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sergejj [24]
3 years ago
13

Question One

Mathematics
1 answer:
Alex3 years ago
7 0

Answer:

Step-by-step explanation:

Set up proportions. Make sure length and widths line up!

\frac{12}{5} =\frac{x}{8}

Then do butterfly method.

12(8) = 5x

Solve for x.

96 = 5x\\\\x= 19.2 in

Hope that helped!

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Plz help this is due in like 1 hour
poizon [28]

Answer:

A

Step-by-step explanation:

Since each star is worth 50 points, if 5 stars are zapped that would convert to 250 points. Each level the number of stars increase by 5, level 6 would have 45 stars and so on.. With level 10 having 65 stars to complete.

4 0
3 years ago
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An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
3 years ago
WRITING TO EXPLAIN, HOW COULD YOU USE 5 X 6 =30 TO FIND THE PRODUCT <br> OF 6 X6 ?
Solnce55 [7]
5x6=30
6x6=36

You would need 1 more 6
I can't explain this very well, sorry.
8 0
3 years ago
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Help xd <br> -(x+2) =<br> 3(7x - 8)=<br> 7(x+3)-5x<br><br> Thank you!
HACTEHA [7]

Answer:

x=−5

x= 31/21 =1 10/21= 1.476190476

2x+21

hope this helps

3 0
3 years ago
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Please help me. i don't remember how to do this stuff so please explain it.
AVprozaik [17]

Answer:

which class are you read in?

actually don't know ans

Step-by-step explanation:

hope it's helpful to you

gud nyt

sweet dreams

8 0
3 years ago
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