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olganol [36]
3 years ago
7

-36 • (-1/3) divided by 4 A. -48 B. -3 C. 2 D. 3

Mathematics
2 answers:
Alexxx [7]3 years ago
3 0

       ➣ Assignment: \bold{Solve \ Equation: \ -36\left(-1/3\right)\div \:4}

↔↔↔↔↔↔↔↔

       ➣ Answer: \boxed{\bold{3}}

↔↔↔↔↔↔↔↔

Explanation: ↓↓↓↓↓

↔↔↔↔↔↔↔↔

       ➤ [ Step One ] Follow PEMDAS Order Of Operations

       ➤ Note: \bold{Parenthesis, \ Exponents, \ Multiply, \ Divide, \ Add, \ Subtract}

\bold{Apply \ Rule: \ -a\cdot \left(-b\right)\:=\:a\cdot \:b}

\bold{-36\left(-1/3\right)=36\cdot \:1/3=12}

↔↔↔↔↔↔↔↔

\bold{12 \ \div \ 4}

       ➤ [ Step Two ] Solve

\bold{3}

↔↔↔↔↔↔↔↔

\bold{\rightarrow Mordancy \leftarrow}

garik1379 [7]3 years ago
3 0

D. Because -36 times -1/3 is 12 and 12 divided by 4 is 3
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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

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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}

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(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}

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(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}

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(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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