Question

answer these questions 15 18 22 19 (15 is in previous question i posted) rest are here plz answer fast it's urgent

Answer 1

Answer:

18) k = 8 / 3

19)  (i)  5 / 12     (ii)  1 / 2

22) see below

Hello.  Hope this helps!!!

Step-by-step explanation:

18)  We need to know the factorization:

• x^n - a^n = ( x - a ) ( x^{n-1} + x^{n-2} a + x^{n-3} a² + ... + a^{n-1} )

In the special cases here, this means we'll use:

• x⁴ - 1 = ( x - 1 ) ( x³ + x² + x + 1 )
• x³ - k³ = ( x - k ) ( x² + xk + k² )
• x² - k² = ( x - k ) ( x + k )

First

$\displaystyle \lim_{x\rightarrow 1}\frac{x^4-1}{x-1}=\lim_{x\rightarrow 1} (x^3+x^2+x+1) = 1^3+1^2+1+1 = 4$

Next

$\displaystyle\lim_{x\rightarrow k}\frac{x^3-k^3}{x^2-k^2}\\=\lim_{x\rightarrow k}\frac{(x-k)(x^2+xk+k^2)}{(x-k)(x+k)}\\=\lim_{x\rightarrow k}\frac{x^2+xk+k^2}{x+k}\\= \frac{3k^2}{2k} = \frac{3k}{2}$

So for these two limits to be equal, we must have

3k / 2 = 4  =>  k = 8 / 3

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19) # ways of choosing 5 bulbs from 10

$\displaystyle=\binom{10}{5} = \frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1} = 2^2\times3^2\times7^1$

( I prefer to keep it in small factors rather than put "252" since this is just going to be a denominator in our probabilities and I expect (hope?) there'll be some cancellation.)

(i) # ways of choosing 5 bulbs with 2 of them being bad

= ( # ways of choosing 2 bad bulbs from 3 ) × ( # ways of choosing 3 good bulbs from 7 )

$=\displaystyle\binom{3}{2}\binom{7}{3} = \frac{3\times2}{2\times1}\times\frac{7\times6\times5}{3\times2\times1}=3\times5\times7$

So

P( getting exactly 2 bad bulbs )

= ( # ways of choosing 5 bulbs with 2 of them being bad ) / ( # ways of choosing 5 bulbs )

= ( 3 × 5 × 7 ) / ( 2² × 3² × 7 )

= 5 / 12

(ii) As before:

# ways of choosing 5 bulbs with no bad ones

= # ways of choosing 5 good bulbs from 7

$\displaystyle=\binom{7}{5} = \binom{7}{2} = \frac{7\times6}{2\times1} =21$

# ways of choosing 5 bulbs with 1 bad bulb

= ( # ways of choosing 1 bad bulbs from 3 ) × ( # ways of choosing 4 good bulbs from 7 )

$\displaystyle=\binom{3}{1}\binom{7}{4} = \binom{3}{1}\binom{7}{3}= 3\times\frac{7\times6\times5}{3\times2\times1}=5\times21$

So

# ways of choosing at most 1 bad bulb

= ( # ways of choosing 0 bad bulbs ) + ( # ways of choosing 1 bad bulb )

= 21 + 5 × 21 = 6 × 21

Finally

P( getting at most 1 bad bulb )

= ( # ways of choosing at most 1 bad bulb ) / ( # ways of choosing 5 bulbs )

= ( 6 × 21 ) / ( 2² × 3² × 7 )

= 1 / 2

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22)  Not sure what is being expected from you here.  It depends what you've been learning.

Perhaps you're expected to use the binomial expansion:

$\displaystyle( 1 + x )^n = \sum_{k=0}^n \binom{n}{k}x^k$

and the recurrence relation for binomial coefficients:

$\displaystyle\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

( this is the property that you use to write down Pascal's Triangle ).

The middle term of ( 1 + x )^{2n} is then $\binom{2n}{n}x^n$

while the two middle terms of ( 1 + x )^{2n-1} are $\binom{2n-1}{n-1}x^{n-1}$ and $\binom{2n-1}{n}x^n$.

Substituting "n" for "k" in the recurrence relation, and "2n" for "n", the recurrence relation tells us that

$\displaystyle\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$

The terms in this sum though are precisely the coefficients of the terms that we identified above in the expansions of ( 1 + x )^{2n} and ( 1 + x )^{2n-1}.

So the coefficient of the middle term of ( 1 + x )^{2n} is the sum of the coefficients of the two middle terms of ( 1 + x )^{2n-1}.

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If you're actually expected to do it more directly, then perhaps this is more what you want.

Let the middle coefficient of ( 1 + x )^{2n} be A and the middle two coefficients of ( 1 + x )^{2n-1} be B and C.

So

( 1 + x )^{2n} = 1 + ..... + A x^n + ..... + x^{2n}            (**)

and

( 1 + x )^{2n-1} = 1 + ..... + B x^{n-1} + C x^n + .... + x^{2n-1}

Multiplying the second of these by 1+x give

( 1 + x )^{2n} = ( 1 + x ) ( 1 + .... + B x^{n-1} + C x^n + .... + x^{2n-1} )

= 1 + ... + B x^{n-1} + C x^n + .... + x^{2n-1}

+ x + .... + B x^n + C x^{n+1} + .... + x^{2n}

= 1 + ..... + ( B + C ) x^n + .... + x^{2n}

Comparing this with (**), we conclude that A = B + C.

That is, the coefficient of the middle term of ( 1 + x )^{2n} is the sum of the coefficients of the two middle terms of ( 1 + x )^{2n-1}.