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ziro4ka [17]
2 years ago
15

Solve system of equation by elimination:

Mathematics
2 answers:
777dan777 [17]2 years ago
6 0

Answer:

its b

Step-by-step explanation:

mars1129 [50]2 years ago
4 0

Answer:

a. (4, 1)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Solving systems of equations using substitution/elimination

Step-by-step explanation:

<u>Step 1: Define Systems</u>

-x + 4y = -8

-5x - 4y = -16

<u>Step 2: Solve for </u><em><u>x</u></em>

<em>Elimination</em>

  1. Combine equations:                     -6x = -24
  2. Isolate <em>x</em>:                                        x = 4

<u>Step 3: Solve for </u><em><u>y</u></em>

  1. Define equation:                   -x + 4y = -8
  2. Substitute in <em>x</em>:                      -4 + 4y = -8
  3. Isolate <em>y</em> term:                       4y = -4
  4. Isolate <em>y</em>:                                y = -1
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Step-by-step explanation:

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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)

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  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

3 0
3 years ago
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Answer:

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now we are just the top and leave the denominator the same as before

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