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3 years ago

|r|=-2 solve the equation and graph the solution if possible

1 answer:

6 0

That bracket symbol, with an r in it, means "the absolute value of r".
So r's absolute value is -2 meaning r is two units from zero
r= 2

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2 years ago

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Round 6.071981 to 3 significant figures

natita [175]

Answer:

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3 years ago

Rewrite in simplest radical form! Show your work.

vazorg [7]

$x^{\frac{1}{\frac{-3}{6} }}$

First, let's deal with the fraction in the denominator of the exponent. Multiply the top and bottom of the exponent by 6.

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3 years ago

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Mrac [35]

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3 years ago

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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

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

3 0

3 years ago