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Jobisdone [24]
2 years ago
12

There are five candies in a bag. Three are chewy, and two are chocolate. You draw a candy, do not replace it, and then draw anot

her. What is the probability of choosing two chocolate candies?
Mathematics
1 answer:
castortr0y [4]2 years ago
4 0

It has been computed that the probability of choosing two chocolate candies is 2/25.

<h3>How to calculate probability</h3>

The probability of choosing the first chocolate candy will be 2/5.

The probability of choosing the second chocolate candy will be 1/5.

Therefore, the probability of choosing two chocolate candies will be:

= 2/5 × 1/5

= 2/25

Learn more about probability on:

brainly.com/question/25870256

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zubka84 [21]

Answer:

2y^4√5

Step-by-step explanation:

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

hope i helped :) !!!!!!!

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8 0
3 years ago
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How do you solve this?
Alchen [17]
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3 years ago
8 divided by 2/7 = help me
Ad libitum [116K]
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6 0
3 years ago
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How to prove it equal 8?
iogann1982 [59]

Answer:

The answer equals -8

Step-by-step explanation:

Order of Operations: BPEMDAS

FOIL - First, Outside, Inside, Last

Step 1: Write out expression

(\sqrt{4-3i} -\sqrt{4+3i})^6

Step 2: Expand

(\sqrt{4-3i} -\sqrt{4+3i})(\sqrt{4-3i} -\sqrt{4+3i})(\sqrt{4-3i} -\sqrt{4+3i})(\sqrt{4-3i} -\sqrt{4+3i})(\sqrt{4-3i} -\sqrt{4+3i})(\sqrt{4-3i} -\sqrt{4+3i})

Step 3: FOIL first 2

(\sqrt{4-3i} -\sqrt{4+3i})^2 = -2

Step 4: Replace square roots with -2

-2(-2)(-2) = (-2)³ = 8

4 0
3 years ago
WILL GIVE BRAINLIEST if you find area of figure
Lyrx [107]

Look at the picture.

A_1=2\ in\cdot3\ in=6\ in^2\\\\A_2=(6\ in-2\ in)\cdot5\ in=4\ in\cdot5\ in=20\ in^2

Area of the figure:

A=A_1+A_2\to A=6\ in^2+20\ in^2=26\ in^2

5 0
3 years ago
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