there are 5280 feet in a mile. what part of a mile, in decimal form, will you drive until you reach the exit? pls someone help

Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno

Roman55 [17]

Answer:

a) P(X>np+3 $\sqrt{np(1-p)}$ =0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= $\sqrt{0.198}$ ≈ 0.445

a) P(X>np+3 $\sqrt{np(1-p)}$ =P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

$\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}$

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - $\frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}$ - $\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}$

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - $\frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)}$ - $\frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}$

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)= $\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}$ =0.349

P(Y>1)=1-0.349=0.651

3 0

3 years ago

In a carnival game, the player selects a ball one at a time, without replacement, from an urn containing twotwo purplepurple

dezoksy [38]

The urn contains 2 purple balls and 4 white balls. The player pay $4 for start the game and get $1.5 for every ball drawn until one purple ball is drawn. The maximal revenue would be $7.5 when 4 white balls and 1 purple balls are drawn.
If the purple ball is p and white ball is w, the possible sample space of drawings are {p, wp, wwp, wwwp, wwwwp}

1. Write down the probability distribution for the player earning

The player earning for each event depends on the number of balls drawn subtracted the ticket price.
p= 2/6
The player earnings would be: 1*$1.5 -$4= - $2.5
wp= (4*2)/(6*5) = 4/15
The player earnings would be: 2*1.5- $4= - $1
wwp= (4*3*2)/(6*5*4)= 1/5
The player earnings would be: 3*$1.5 -$4= $0.5
wwwp= (4*3*2*2)/(6*5*4*3*2)= 2/15
The player earnings would be: 4*$1.5 -$4= $2
wwwwp= (4*3*2*2*1)/(6*5*4*3*2*1) = 1/15
The player earnings would be: 5*$1.5 -$4= $3.5

2. Find its expected value

The expected value would be:
chance of event * earning
You need to combine the 5 possible outcomes from the number 1 to get the total expected value.

Total expected value= (1/3 * - 2.5)+ (4/15*-1) + (1/5*0.5) + (2/15 *2) + ( 1/15 *3.5)=
(-12.5 -4 + 1.5 + 4 + 3.5) /15= -$7.5
This game basically a rip off.

6 0

3 years ago

Evaluate the integral Integral ∫ from (1,2,3 ) to (5, 7,-2 ) y dx + x dy + 4 dz by finding parametric equations for the line seg

n200080 [17]

$\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k$

is conservative if there is a scalar function $f(x,y,z)$ such that $\nabla f=\vec F$ . This would require

$\dfrac{\partial f}{\partial x}=y$

$\dfrac{\partial f}{\partial y}=x$

$\dfrac{\partial f}{\partial z}=3$

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

$f(x,y,z)=xy+g(y,z)$

$\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$f(x,y,z)=xy+h(z)$

$\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C$

$f(x,y,z)=xy+3z+C$

so $\vec F$ is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints: