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valina [46]
3 years ago
12

1. From a circular sheet of radius 18 cm, two circles of radii 4.5 cm and a rectangle of length 4 cm and breadth 1 cm are remove

d; find the area of the remaining sheet. [π = 22/7 ]
Mathematics
1 answer:
telo118 [61]3 years ago
4 0

Area of a circle = PI x r^2

Area of sheet = 22/7 x 18^2 = 1,018.29 square cm.

Area of small circle = 22/7 x 4.5^2 = 63.64 square cm.

There are 2 small circles so total area of the circles are 63.64 x 2 = 127.28 square cm.

Area of rectangle = l x w = 4 x 1 = 4 square cm.

Total area of cutouts = 127.28 + 4 = 131.28 square cm.

Area of sheet left, subtract area of cutouts fro area of sheet:

1018.29 - 131.28 = 887.01 square cm.

Round everything as needed.

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Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

= \frac{20!}{(20-5)! 5!)}

= \frac{20*19*18*17*16}{5*4*3*2*1}

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

= \frac{20}{320} = \frac{1}{19}

P(A5) = ⁸C₅ / ²⁰C₅

= \frac{56}{15540} = \frac{7}{1938}

P(A4) = ⁷C₄ / ²⁰C₄

= \frac{35}{4845} = \frac{7}{969}

c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

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