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masya89 [10]
2 years ago
15

The point P(8, −3) lies on the curve y = 3/(7 − x). (a) If Q is the point (x, 3/(7 − x)), use your calculator to find the slope

mPQ of the secant line PQ (correct to six decimal places) for the following values of x. (i) 7.9 mPQ =
Mathematics
1 answer:
yan [13]2 years ago
7 0

Answer:

Slope of the line PQ is -63.434948.

Step-by-step explanation:

Given that,

The point P(8,-3) lies on the curve y=\frac{3}{7-x}.

If Q is the point lies on (x,\frac{3}{7-x} ).

To find:- Find the slope of line PQ.

So,  

The coordinates of point Q when it lies on (x,\frac{3}{7-x} )

        if x=1 then y= \frac{3}{7-1} =\frac{3}{6} =\frac{1}{2}

       So,   Q ≡ (1,\frac{1}{2} ) and many points can be calculated by given Equation.

Using the formula when two points (x_{1} ,y_{1} ) \& (x_{2}, y_{2}  ).

                Slope=Tan\theta = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

Then, substituting the coordinates we get,

             Slope = \frac{1-8}{\frac{1}{2}-(-3) }

             Slope = \frac{-7}{\frac{1}{2}+3 } = \frac{-7}{\frac{7}{2} }

             Slope = \frac{-14}{7}=-2

              tan\theta=-2   ⇒  \theta = tan^{-1} (-2)

               \theta= -63.434948

Therefore,

Slope of the line mPQ is -63.434948.

                     

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Memory module consists of 9 chips. The device is designed with redundancy so that it works even if one of its chips is defective
soldier1979 [14.2K]

Answer:

a) P[C]=p^n

b) P[M]=p^{8n}(9-8p^n)

c) n=62

d) n=138

Step-by-step explanation:

Note: "Each chip contains n transistors"

a) A chip needs all n transistor working to function correctly. If p is the probability that a transistor is working ok, then:

P[C]=p^n

b) The memory module works with when even one of the chips is defective. It means it works either if 8 chips or 9 chips are ok. The probability of the chips failing is independent of each other.

We can calculate this as a binomial distribution problem, with n=9 and k≥8:

P[M]=P[C_9]+P[C_8]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)\\\\P[M]=p^{8n}(p^{n}+9(1-p^n))\\\\P[M]=p^{8n}(9-8p^n)

c)

P[M]=(0.999)^{8n}(9-8(0.999)^n)=0.9

This equation was solved graphically and the result is that the maximum number of chips to have a reliability of the memory module equal or bigger than 0.9 is 62 transistors per chip. See picture attached.

d) If the memoty module tolerates 2 defective chips:

P[M]=P[C_9]+P[C_8]+P[C_7]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1+\binom{9}{7}P[C]^7(1-P[C])^2\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])+36P[C]^7(1-P[C])^2\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)+36p^{7n}(1-p^n)^2

We again calculate numerically and graphically and determine that the maximum number of transistor per chip in this conditions is n=138. See graph attached.

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