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ASHA 777 [7]
3 years ago
11

What length is the shortest path from A to G in the graph below?

Mathematics
1 answer:
umka2103 [35]3 years ago
3 0

Answer:

graph a

Step-by-step explanation:

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Help me with this question <br>I will mark brainlist <br>no web link <br><br>Solve :-<br>9n²-64=0​
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The Kendal Company manufactures automobile engine parts. Running the manufacturing operations requires 12 watts (units of electr
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The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medic
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Answer:

(a) 0.00605

(b) 0.0403

(c) 0.9536

(d) 0.98809

Step-by-step explanation:

We are given that 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012) and suppose ten first-round appeals have just been received by a Medicare appeals office.

This situation can be represented through Binomial distribution as;

P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....

where,  n = number of trials (samples) taken = 10

            r = number of success

            p = probability of success which in our question is % of first-round

                   appeals that were successful, i.e.; 40%

So, here X ~ Binom(n=10,p=0.40)

(a) Probability that none of the appeals will be successful = P(X = 0)

     P(X = 0) = \binom{10}{0}0.40^{0}(1-0.40)^{10-0}

                   = 1*0.6^{10} = 0.00605

(b) Probability that exactly one of the appeals will be successful = P(X = 1)

     P(X = 1) = \binom{10}{1}0.40^{1}(1-0.40)^{10-1}

                  = 10*0.4^{1} *0.6^{10-1} = 0.0403

(c) Probability that at least two of the appeals will be successful = P(X>=2)

    P(X >= 2) = 1 - P(X = 0) - P(X = 1)

                     = 1 - \binom{10}{0}0.40^{0}(1-0.40)^{10-0} - \binom{10}{1}0.40^{1}(1-0.40)^{10-1}

                     = 1 - 0.00605 - 0.0403 = 0.9536

(d) Probability that more than half of the appeals will be successful =             P(X > 0.5)

  For this probability we will convert our distribution into normal such that;

   X ~ N(\mu = n*p=4,\sigma^{2}= n*p*q = 2.4)

  and standard normal z has distribution as;

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

  P(X > 0.5) = P( \frac{X-\mu}{\sigma} > \frac{0.5-4}{\sqrt{2.4} } ) = P(Z > -2.26) = P(Z < 2.26) = 0.98809

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