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skad [1K]
3 years ago
8

Find the exact value of the 8th term of the geometric sequence: 648,216,72,24

Mathematics
1 answer:
VLD [36.1K]3 years ago
6 0
The sequence multiplies by 1/3.
Using the formula for a geometric sequence:
a(n) = a(1) * r^(n-1)
So...
a(n) = 648 * (1/3)^(n-1)
Plugging in 8 for n produces:
a(n) = 648 * (1/3)^(7)
Using a calculator, I find the answer to be 0.296296296... 
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Step-by-step explanation:

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45 POINTS Evaluate the limit as h goes to 0 of the quotient of the quantity the 4th power of 3 plus h minus 81 and h. SEE PICTUR
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Answer:

108

Step-by-step explanation:

View Image.

Whenever you have a limit and the answer to that limit is one of the indeterminant form, you can use the L'Hopital's Rule to solve for the limit.

<u>Indeterminant forms:</u>

\frac{0}{0} , \frac{\infty}{\infty}, 0\cdot \infty, \infty - \infty, 0^{0}, \infty^{0}, 1^{\infty}

<u>L'Hopitals Rule:</u>

L'Hopitals rule is just taking the derivative of the numerator, and the derivative of the denominator separately.

After doing the L'Hopital's rule, take the limit again. If the result is an actual number then you got your answer.

If the result is another indeterminant form then use L'Hopitals rule again and check the limit again after that. Repeat the process until you get your answer.

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Bill and George go target shooting together. Both shoot at atarget at the same time. Suppose Bill hits the target withprobabilit
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Answer: (a) \frac{2}{9}       (b) \frac{6}{41}

Step-by-step explanation:

(a) P( Bill hitting the target) = 0.7        P( Bill not hitting the target) = 0.3

    P( George hitting the target) = 0.4     P(George not hitting the target) = 0.6

Now the chances that exactly one shot hit the target is = 0.7 x 0.6 + 0.4 x 0.3

                                                                                            = 0.54

Chances that George hit the target is = 0.4 x 0.3 = 0.12

So given that exactly one shot hit the target, probability that it was George's shot = \frac{0.12}{0.54} = \frac{2}{9} .

(b) The numerator in the second part would be the same as of (a) part which is 0.12.

The change in the denominator will be that now we know that the target is hit so now in denominator we include the chance of both hitting the target at same time that is 0.4 x 0.7 and the rest of the equation is same as above i.e.

Given that the target is hit,probability that George hit it =                                                 \frac{0.4\times 0.3}{0.7\times 0.6 +0.4\times 0.3+0.4\times 0.7}  = = \frac{6}{41}                                                                        

                                                                                           

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