I really appreciate the people helping and I was freaking out

Why must the value of the sine ratio for an acute angle

Fynjy0 [20]

Answer:

Because of the range of the sine function.

Step-by-step explanation:

The reason why the value of the sine ratio for an acute angle

of a right triangle must always be a positive value less

than 1 has to do with the the range of the sine function.

The range of the sine function is

$- 1 \leqslant y \leqslant 1$

But in the first quadrant all y-values are positive and acute angles are less than 90°.

Therefore the range will now be 0≤y<1 for acute angles.

4 0

3 years ago

Emerson is making costumes for the

Elis [28]

9514 1404 393

Answer:

13 costumes

Step-by-step explanation:

The amount of fabric used is the product of the amount per costume and the number of costumes.

A = na . . . . . A = total area; n = number of costumes; 'a' = area per costume

46 2/3 = n(3 1/2)

n = (46 2/3)/(3 1/2) = (140/3)/(7/2) = (140/3)(2/7) = 40/3 = 13 1/3

Apparently there is some waste, as Emerson used more fabric than necessary to make 13 costumes.

6 0

2 years ago

Three consecutive odd integers have a sum of 3.what are the number

lapo4ka [179]

X = smallest odd
x + 2 = second odd
x + 4 = third odd

x + x + 2 + x + 4 = 3
3x + 6 = 3
3x = -3
x = -1

The numbers are -1, 1, 3. Hope this helps!

6 0

3 years ago

Ill mark u brainliest please help me!

dem82 [27]

Answer:

C

Step-by-step explanation:

Line AC is equal to line DF.

6 0

3 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

3 years ago