A tablecoth is 82 inches long. What is the length in feet and inches

Comparing and Extending Ratios

son4ous [18]

I think she is correct because if u divide 100 to 5 u'll get 20 and if u divide 40 to 2 u'll get 20 so yea

6 0

3 years ago

The zeros of f(x)= 3x^3+16x^2+18x-4

Otrada [13]

After trial and error, you discover that - 2 works.

-2 || 3 16 18 -4

-6 -20 4

====================================

3 10 - 2 0

What you have left is a quadratic

3x^2 + 10x - 2 You can just use the quadratic formula to solve for the other 2 roots.

x1 = [ - 10 +/- sqrt(10^2 - 4*3*(-2) ) ] / 6

x1 = [ -10 +/- sqrt (100 + 24) ) /6

x1 = [ - 10 +/- sqrt(124) ) / 6

x1 = [ - 10 +/- 11.14 ] / 6 = 0.1893

x2 = [ - 21.14 ] / 6 = - 3.522

Answer

x1 = - 2

x2 = 0.1893

x3 = - 3.522

4 0

3 years ago

A rectangle has a perimeter of 64 inches and a length of 22 inches. Which equation can you solve to find the width?

ella [17]

It's B because if one length is 22 then both lengths combined is 44 + the width. Since there are 2 of those, each are represented with w so basically you're adding 2w to the total length,44 which will end up to look like 44+ 2w

6 0

3 years ago

Read 2 more answers

1/3(n 4)=9 how do you work this problem out

SIZIF [17.4K]

1/3(n4)=9
I assume that it's a multiplication between n and 4
then solving the equation
4n= 9/1/3
4n= 27
n=27/4=6.75

4 0

3 years ago

Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let X(t) be the number of custom

mash [69]

Answer:

Step-by-step explanation:

Given that:

X(t) = be the number of customers that have arrived up to time t.

$W_1,W_2$ ... = the successive arrival times of the customers.

(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

$E(W_!|X(t)=2) = \int\limits^t_0 {X} ( \dfrac{d}{dx}P(X(s) \geq 1 |X(t) =2))$

$= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}$

$= 1 - \dfrac{P(X(s) \leq 0 , 1 \leq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

$= 1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

Now $P(X(s) \leq 0) = P(X(s) = 0)$

(b) We can Determine the conditional mean E[W3|X(t)=5] as follows;

$E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\ = 1- P (X(s) \leq 2 | X (t) = 5 ) \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}$

Now; $P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)$

(c) Determine the conditional probability density function for W2, given that X(t)=5.

So ; the conditional probability density function of $W_2$ given that X(t)=5 is:

$f_{W_2|X(t)=5}}= (W_2|X(t) = 5) \\ \\ =\dfrac{d}{ds}P(W_2 \leq s | X(t) =5 ) \\ \\ = \dfrac{d}{ds}P(X(s) \geq 2 | X(t) = 5)$

7 0

4 years ago