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ollegr [7]
3 years ago
15

The sum of three consecutive even integers is -78, what are the integers ?

Mathematics
1 answer:
Digiron [165]3 years ago
5 0
First, you need to analize and understand the problem, then you must choose and strategy. There is a wide variety of strategies to solve a mathematical problem, in this case, you can use the following, which is based on the information given above:
 1. You have that t<span>he sum of three consecutive even integers is </span>-78<span>, therefore, you can given the variable </span>x&#10;<span> to the first integer, the second even integer is </span>x+2<span> and the third one is </span>x+4&#10;<span>.
 2. Calculate x:
 </span>x+(x+2)+(x+4)=-78<span>
 </span>3x+6=-78
<span> </span>x=-28&#10;<span>
 </span>x+2=-28+2=-26&#10;<span>
 </span>x+4=-28+4=-24&#10;<span>
 Therefore, the answer is:</span>-28,-26,-24&#10;
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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
Ymorist [56]

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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Answer:

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Step-by-step explanation:

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