What is the perimeter of the rectangle below 8 in 6 in 8 in 6 in

Can someone help me with the 2 highlighted questions?

dexar [7]

Answer:

2. domain: {3,4,5,6,7}; range: {2,3,4,5,6}

4. IS a function

Step-by-step explanation:

2. The domain is the list of first numbers of each pair.

The range is the list of second numbers.

4. If each value on the left only has one arrow coming from it, the relation is a function. That is the case here.

__

Contrast that to the problem above (3.). In that one, the number 4 1/2 has two arrows coming from it. That relation is NOT a function.

4 0

3 years ago

Find the area of the polygon. A. 1,922 ft2 C. 1,932 ft2 B. 932 ft2 D. 922 ft2

r-ruslan [8.4K]

Answer:

1932

Step-by-step explanation:

just multiply

7 0

4 years ago

Read 2 more answers

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

See the picture below!!

Darya [45]

Answer:

Q1) x = 17.3

y = 8.7

Q2) d = 19

Step-by-step explanation:

Question 1

Angle = 60°

Hypotenuse = x

Adjacent = y

Opposite = 15

Step 1: Find x

Sin (angle) = opposite/hypotenuse

Sin (60°) = 15/x

x = 10√3 = 17.3

Step 2: Find y

Tan (angle) = opposite/adjacent

Tan (60°) = 15/y

y = 5√3 = 8.7

Question 2

Step 1: Find the diagonal of the base

c² = a²+b²

c² = 6² + 10²

c = 2√34 = 11.7

Step 2: Find diagonal d

c² = a² + b²

d² = 11.7² + 15²

d = 19

!!

7 0

3 years ago

How do I know whether something is a function or not? Answer the worksheet please

djyliett [7]

Because you if it changes its a function

6 0

3 years ago