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Harman [31]
2 years ago
13

1. Each cube has 6 faces. If Tandra

Mathematics
2 answers:
Ede4ka [16]2 years ago
6 0

1 cube = 6 Faces

2 cubes = 10 Faces

7 cubes = 36 Faces

2÷7 = 3 × 10 = 30 + 6 = 36

The Answer is <u>3</u><u>6</u>

Setler [38]2 years ago
3 0

Answer:

2 cubes = 10 faces

3 cubes = 14 faces

4 cubes = 18 faces

5 cubes = 22 faces

6 cubes = 26 faces

7 cubes = 30 faces

10 faces = 2 cubes

14 faces = 3 cubes

18 faces = 4 cubes

Step-by-step explanation:

We know that 1 cube shows 6 faces, and 2 cubes shows 10 faces. With 3 cubes, it shows 14 faces.

With this pattern, we can conclude that

f(x) = 2 + 4x

f(x) is a function for where f is the value of faces and x is the amount of cubes connected.

To find the amount of faces for the cubes used

f(2) = 2 + 4(2) = 2 + 8 = 10

f(3) = 2 + 4(3) = 2 + 12 = 14

f(4) = 2 + 4(4) = 2 + 16 = 18

f(5) = 2 + 4(5) = 2 + 20 = 22

f(6) = 2 + 4(6) = 2 + 24 = 26

f(7) = 2 + 4(7) = 2 + 28 = 30

For the amount of cubes used given the amount of faces

10 = 2 + 4x

-2   -2

8/4 = 4x/4

2 = x

x= 2

14 = 2 + 4x

-2    -2

12/4 = 4x/4

3 = x

x = 3

18 = 2 + 4x

-2   -2

16/4 = 4x/4

4 = x

x = 4

Your welcome! If you have any questions, please post them in the comments section of this answer! If you could mark this answer as the brainliest, I would greatly appreciate it! :D

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Answer:

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In Vancouver, British Columbia, the probability of rain during a winter day is 0.42, for a spring day is 0.23, for a summer day
nadezda [96]

Answer:

31.82% probability that this day would be a winter day

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

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Intersection:

Rain on a winter day, which is 0.42 of 0.25. So

P(A \cap B) = 0.42*0.25 = 0.105

If you were told that on a particular day it was raining in Vancouver, what would be the probability that this day would be a winter day?

P(B|A) = \frac{0.105}{0.33} = 0.3182

31.82% probability that this day would be a winter day

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3 years ago
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