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Eva8 [605]
3 years ago
11

You roll a six-sided number cube and flip a coin. What is the probability of rolling a number less than 2 and flipping heads? *

Mathematics
2 answers:
leonid [27]3 years ago
7 0

Answer:

1/12

Step-by-step explanation:

First, you should already be able to understand that the only number less than 2 on this cube is 1. So, you can understand that there is a 1/6 chance because there is a total of 6 numbers, and out of those, only 1 of those are your answer.

Then, you will note that a coin only has 2 sides, so there is a 50/50 chance of getting heads, or 50%. Now, you will turn that into a fraction, which is 1/2.

Next, you will multiply to make the denominators the same. And what you do to the denominator must happen to the numerator. (1/6 * 2--> 1/12 and 1/2 * 6 = 1/12)

Hope this helps! :)

katrin [286]3 years ago
7 0
1/12 that the answer
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The problem can be represented by the the exponential growth formula which is :
P(t) = A * r^{t}
Where: t ⇒ time , A ⇒ initial amount , r ⇒ rate of increase 
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For the given problem:
initial amount = A = $278,640
<span>predicted increase in value per year = 4% =0.04
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<span>∴ r = 1 + 0.04 = 1.04
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<span>for t = 18 years
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∴ P(t) = A * r^{t} =278,640 * 1.04^{18}=\framebox{564,473.5}<span />
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Step-by-step explanation:

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According to a recent Census Bureau report, 12.7% of Americans live below the poverty level. Suppose you plan to sample at rando
larisa [96]

Answer:

a) P(X=10)=(100C10)(0.127)^{10} (1-0.127)^{100-10}=0.0928

b) P(X \leq 10) = 0.2614

c) (1-0.127)^7 (0.127) =0.0491

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=100, p=0.127)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

a. What is the probability that you count exactly 10 in poverty?

For this case we want this probability P(X=10)

P(X=10)=(100C10)(0.127)^{10} (1-0.127)^{100-10}=0.0928

b. What is the probability that you count 10 or less in poverty?  .2614

For this case we want this probability P(X=\leq10)

P(X\leq10)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)

And we can find the individual probabilities like this:

P(X=0)=(100C0)(0.127)^{0} (1-0.127)^{100-0}=1.263x10^{-6}

P(X=1)=(100C1)(0.127)^{1} (1-0.127)^{100-1}=1.837x10^{-5}

P(X=2)=(100C2)(0.127)^{2} (1-0.127)^{100-2}=0.000132

P(X=3)=(100C3)(0.127)^{3} (1-0.127)^{100-3}=0.000629

P(X=4)=(100C4)(0.127)^{4} (1-0.127)^{100-4}=0.00222

P(X=5)=(100C5)(0.127)^{5} (1-0.127)^{100-5}=0.00620

P(X=6)=(100C6)(0.127)^{6} (1-0.127)^{100-6}=0.0143

P(X=7)=(100C7)(0.127)^{7} (1-0.127)^{100-7}=0.0279

P(X=8)=(100C8)(0.127)^{8} (1-0.127)^{100-8}=0.0471

P(X=9)=(100C9)(0.127)^{9} (1-0.127)^{100-9}=0.0701

P(X=10)=(100C10)(0.127)^{10} (1-0.127)^{100-10}=0.0928

And then repplacing we got:

P(X \leq 10) = 0.2614

c. What is the probability that you start taking the random sample and you find the first person in poverty on the 8th person selected?  .0491

For this case we need after 7 people , 1 in poverty so we can find this probability like this:

(1-0.127)^7 (0.127) =0.0491

7 0
3 years ago
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