A geometric sequence, g(n) starts 20, 60, .... define g for the nth term

when multiplie discount are used if an item is first discount by 40%,then again by 30%.then again by 10%,the final or cumulative

nadezda [96]

The answer is : 40+30=70
70+10=80 therfore the final or cumulative discount is 80%

4 0

3 years ago

The volume of a cylinder is pi times the radius r squared multiplied by the height h. Write an expression for the volume.

Paul [167]

V= Area of the circle at the base • height =
Pi•r^2 •h

7 0

3 years ago

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

Please help me, i'm trying to graduate!!!

Svet_ta [14]

If you use this equation then you say that the ground is h=0 and solve as a quadratic.
The quadratic formula is (-b±<span>√(b^2-4ac))/2a when an equation is in the form ax^2 + bx + c
So the equation you have been given would be -16t^2-15t-151 = 0
This equation has no real roots which leads me to believe it is incorrect.
This is probably where your difficulty is coming from, it's a mistake.

The equation is formed from S=ut+(1/2)at^2+(So) where (So) is the initial height and S is the height that you want to find.
In this case you want S = 0.
If the initial height is +151 and the initial velocity and acceleration are downwards (negative) and the initial velocity (u) is -15 and the initial acceleration is -32 then you get the equation S=-15t-16t^2+151

Solving this using the quadratic formula gives you t = 2.64 or t = -3.58
Obviously -3.58s can't be the answer so you're left with 2.64 seconds.

Hope this makes sense.
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3 0

4 years ago

Read 2 more answers

Twice x, plus 8, is the sum as -10

Fofino [41]

Answer:

×=-9

Step-by-step explanation:

2x+8=-10

2x=-18

×=-9

3 0

4 years ago

Read 2 more answers