 0.0002.

(b)

Compute the probability that Bob receives at least 4 tweets during his lunch hour as follows:

P (X ≥ 4) = 1 - P (X < 4)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

              =1-\frac{e^{-7}(7)^{0}}{0!}-\frac{e^{-7}(7)^{1}}{1!}-\frac{e^{-7}(7)^{2}}{2!} - \frac{e^{-7}(7)^{3}}{3!}\\=1-0.0002-0.0064-0.0223-0.0521\\=0.9190

Thus, the probability that Bob receives at least 4 tweets during his lunch hour is 0.9190.

(c)

The average number of tweets in 60 minutes is 7.

Then the average number of tweets in 1 minute is, \frac{7}{60}.

Hence, the average number of tweets during 30 minutes is, \frac{7}{60}\times30=3.5

Thus, the expected number of tweets Bob receives during the first 30 minutes of his lunch hour is 3.5.

(d)

Compute the probability that Bob receives no tweets during the first 30 minutes of his lunch hour as follows:

P(No\ tweets) = \frac{e^{-3.5}(3.5)^{0}}{0!}=\frac{0.0302\times1}{1} =0.0302

Thus, the probability that Bob receives no tweets during the first 30 minutes of his lunch hour is 0.0302.

3 0
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Round the number 78.372 to the nearest tenth.
notka56 [123]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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It said that a poll found that 61% of a random sample of 1078 adults said that they believe in ghosts which gives a less believable percentage they are looking for a 90%  of a random sample of 1078 adults said that they believe in ghosts.

Hope this helped!

Can I have brainliest please?

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