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luda_lava [24]
4 years ago
4

Given a real number x and a positive integer k, determine the number of multiplications used to find x2k starting with x 24 2k a

nd successively squaring (to find x , x , and so on). Is this a more efficient way to find x than by multiplying x by itself the appropriate number of times
Mathematics
1 answer:
astra-53 [7]4 years ago
7 0

Answer:

In 1st case the multiplication is to be done for the k times, in the second case the number of multiplication is given as 2^k-1 times.

As the value of k will always be less than that of  2^k-1, thus the case 1 is an efficient way of finding the values.

Step-by-step explanation:

Case No 1 is given as:

In this case  the value of {x^2}^k is found by subsequent squaring such that

{{x^2}^k}=x(x^2.x^4.x^8......{x^2}^k)\\{{x^2}^k}=x({x^2}^1.{x^2}^2.{x^2}^3......{x^2}^k)\\

So in this case the multiplication is to be done for the k times.

Case No 2 is given as

{x^2}^k=x(x.x.x.x.x...x)  2^k-1 times

So in this case the number of multiplication is given as 2^k-1 times.

As the value of k will always be less than that of  2^k-1, thus the case 1 is an efficient way of finding the values.

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lisov135 [29]

Answer:

(a) 0.50928

(b) 0.857685.

Step-by-step explanation:

We are given that an engineer is going to redesign an ejection seat for an airplane. The new population of pilots has normally distributed weights with a mean of 155 lb and a standard deviation of 29.2 lb i.e.;                                                         \mu = 160 lb  and \sigma = 27.5 lb

(A) We know that Z = \frac{X - \mu}{\sigma} ~ N(0,1)

Let X = randomly selected pilot  

If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)

P(150 < X < 201) = P(X < 201) - P(X <= 150)

P(X < 201) = P( \frac{X - \mu}{\sigma} < \frac{201 - 155}{29.2} ) = P(Z < 1.57) = 0.94179

P(X <= 150) = P( \frac{X - \mu}{\sigma}  < \frac{150 - 155}{29.2} ) = P(Z < -0.17) = 1 - P(Z < 0.17) = 1 - 0.56749

                                                                                                   = 0.43251

Therefore, P(150 < X < 201) = 0.94179 - 0.43251 = 0.50928 .

(B) We know that for sampling mean distribution;

           Z = \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);

 P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)

P(X bar < 201) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{201 - 155}{\frac{29.2}{\sqrt{39} } } ) = P(Z < 9.84) = 1 - P(Z >= 9.84)

                                                                                  = 0.999995

P(X bar <= 150) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{150 - 155}{\frac{29.2}{\sqrt{39} } } ) = P(Z < -1.07) = 1 - P(Z < 1.07)

                                                                                   = 1 - 0.85769 = 0.14231

Therefore,  P(150 < X bar < 210) = 0.999995 - 0.14231 = 0.857685.

C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.

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