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Makovka662 [10]
2 years ago
11

Write down the reciprocal of 1/3

Mathematics
2 answers:
dem82 [27]2 years ago
8 0

\large \sf  \underline \red{RECIPROCAL}

\\

\underline \mathbb{ANSWER}{: }

  • \large \boxed { \large \sf \green{3}}

\mathbb{EXPLANATION}{:}

The question is divided into two parts.

the reciprocal of ¹/3

  • \boxed{ \tt1÷   \frac{1}{5}  =3}

Four times of 3

  • \large \boxed{ \sf4  \: x \:  3  \: =  \: 12}

So Four times the reciprocal of 1/3 is 12

Leokris [45]2 years ago
6 0
<h2><u>RECIPROCAL</u></h2>

<h3>ANSWER: </h3>
  • \boxed{ \bold{3 \: \: \:  or  \: \:  \:  \frac{3}{1} }}

<h3>EXPLANATION:</h3>

As we can see above, it's clear that 3 and 1/3 are reciprocals of each other.

And we can find the reciprocal of 1/3 by forming a fraction and to multiply the numerator and denominator by 3.

  • \\ \large  \tt \:  \frac{1 \times 3}{ \frac{1}{3} \times 3 }  =  \bold{ \frac{3}{1}  = 3} \\ \\

Hence, 3/1 or 3 is the correct answer.

_______________∞_______________

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x            y
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3 years ago
Which events are independent? Select each correct answer.
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Answer:

Independent events are those events when the happening of one event does not affect the happening the other event.

In the given example we have the following events as independent:

A) A spinner is spun and a coin is flipped

C) Two number cubes are rolled at the same time.

Both these events are independent because the happening of one event does not affect the other in both the cases.

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4 years ago
Read 2 more answers
Suppose that 3 balls are chosen without replacement from an urn consisting of 3 white and 7 red balls. Assume moreover that the
zimovet [89]

Answer:

P(X1=1, X2=1) = 1/15

P(X1=1, X2=0) = 7/30

P(X1=0, X2=1) = 7/30

P(X1=0, X2=0) = 7/15

Step-by-step explanation:

Let Xi = 1 if the i-th white ball is selected. In this question the 3 white balls are marked 1,2 and 3.

We need to know the possible combination between X1 and X2 i.e. for the white ball 1 and 2 being chosen in the event.

We also need to note that the event is dependent which that after a ball is being chosen, it will not be put back hence affecting the probability of picking the next ball.

Consider all the possible combination between X1 and X2

a) both being chosen P(X1=1, X2=1)

= (3/10) x (2/9) = 1/15

Note that the first probability is the probability before any ball is being picked. The chances for ball white to be pick is 3/10 (3 white ball from the total 10 balls).

After 1 white ball being selected, that ball is not again out back into the urn making white ball 2 and total ball 9. Hence probability of picking another white ball is 2/9

b) only X1 chosen P(X1=1, X2=0)

= (3/10) x (7/9) = 7/30

After the white ball was picked, the probability of white not being pick again is the same as red being picked. Since there is still 7red balls and a total of 9 balls, the probability is 7/9

c) only X2 chosen P(X1=0, X2=1)

= (7/10) x (3/9) = 7/30

The white is not being picked first, making the probability of picking red is 7/10. Then the probability of white being picked is 3/9

d) both not chosen

P(X1=0, X2=0)

= (7/10) x (6/9) = 7/15

In other word only red being chosen. So the first probability is 7 red out of 10 balls (7/10), and the next red ball being picked next is 6/9

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