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vovangra [49]
2 years ago
5

Apr 23, 8:46:54 AM What is the missing term?

Mathematics
1 answer:
kipiarov [429]2 years ago
7 0
Answer: -10x
Explanation: Multiply -2 by 5x
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2x -¬ y = 6
artcher [175]
The lines are perpendicular

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3 years ago
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The probability that it will snow on the last day of January is 85%. If the probability remains the same of the first eight day
MrRa [10]

Answer:

Here, we have:

P(5 days snow in this 8 days) = 8C5 x (0.85)^5 x (1 - 0.85)^3 = 0.084

P(6 days snow in this 8 days) = 8C6 x (0.85)^6 x (1 - 0.85)^2 = 0.238

P(7 days snow in this 8 days) = 8C7 x (0.85)^7  x (1 - 0.85)^1  = 0.385

P(8 days snow in this 8 days) = 8C8 x (0.85)^8 x (1 - 0.85)^0 = 0.272

Add up those above, then the probability that it will snow AT LEAST five of those days in February:

P = 0.084+ 0.238 + 0. 385 + 0.272 = 0.979

Hope this helps!

:)

5 0
3 years ago
Karen is starting a career as a professional wildlife photographer and plans to photograph Canadian Geese at one of the staging
balandron [24]

Answer:

$5000*0.816 = $4082

Step-by-step explanation:

It's a strange question, but based on the statement and the question it sounds like it's a poisson distribution:

* For 3 days she was able to get 2 good shots (<em>typical of that time of the year</em>)

* Good shots happen randomly

* Each day is independent of another

Let's call 'p' the probability that she makes a good shot per day

Let's call 'n' the number of days Karen is taking shots.

So, if in 3 days he got 2 good shots and that is typical at that time of the year, then the expected value for the number of good shots (X) is:

E(x)=\frac{2}{3}

For a Poisson distribution E (x)=\lambda\\\lambda= np

So:

\lambda =\frac{2}{3}

For a Poisson distribution the standard deviation is:

\sigma = \sqrt{\lambda}\\\sigma = \sqrt{\frac{2}{3}}

\sigma = 0.816 this is the standard deviation for the number of buentas taken.

So the standard deviation for income is the price of each shot per sigma

$5000*0.816 = $4082, which is the desired response.

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