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HACTEHA [7]
2 years ago
10

Please help me solve this :)

Mathematics
1 answer:
Svetradugi [14.3K]2 years ago
7 0

In given picture we see that there are total 8 boxes out of which only 5 are shaded.

So the number that represents shaded part will be 5/8.

But we have to find which choice doesn't indicate shaded part.

So we will pick those choices which are not equivalent to 5/8.

5/8=0.625

if you convert that into percent then we get (5/8)100%=62 1/2 %

that means 5/8, 0.625, 62 1/2 % all represents same thing and represent shaded part.

So the remaining choice a) 0.582 is the final answer .

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Measurements of regular hexagon
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The regular hexagon has both reflection symmetry and rotation symmetry.

Reflection symmetry is present when a figure has one or more lines of symmetry. A regular hexagon has 6 lines of symmetry. It has a 6-fold rotation axis.

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Rotation symmetry is present when a figure can be rotated (less than 360°) and still look the same as before it was rotated. The center of rotation is a point a figure is rotated around such that the rotation symmetry holds. A regular hexagon can be rotated 6 times at an angle of 60°

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The residents of a city voted on whether to raise property taxes. The ratio of yes votes to no votes was 5 to 7. If there were 6
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Answer:

4046

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)

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3 years ago
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The answer i D because 11-6=5
8 0
2 years ago
What is the probability if rolling an 8 with one regular die if you roll the die just once?
lidiya [134]

Answer:

the answer to the probability question should be zero because there isn't an eight on a normal die. if you were to only roll once you would never get 8

8 0
3 years ago
Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The
Marina86 [1]

Answer:

Step-by-step explanation:

Given that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.

G = card drawn is green

Y = card drawn is yellow

E = card drawn is even-numbered

List:

Sample space = {G1, G2, G3, G4, G5, Y1, Y2, Y3}

2) P(G) = 5/8

3) P(G/E) = P(GE)/P(E)

GE = {G2, G4}

Hence P(G/E) = 2/5

4) GE = {G2, G4}

P(GE) = 2/8 = 1/4

5) P(G or E) = P(G)+P(E)-P(GE)

= 5/8 + 3/8-2/8 = 3/5

6) No there is common element as G2 and G4

Cannot be mutually exclusive

6 0
3 years ago
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