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12345 [234]
3 years ago
5

A zoo gift shop sells souvenir T-shirts. The line graph shows the average number of T-shirts sold on different days of the week.

Based on the information shown on the line graph which value is most likely the difference between the number of T-shirts the zoo gift shop will sell on Sunday in the number of T-shirts the zoo gift shop will sell on Tuesday in the next eight weeks? A.) 20 B.)160 C.) 200 D.) 480

Mathematics
2 answers:
katen-ka-za [31]3 years ago
5 0

Answer:

get smart

Step-by-step explanation:iuwiufgy

ivolga24 [154]3 years ago
3 0

Answer:

Its 160

Step-by-step explanation:

Trust me :)

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The triangle is isosceles since two of its legs coincide with the radius of the circle, and the angle between these sides measures 120°, same as the arc it subtends. If b is the length of the third side in the triangle, then by the law of cosines

b² = 2 • (96 cm)² - 2 (96 cm)² cos(120°)   ⇒   b = 96√3 cm

Call b the base of this triangle.

The vertex angle is 120°, so the other two angles have measure θ such that

120° + 2θ = 180°

since the interior angles of any triangle sum to 180°. Solve for θ :

2θ = 60°

θ = 30°

Draw an altitude for the triangle that connects the vertex to the base. This cuts the triangle into two smaller right triangles. Let h be the height of all these triangles. Using some trig, we find

tan(30°) = h / (b/2)   ⇒   h = 48 cm

Then the area of the triangle is

1/2 bh = 1/2 • (96√3 cm) • (48 cm) = 2304√3 cm²

and the area of the shaded segment is the difference between the area of the sector and the area of the triangle:

3072π cm² - 2304√3 cm² ≈ 5660.3 cm²

7 0
2 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)

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