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NeX [460]
4 years ago
7

One half the sum of four consecutive multiples of 6 is 6 more than twice the 3rd highest of the multiples. What is the sum of th

e two highest multiples?
Mathematics
1 answer:
MAXImum [283]4 years ago
3 0
I solved this by hand first, then cymath, then Wolfram alpha. we all agree the solution to the equation involved is x=-3 however -3 is not a multiple of 6. if you plug it in you get a solution of 24 but I don't consider this to be a valid question.
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Answer:

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Step-by-step explanation:

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4X=360-90

4X=270

X=67.5

7 0
3 years ago
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bulgar [2K]
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,0)0.1^0(1-0.1)^{3-0}
=1(1)(0.729)
=0.729

(b) x=2
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,2)0.1^2(1-0.1)^{3-2}
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(c) x ≥ 2
P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}
=P(2)+P(3)
=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}
=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}
=3(0.01)(0.9)+1(0.001)1
=0.027+0.001
=0.028


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Answer:

\sqrt{x} with x being square rooted.

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Answer:

x= -3 is the answer

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