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ser-zykov [4K]
2 years ago
15

Suppose that for a recent admissions class, an Ivy League college received 2,851 applications for early admission. Of this group

, it admitted 1,033 students early, rejected 854 outright, and deferred 964 to the regular admission pool for further consideration. In the past, this school has admitted 18% of the deferred early admission applicants during the regular admission process. Counting the students admitted early and the students admitted during the regular admission process, the total class size was 2,375. Let E, R, and D represent the events that a student who applies for early admission is admitted early, rejected outright, or deferred to the regular admissions pool.
Use the data to estimate P(E ), P(R), and P(D). If required, round your answers to four decimal places.

(a)

P(E) =

P(R) =

P(D) =

(b) Are events E and D mutually exclusive? Find P(E ∩ D). If your answer is zero, enter "0".

(c) For the 2,375 students who were admitted, what is the probability that a randomly selected student was accepted during early admission? If required, round your answer to four decimal places.

(d) Suppose a student applies for early admission. What is the probability that the student will be admitted for early admission or be deferred and later admitted during the regular admission process? If required, round your answer to four decimal places.
Mathematics
1 answer:
stepan [7]2 years ago
3 0

Answer:

A) The probability is a measure of the likelihood than an event is going to occur. In this case we can calculate the probability of X as:

P(X) = X/N where X is the number of ways X occurs and N is the total number of events.

P(E) = E/N = 1033/2851 = 0.3623

P(R) = R/N = 854/2851 = 0.2995

P(D) = D/N = 964/2851 = 0.3381

B)

E and D are mutually exclusive because the students that were admitted early where not deffered to regular admission pool. There is no chance that there is a student that were admitted erarly and deffered at the same time, so the intersection between E and D (E∩D) is 0.

C) The number of students that were early accepted is 1033 and the total number of students accepted is 2375. The probability is going to be:

P = 1033/2375 = 0.4349

D)

In this case we can reformulate the question as: What is the probability of being accepted if you apply for early admission? This

Since 18% of the students deffered to regular admission pool 0.18 × 964 = 174 were admitted at the end.

The probability of being deffered and then accepted is going to be:

P(DA) = 174/2831 = 0.0610

The probability of ramdomly selecting a students that has been early accepted or deffered and then accepted is going to be:

P(E or DA) = 0.0610 + 0.3623 = 0.4233

This is the addition rule.

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Henry wants to double a cake recipe that uses 2 cups and 10 tbsp of flour. How much flour will he need?
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Answer:

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10 tbsp of flour

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5 0
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What type of polygon would a slice of a hexahedron at a vertex create? Explain. What type of polygon would a slice of an icosahe
harina [27]

Answer:

  • hexahedron: triangle or quadrilateral or pentagon
  • icosahedron: quadrilateral or pentagon

Step-by-step explanation:

<u>Hexahedron</u>

A hexahedron has 6 faces. A <em>regular</em> hexahedron is a cube. 3 square faces meet at each vertex.

If the hexahedron is not regular, depending on how those faces are arranged, a slice near a vertex may intersect 3, 4, or 5 faces. The first attachment shows 3- and 4-edges meeting at a vertex. If those two vertices were merged, then there would be 5 edges meeting at the vertex of the resulting pentagonal pyramid.

A slice near a vertex may create a triangle, quadrilateral, or pentagon.

<u>Icosahedron</u>

An icosahedron has 20 faces. The faces of a <em>regular</em> icosahedron are all equilateral triangles. 5 triangles meet at each vertex.

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3 0
2 years ago
If θ=0rad at t=0s, what is the blade's angular position at t=20s
babunello [35]
The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
              ω = 250 (rpm) = 250 * (2π/60) = (25/3)π    rad/s
              ω = 100 (rpm) = 100 * (2π/60) = (10/3)π    rad/s

and from the figure it is clear that the operation is at constant speed but with variable levels
            ⇒   ω = dθ/dt   ⇒   dθ = ω dt

            ∴    θ = ∫₀²⁰  ω dt  
 
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
       ω = 0                                          from t = 0  to t = 5
       ω = 250 (rpm) = (25/3)π            from t = 5   to t = 15
       ω = 100 (rpm) = (10/3)π            from t = 15 to t = 20

∴ θ = ∫₀⁵  (0) dt   + ∫₅¹⁵  (25/3)π dt + ∫₁₅²⁰  (10/3)π dt
     
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π

∴ θ = 0 + (250/3)π + (50/3)π = 100 π

while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad

Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50
















3 0
2 years ago
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