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vesna_86 [32]
3 years ago
7

PLEASE ANSWER ALLL QUESTIONS BRAINLEST!!! STEP BY STEP PLEASE

Mathematics
1 answer:
Slav-nsk [51]3 years ago
4 0

Step-by-step explanation:

/qiq-iorx-dxr

Meet

anyone come

You might be interested in
The answer to this question
Ilia_Sergeevich [38]

Answer:

The fourth one J is the answer

Step-by-step explanation:

we only need to find one point(0,-2)

the following selections only the fourth one

when x=0 ,y=-2

3 0
3 years ago
Brianna asked 45 students if they would vote for her to be student council president. She used her calculator to compare the num
lisabon 2012 [21]
0.6222... is a recurring decimal at the digit '2'

let x = 0.62222....

multiply x by 10 ⇒ 10x = 6.2222.... [we multiply by 10 to leave the recurring digits after decimal point]

multiply x by 100 ⇒ 100x = 62.2222... [we multiply by 100 so that we can again leave the recurring digit after decimal point]

subtract 10x from 100x
100x - 10x = 62.2222... - 6.2222... 
90x = 56
x = 56/90
x = 28/45

-----------------------------------------------------------------------------------------------------------

There are 28 students would vote for Brianna


3 0
4 years ago
Is this correct? answer quick please!
scZoUnD [109]

Answer:

yessssss

Step-by-step explanation:

...........

8 0
3 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
Please help me I'll give brainliest
Olenka [21]
Hey, there isn't a question, but I'd love to help!
4 0
3 years ago
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