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3 years ago

PLEASE HELP NO LINKS OR UNNECESSARY COMMENTS.

Mathematics

1 answer:

DENIUS [597]3 years ago

3 0

To find Interquartile rage you subtract quartile 1 from quartile 3

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Adding Unlike denominators 1/11 + 1/4

castortr0y [4]

Answer: 15/44

Step-by-step explanation:

You have to make the denominators match so:

1/11(4) + 1/4(11)=

4/44 + 11/44=

15/44

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3 years ago

(7th grade work) How many gallons of water were in the bucket after 1 second? Explain your thinking.

Veseljchak [2.6K]

Answer:0.5, half a gallon

Step-by-step explanation:

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3 years ago

Read 2 more answers

ILL GIVE YOU BRAINLIST !!! SHOW WORK

alexgriva [62]

Answer:

-0.004 repeating

Step-by-step explanation:

5 times negative three is -15 and that to the negative second power is -0.004 with the 4 repeating.

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2 years ago

What are some things you like about science, or about science experiments?

liraira [26]

I think this question is for you to answer it with your own opinion, think about it and just answer it with what you think :)

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3 years ago

Messages arrive to a computer server according to a Poisson distribution with a mean rate of 11 per hour. Determine the length o

Zarrin [17]

Answer:

The length of the interval during which no messages arrive is 90 seconds long.

Step-by-step explanation:

Let X = number of messages arriving on a computer server in an hour.

The mean rate of the arrival of messages is, λ = 11/ hour.

The random variable X follows a Poisson distribution with parameter λ = 11.

The probability mass function of X is:

$P(X=x)=\frac{e^{-11}11^{x}}{x!};\ x=0,1,2,3....$

It is provided that in t hours the probability of receiving 0 messages is,

P (X = 0) = 0.76

Compute the value of t as follows:

$P(X=0)=\frac{e^{-11\times t}(11\times t)^{0}}{0!}\\0.76=e^{-11\times t}\\\ln(0.76)=-11\times t\\t=-\frac{\ln(0.76)}{11} \\=0.025\ hours\\\approx90\ seconds$

Thus, the length of the interval during which no messages arrive is 90 seconds long.

6 0

3 years ago