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ohaa [14]
3 years ago
5

I need help with math please​

Mathematics
1 answer:
Law Incorporation [45]3 years ago
5 0

Answer:

What is the question ?

Step-by-step explanation:

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5.916 in word and expanded form
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Five. nine thousand and sixteen
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A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude
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Answer:

a) 20.61%

b) 21.82%

c) 42.36%

d) 4 withdrawals

Step-by-step explanation:

This situation can be modeled with a binomial distribution, where p = probability of “success” (completing the course) equals 80%  = 0.8 and the probability of “failure” (withdrawing) equals 0.2.

So, the probability of exactly k withdrawals in 20 cases is given by

\large P(20;k)=\binom{20}{k}(0.2)^k(0.8)^{20-k}

a)

We are looking for

P(0;20)+P(0;1)+P(0;2) =  

\large \binom{20}{0}(0.2)^0(0.8)^{20}+\binom{20}{1}(0.2)^1(0.8)^{19}+\binom{20}{2}(0.2)^2(0.8)^{18}=

0.0115292150460685 + 0.0576460752303424 + 0.136909428672063 = 0.206084718948474≅ 0.2061 or 20.61%

b)

Here we want P(20;4)

\large P(20;4)=\binom{20}{4}(0.2)^4(0.8)^{16}=0.218199402\approx 0.2182=21.82\%

c)

Here we need

\large \sum_{k=4}^{20}P(20;k)=1-\sum_{k=1}^{3}P(20;k)

But we already have P(0;20)+P(0;1)+P(0;2) =0.2061 and

\large \sum_{k=1}^{3}P(20;k)=0.2061+P(20;3)=0.2061+0.205364 \approx 0.4236=42.36\%

d)

For a binomial distribution the <em>expectance </em>of “succeses” in n trials is np where p is the probability of “succes”, and the expectance of “failures” is nq, so the expectance for withdrawals in 20 students is 20*0.2 = <em>4 withdrawals.</em>

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3 years ago
Which expression correctly shows x^6+2x^3+1 factored completely over the integers?
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Answer:

Option D, (x + 1)^2(x^2 - x + 1)^2

Step-by-step explanation:

<u>Step 1:  Factor</u>

x^6 + 2x^3 + 1

<em>(x + 1)^2(x^2 - x + 1)^2</em>

<em />

Answer:  Option D, (x + 1)^2(x^2 - x + 1)^2

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