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7nadin3 [17]
3 years ago
15

This interactive glossary contains all of the postulates and Theorems from geometry. Explain how you could use this glossary as

a tool throughout the course.
Mathematics
1 answer:
pshichka [43]3 years ago
3 0
I can not only easily see the material I will need in the future, I can also start learning on my own, and also use it to look up stuff we are using in class.
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If you guess an answer on two multiple choice questions with the options a, b, or c, what is the probability of you guessing the
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First, you find the probability of getting each question right individually, then you multiply the two probabilities together.

(1/3) x (1/3) = 1/9
5 0
3 years ago
Can someone explain how this is done?
Musya8 [376]

Answer:

Just put in the values in the equation for example 4(1)-5<7 and solve it if the eqution is true at the end than that the right answer

Step-by-step explanation:

3 0
3 years ago
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Please please please tell the answer​
OlgaM077 [116]

Answer:

a. 0

b. 4^1

c. 14

d. 8

Step-by-step explanation:

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Jorge finish 1/4 of a level of his computer game in 1/3 of the air at this rate how many hours will it take to complete one leve
aliya0001 [1]
To solve this, we'd need an answer with a whole level, which means we'll multiply everything by 4. This would mean that 1/3 is also multiplied by 4, making 4/3. Which means a whole level is made in 4/3 hours or 1 1/3 hours.
5 0
3 years ago
2 cards are drawn from a standard deck without replacement. What is the probability that at least one of the cards drawn is an a
olya-2409 [2.1K]

Answer:

<h2>The answer is 0.1493.</h2>

Step-by-step explanation:

In a standard deck there are 52 cards in total and there are 4 aces.

Two cards can be drawn from the 52 cards in ^{52}C_2 = \frac{52!}{50!\times2!} = 51\times26 ways.

There are (52 - 4) = 48 cards rather than the aces.

From these 48 cards 2 cards can be drawn in ^{48}C_2 = \frac{48!}{46!\times2!} = 47\times24 ways.

The probability of choosing 2 cards without aces is \frac{47\times24}{51\times26} = \frac{188}{221}.

The probability of getting at least one of the cards will be an ace is 1 - \frac{188}{221} = \frac{33}{221} = 0.1493.

6 0
3 years ago
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