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Nonamiya [84]
3 years ago
11

What is wrong with this "proof"? "Theorem" For every positive integer n,n i =1 i = (n + 1 2 ) 2 /2. Basis Step: The formula is t

rue for n = 1. Inductive Step: Suppose thatn i=1 i = (n + 1 2 ) 2 /2. Then n+1 i=1 i = ( n i=1 i) + (n + 1). By the induc-tive hypothesis, n+1 i=1 i = (n + 1 2 ) 2 /2 + n + 1 = (n 2 + n + 1 4 )/2 + n + 1 = (n 2 + 3n + 9 4 )/2 = (n + 3 2 ) 2 /2 =[ (n + 1) + 1 2 ] 2 /2, completing the induc-tive step.
Mathematics
1 answer:
creativ13 [48]3 years ago
7 0

Answer:

The basis step is not even true.

Step-by-step explanation:

The basis step would be for     n = 1.

If the basis step was true then you would have that.

\sum\limits_{i=1}^{1} i  =  \frac{(1+1/2)^2}{2}

Now. When you say

                                            \sum\limits_{i=1}^{1}  i            

That's just a fancy way of writing    1.  

On the other hand    

\frac{(1+1/2)^2}{2}  = \frac{9}{4} = 2.25

Therefore  

  1 \neq 2.25

And the basis hypothesis is false.

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Can someone help me find the equivalent expressions to the picture below? I’m having trouble
miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is \frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}.

Now we will solve this expression with the help of law of exponents.

\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}

           =2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}

           =2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}

           =2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}

           =2^{\frac{2}{9}}\times 3^{-\frac{2}{3} } [Option 2]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2 [Option 1]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2

                =(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }

                =\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} } [Option 3]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }

               =\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}} [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

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Find the indicated probability.
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Answer:

C) 0.179

Step-by-step explanation:

Since the trials are independent, this is a binomial distribution:

<u>Recall:</u>

  • Binomial Distribution --> P(k)={n\choose k}p^kq^{n-k}
  • P(k) denotes the probability of k successes in n independent trials
  • p^k denotes the probability of success on each of k trials
  • q^{n-k} denotes the probability of failure on the remaining n-k trials
  • {n\choose k}=\frac{n!}{(n-k)!k!} denotes all possible ways to choose k things out of n things

<u>Given:</u>

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  • {n\choose k}={10\choose 4}=\frac{10!}{(10-4)!4!}=210

<u>Calculate:</u>

  • P(4)=(210)(0.53^4)(0.47^6)=0.1786117069\approx0.179

Therefore, the probability that the archer will get exactly 4 bull's-eyes with 10 arrows in any order is 0.179

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35 divided by 2 1/2

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35 * 2/5

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