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3 years ago

I need a answer right now I’ll give you points!

Mathematics

2 answers:

Kipish [7]3 years ago

8 0

Answer:

100% right answer

satela [25.4K]3 years ago

3 0

Answer:

G. 118°

Step-by-step explanation:

Imagine a line, parallel to AD, that passes through B to a point x on the line CD.

∠CBX=∠CAD=28 (corresponding angles)

∠EBX=90 (BX║ED, BE║XD)

∠CBE=∠CBX+EBX

∠CBE=28+90

∠CBE=118°

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Suppose we have an urn containing 9 marbles. Two are red, three are green, and four are blue. We randomly select 5 marbles from

Aleonysh [2.5K]

Answer:

probability of selecting 3 green marbles, 1 red marble, and 1 blue marbles

$P(E) =\frac{8}{126} = 0.0634$

Step-by-step explanation:

Given data urn containing '9' marbles

given red marbles = 2

green marbles =3

blue marbles = 4

Five marbles can be selected at a time from '9' marbles in $9_{C_{5} }$ ways

by using formula $n_{C_{r} }=\frac{n!}{(n-r)!r!}$

$9_{C_{5} }=\frac{9!}{(9-5)!5!} = \frac{9X8X7X6X5!}{4!5!}$

After simplification , we get $9_{C_{5} } = 126$

total number of ways n(S) = 126

The probability of selecting '3' green marbles ,one red marble and one blue marble with replacement.

let ' E' be the event of selecting '3' green marbles ,one red marble and one blue marble with replacement.

n(E) = $3_{C_{3} } X2_{C_{1} }X 4_{C_{1} }$ ways

The required probability $P(E) = \frac{n(E)}{n(S)}$

$p(E) = \frac{3_{C_{3} } X2_{C_{1} }X 4_{C_{1} }}{9_{C_{5} } }$

on simplification , we get

$P(E) = \frac{1X2X4}{126} =\frac{8}{126}$

$P(E) =\frac{8}{126} = 0.0634$

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3 years ago