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Mekhanik [1.2K]
3 years ago
9

EXPLAIN ANSWER PLZ!

Mathematics
1 answer:
Dmitriy789 [7]3 years ago
6 0

height,h = 3"

base, b = length, l = 2

area of triangle, t = bxh/2...eqn 1

area of square, s = l^2...eqn 2

area of pyramid, p =4t + s...eqn 3

subst h = 3 and b = 2 in eqn 1

t = 3x2/2 = 3

subst l =2 in eqn 2

s = 2^2 = 4

subst t = 3 & s = 4 in eqn 3

p = 4(3) + 4 = 16sq inch

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Mean=total/the number of values =>total=mean*the number of values
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3 years ago
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Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> =  Mike will show up.

Given:

P(D^{c})=0.55\\P(M^{c})=0.45

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

                                                   =1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25

Thus, the probability that neither Dave nor Mike will show up is 0.25.

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Answer:

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After rotation by 360° about the center, we always get an onto figure.

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