With out calculation the answer, is 1/3 x 7/15 greater or less than 1/3?

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

3 0

3 years ago

What is the sum of the roots of the polynomial shown below? f(x) = x^3 + 2x^2 - 11x-12

svp [43]

Answer:

Sum of the roots of the polynomial $x^{3}+2 x^{2}-11 x-12 \text { is }-2$

Solution:

The general form of cubic polynomial is $a x^{3}+b x^{2}+c x+d=0$ ---- (1)

If we have any cubic polynomial $a x^{3}+b x^{2}+c x+d=0$ having roots $\alpha , \beta , \theta$

Sum of roots $\alpha + \beta + \theta$ = $\frac{-b}{a}$ ---(2)

From question given that,

$x^{3}+2 x^{2}-11 x-12$ --- (3)

On comparing equation (1) and (3), we get a = 1, b = 2, c = -11 and d = -12

Hence the sum of roots using eqn 2 is given as,

= $\frac{-2}{1}$

= -2

Hence the sum of the roots of the polynomial $x^{3}+2 x^{2}-11 x-12 \text { is }-2$

4 0

2 years ago

Find the domain and range of the function f(x) = x -|2 cos x|

adell [148]

Answer:

6

Step-by-step explanation:

3 0

2 years ago

Hi, I need the are to this question, I just need to show my work btw

Nana76 [90]

What is the question?

6 0

3 years ago

Evaluate X + Y when X = 13.1 and Y = 2.2 A. 10.9 B. -10.9 C. 35.1 D. 15.3

bazaltina [42]

X +Y = 13.1 + 2.2
= 15.3

answer
D. 15.3

4 0

3 years ago