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goldenfox [79]
4 years ago
6

You are asked to distribute 10 balls into three groups: 2 balls, 3 balls, and 5 balls. (a) if the 10 balls are indistinguishable

, how many different arrangements can you have
Mathematics
1 answer:
Valentin [98]4 years ago
7 0
<span>In how many ways can 20 identical balls be distributed into 4 distinct boxes subject ? a) With no constraints. b) Each box gets at least two balls. c) Box 1 gets exactly one ball and box 2 gets at least one ball. d) Each box gets an even number of balls.</span>
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Answer please! I need answers
blagie [28]

A light-year is how astronomers measure distance in space. It’s defined by how far a beam of light peregrinates in one year – a distance of six trillion miles. Cerebrate of it as the more sizably voluminous, badder cousin of the inch, the mile, the kilometer, and the furlong. If you relish to keep up with what’s going on in astronomy, it’s worth spending scarcely of time understanding what the deal is with this hysterical unit of quantification. Your answer is  C 6 x 10 12

8 0
4 years ago
pi/3 is the reference angle for: A. 2pi/3 B. 15pi/3 C. 7pi/3 D. 19pi/3 **its CHOOSE ALL THAT APPLY!!!:) my answers: C. 7pi/3 and
inn [45]

Answer:

yes those are right but 2pi/3 is also right.

Step-by-step explanation:

6 0
4 years ago
Read 2 more answers
Huck and Jim are waiting for a raft. The number of rafts floating by over intervals of time is a Poisson process with a rate of
Harrizon [31]

Answer: 0.0081

Step-by-step explanation:

Let X be the number of rafts.

Given : The mean number of rafts floating : \lambda=0.4 rafts per day .

Then , for 7 days the number of rafts = \lambda_1=\lambda\times 7=0.4\times7=2.8 rafts per day .

The formula to calculate the Poisson distribution is given by :_

P(X=x)=\dfrac{e^{-\lambda_1}\lambda_1^x}{x!}

Now, the  probability that they will have to wait more than a week is given by :-

P(X>7)=1-P(X\leq7)\\\\=1-(P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(5)+P(6)+P(7))\\\\=1-(\dfrac{e^{-2.8}2.8^{0}}{0!}+\dfrac{e^{-2.8}2.8^{1}}{1!}+\dfrac{e^{-2.8}2.8^{2}}{2!}+\dfrac{e^{-2.8}2.8^{3}}{3!}+\dfrac{e^{-2.8}2.8^{4}}{4!}+\dfrac{e^{-2.8}2.8^{5}}{5!}+\dfrac{e^{-2.8}2.8^{6}}{6!}+\dfrac{e^{-2.8}2.8^{7}}{7!})\\\\=1-0.991869258012=0.008130741988\approx0.0081

Hence, the required probability : 0.0081

4 0
3 years ago
Hey I need help on how to subtract 56,853 - 26,586
Aneli [31]

Answer:

The answer is 30,267.

Step-by-step explanation:

I got this by simply putting it in a calculator, just like you wrote it. Hope I helped!

7 0
4 years ago
Plz help it’s due tonight i rlly need help thanks
belka [17]
i think we need some more information or something is there any other information that they give you for this question?
7 0
3 years ago
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