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gladu [14]
4 years ago
5

What are the coordinates of a point that is 6 units from (2,5)​

Mathematics
1 answer:
Vesna [10]4 years ago
7 0
Answer: (8,6)

(; ya yaya
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Perform the indicated operation -3+14= ?
Zigmanuir [339]
The answer is -3+14=11
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4 years ago
People with severe hearing loss were given a sentence and word recognition test six months after they got implants in their ears
Elodia [21]
.82

When changing a percentage to a decimal you will divide the number by 100. 82/100= .82

To change it back to percentage you will take .82 and multiply by 100 or simply move the decimal two stops to the right, making 82. or 82% 
4 0
3 years ago
What is 1 and 1/4 divided by 1 and 2/5
Sati [7]

Answer:

25/28

Step-by-step explanation:

6 0
3 years ago
In a certain Algebra 2 class of 30 students, 14 of them play basketball and 10 of them play baseball. There are 14 students who
Veseljchak [2.6K]

Answer:

In a certain Algebra 2 class of 30 students, 22 of them play basketball and 18 of them play baseball. There are 3 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

I know how to calculate the probability of students play both basketball and baseball which is 1330 because 22+18+3=43 and 43−30 will give you the number of students plays both sports.

But how would you find the probability using the formula P(A∩B)=P(A)×p(B)?

Thank you for all of the help.

That formula only works if events A (play basketball) and B (play baseball) are independent, but they are not in this case, since out of the 18 players that play baseball, 13 play basketball, and hence P(A|B)=1318<2230=P(A) (in other words: one who plays basketball is less likely to play basketball as well in comparison to someone who does not play baseball, i.e. playing baseball and playing basketball are negatively (or inversely) correlated)

So: the two events are not independent, and so that formula doesn't work.

Fortunately, a formula that does work (always!) is:

P(A∪B)=P(A)+P(B)−P(A∩B)

Hence:

P(A∩B)=P(A)+P(B)−P(A∪B)=2230+1830−2730=1330

7 0
3 years ago
How many words can be formed by using the W,X,Y,Z if repetitions is not allowed?
Setler [38]

Answer:

24

Step-by-step explanation:

What you have here is a permutation, seeing as each element can only be used once.

We have 4 letters initially, so we can choose any 1 as our first letter. We have 4 choices for our first letter

However, once we choose our first letter, we can't use it anymore, so, for our second letter, we can only choose from the remaining 3 letters.

Furthermore, once we choose our second letter, we can only choose our 3rd letter from the remaining two letters we didn't choose yet.

Finally, our last letter will always be the one we didn't choose the last 3 times. So there is only one choice here.

Going off of this, we have four choices for the 1st letter, three choices for the 2nd letter, two choices for the 3rd letter, and one choice for the 4th letter

The way to calculate how many permutations we have without repetition is using factorials

N!

Where N is the number of elements you have.

In this case, it would be 4!

4! is 4 * 3 * 2 * 1

Which equals 24

If you notice, each number in 4! is the number of options we have for each choice. 4, then 3, and so on

5 0
3 years ago
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