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AlladinOne [14]
2 years ago
8

Please help!!!!!!!!!

Mathematics
1 answer:
Maru [420]2 years ago
5 0

Answer: A

Step-by-step explanation:

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Assume that a procedure yields a binomial distribution with n=3 trials and a probability of success of p=0.300. use a binomial p
NNADVOKAT [17]
\mathbb P(X=x)=\begin{cases}\dbinom3xp^x(1-p)^{3-x}&\text{for }0\le x\le3\\\\0&\text{otherwise}\end{cases}

\implies\mathbb P(X=2)=\dbinom32(0.3)^2(0.7)^1=0.189=18.9\%
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3 years ago
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vivado [14]

Answer:

Mass

Step-by-step explanation:

"Robin collected data from dropping watermelons of different 'MASS' for the same height"

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2 years ago
Please help and explain please
butalik [34]
The greatest remainder with 3 is 2, the greatest remainder with 8 is 7, and the greatest remainder with 5 is 4, all because if it was one more then it would not be remaining and it would be without a remainder. Hope this helps!
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3 years ago
PLEASE HELP IM TIMED
seropon [69]

Answer:

first one = 15x+30y-220

second one = 2.4x-4.4

Step-by-step explanation:

7 0
2 years ago
In Vancouver, British Columbia, the probability of rain during a winter day is 0.42, for a spring day is 0.23, for a summer day
nadezda [96]

Answer:

31.82% probability that this day would be a winter day

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening

In this question:

Event A: Rain

Event B: Winter day

Probability of rain:

0.42 of 0.25(winter), 0.23 of 0.25(spring), 0.16 of 0.25(summer) or 0.51 of 0.25(fall).

So

P(A) = 0.42*0.25 + 0.23*0.25 + 0.16*0.25 + 0.51*0.25 = 0.33

Intersection:

Rain on a winter day, which is 0.42 of 0.25. So

P(A \cap B) = 0.42*0.25 = 0.105

If you were told that on a particular day it was raining in Vancouver, what would be the probability that this day would be a winter day?

P(B|A) = \frac{0.105}{0.33} = 0.3182

31.82% probability that this day would be a winter day

6 0
3 years ago
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