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Paladinen [302]
3 years ago
13

Need Help Please I'm Out Of Time ;(

Mathematics
1 answer:
vodka [1.7K]3 years ago
8 0

Answer:

Total surface area = 184.86 cm²

Step-by-step explanation:

If we see the the diagram, we can find that the net of this triangular prism includes:

2 triangles each with the dimension of one side 4.3cm, second side 5.2cm and third side 6.75 cm

3 rectangles each with the dimensions of (6.75×10)cm, (5.2×10)cm and (4.3×10)cm

Surface area of triangle with a,b and c side:

s=(a+b+c)/2

Area= √s(s−a)(s−b)(s−c)

Area = 11.18cm²

For 2 triangle:

Area = 22.36cm²

Surface Area of Rectangles:

Area = (6.75×10)cm + (5.2×10)cm + (4.3×10)cm

Area = 162.5 cm²

Total surface area = area of 2 triangle + area of 3 rectangles:

Total surface area = 22.36 + 162.5

Total surface area = 184.86 cm²

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A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi
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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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4 years ago
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Step-by-step explanation:

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Answer:

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