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serg [7]
3 years ago
6

I need help been stuck on this for 3 days

Mathematics
1 answer:
azamat3 years ago
6 0
Try putting 1,7 as ur answer
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Answer please I will give 12<br> points <br> Thanks
arsen [322]

Answer:

A=50.3

Step-by-step explanation:

A= πr²

A= π x 4²

A= 3.142 x 16

A=50.272

A≈50.3

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When you apply for a scholarship, you may have to write an application essay?
Anna [14]

Answer your answer is b

Step-by-step explanation:

to show that you have the writing skills in order to succeed in college

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Help me please ............
olchik [2.2K]
The answer is Choice C: 51 
I Hoped I helped you! (^w^)/ 
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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
Plllssss help me its the last one
DerKrebs [107]
Well I think step 1 is wrong she is suppossed to add 32.75 +32.75 then add she would have 65.50 after she would have to add 70.99 her results form that would be 136.49 after she would need to subtract 12.25 then she ends up with 124.24 and 1/4 of that is 31.06 therefore what she put in her savings was 31.06 I hope this helps!
3 0
4 years ago
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