Which of the following simplifies to a negative number?<br> o4-4<br> о(-4) 4<br> О -4-4

Which expression is equal to

AfilCa [17]

Answer:

4 to the -4 power

Step-by-step explanation:

multiply the 4's and add the exponents. then divide by 4 and divide by that exponent

3 0

3 years ago

What is the solution to the system of equations? ⎧⎩⎨⎪⎪x+3y−z=24x+2y+5z=13x+z=12 (−4, 1, −3) (3, 0, 3) (5, −2, −3) (−1, 2, 3)

snow_tiger [21]

<h3>Answer:</h3>

(5, -2, -3)

<h3>Explanation:</h3>

We assume your equations are ...

x +3y -z = 2
4x +2y +5z = 1
3x +0y +z = 12

These have solution (x, y, z) = (5, -2, -3).

_____

Trying the choices in the last equation eliminates the first and last. Trying the second choice in the first equation eliminates it, leaving the 3rd choice as the answer you're looking for.

We also know because we can ask a graphing calculator to solve the matrix equation (or row-reduce the augmented matrix).

<em>Comment on this problem</em>

The hardest part of this problem is figuring your intent. The equations are not well-separated, so we have a hard time telling what is a constant and what is a coefficient. It took a couple of guesses to sort it out. A little editing before posting would be helpful.

8 0

3 years ago

PLS HELP WILL GIVE BRAINLIEST ANSWER AND LOTS OF POINTS

Brut [27]

Answer: 1:(12^2*10.8)/3=518.4 2: v = 392(3.1415)/3 = 410.5 3: (Πr^2*h) Π*9*15=424

Step-by-step explanation:

3 0

2 years ago

If someone drove 15 miles and is only three fifths of their way home what is the total length of her drive

BabaBlast [244]

3/5 * X = 15
3/5 ÷3/5 * X = 15 ÷ 3/5
X = 25

The total length is 25 miles

3 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

3 years ago

Which of the following simplifies to a negative number? o4-4 о(-4) 4 О -4-4