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SVETLANKA909090 [29]
3 years ago
14

Hector found that the container of orange juice in the refrigerator was full. He poured equal amounts of juice into 6 glasses.

Mathematics
2 answers:
yawa3891 [41]3 years ago
8 0
The answer is c one half because there is 6 glasses so each would get half a glass
gavmur [86]3 years ago
5 0
I think its c 1 2 the answer
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Step-by-step explanation:

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Step-by-step explanation:

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In a parking lot with 250 parking spaces 40% how many of them are vacant and occupied
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3 years ago
Aaron is using Dimensional Analysis to solve the following question:
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Answer:

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3 0
2 years ago
Sali throws an ordinary fair 6 sided dice once.
xenn [34]

a) 1/6

b) 1/36

c)

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

Step-by-step explanation:

a)

The probability of a certain event A to occur is given by

p(A)=\frac{a}{n}

where

a is the number of successfull outcomes, in which event A occurs

n is the total number of possible outcomes

In this problem, the event is

"getting a 6 when throwing a dice once"

We know that the possible outcomes of a dice are six: 1, 2, 3, 4, 5, 6, so we he have

n=6

The successfull outcome in this case is only if we get a 6, so only 1 outcome, therefore

a=1

So, the probability of this event is

p(6)=\frac{1}{6}

b)

In this case instead, we are throwing the dice twice.

The two throws of the dice are independent events (one does not depend on the other): the probability that two independent events A and B occur at the same time is given by the product of the individual probabilities,

p(AB)=p(A)\cdot p(B)

where

p(A) is the probability that event A occurs

p(B) is the probability that event B occurs

Here we have:

- Event A is "getting a 6 in the first throw of the dice". We already calculated this probability in part a), and it is

p(A)=\frac{1}{6}

- Event B is "getting a 6 in the second throw of the dice". Since the dice has not changed, the probability is still the same, so

p(B)=\frac{1}{6}

Therefore, the probability of getting a 6 on both throws is:

p(66)=p(6)\cdot p(6)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}

c)

In this problem, we have:

- A dice that is thrown once

- A coin that is also thrown once

The dice has 6 possible outcomes, as we stated in part a):

1, 2, 3, 4, 5, 6

While the coin has two possible outcomes:

H = head

T = tail

So, in order to find all the outcomes of the two events combined, we have to combine all the outcomes of the dice with all the outcomes of the coin.

Doing so, using the following notation:

1H (getting 1 with the dice, and head with the coin)

The possible outcomes are:

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

So, we have a total of 12 possible outcomes.

4 0
3 years ago
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