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Sergio039 [100]
3 years ago
9

I need help ASAP !! i’ll mark brainliest !

Mathematics
2 answers:
Ainat [17]3 years ago
4 0

I believe your answer would be: C. 64.

Because raise 4 to the power of 3, which equals 64.

Good Luck! =)

Andrej [43]3 years ago
3 0

Answer:

192

Step-by-step explanation:

Multiply 3/4 by 256

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Kyle drew 4 polygons labeled A,B,C, and D. Figure A has twice as many sides as Figure B. Figure B has 3 vertices. Figure C has h
erma4kov [3.2K]

Answer:

Figure A - hexagon

Figure B - triangle

Figure C - rectangle

Figure D - octagon

Step-by-step explanation:

If figure B has 3 vertices (angles), then that means it is a triangle (3 sides)

So that means figure A has 6 sides, (hexagon)

And figure D has 8 vertices, so 8 sides (octagon)

And if C has half as many, that would be 4 (rectangle)

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sharon drops a rubber ball from a height of 160 feet every bounce sends the ball half as high as it was before what is the total
Elanso [62]

Answer:

10 feet.

Step-by-step explanation:

First time: 160/2=80.

Second time: 80/2=40.

Third time: 40/2=20.

Fourth time: 20/2=10.

Boom.

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3 years ago
Question 16 of 17
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2 years ago
Drag each label to the correct location on the image.
Misha Larkins [42]

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is there a picture to this problem?

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3 years ago
A box contains 11 red chips and 4 blue chips. We perform the following two-step experiment: (1) First, a chip is selected at ran
Scilla [17]

Answer:

P(B1) = (11/15)

P(B2) = (4/15)

P(A) = (11/15)

P(B1|A) = (5/7)

P(B2|A) = (2/7)

Step-by-step explanation:

There are 11 red chips and 4 blue chips in a box. Two chips are selected one after the other at random and without replacement from the box.

B1 is the event that the chip removed from the box at the first step of the experiment is red.

B2 is the event that the chip removed from the box at the first step of the experiment is blue. A is the event that the chip selected from the box at the second step of the experiment is red.

Note that the probability of an event is the number of elements in that event divided by the Total number of elements in the sample space.

P(E) = n(E) ÷ n(S)

P(B1) = probability that the first chip selected is a red chip = (11/15)

P(B2) = probability that the first chip selected is a blue chip = (4/15)

P(A) = probability that the second chip selected is a red chip

P(A) = P(B1 n A) + P(B2 n A) (Since events B1 and B2 are mutually exclusive)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/21) + (22/105) = (77/105) = (11/15)

P(B1|A) = probability that the first chip selected is a red chip given that the second chip selected is a red chip

The conditional probability, P(X|Y) is given mathematically as

P(X|Y) = P(X n Y) ÷ P(Y)

So, P(B1|A) = P(B1 n A) ÷ P(A)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(A) = (11/15)

P(B1|A) = (11/21) ÷ (11/15) = (15/21) = (5/7)

P(B2|A) = probability that the first chip selected is a blue chip given that the second chip selected is a red chip

P(B2|A) = P(B2 n A) ÷ P(A)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/15)

P(B2|A) = (22/105) ÷ (11/15) = (2/7)

Hope this Helps!!!

5 0
3 years ago
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