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Citrus2011 [14]
2 years ago
11

Jason went shopping.

Mathematics
1 answer:
melomori [17]2 years ago
8 0

53.55 / 85 = 0.63

0.63 * 100 = 63

63 - 38 = 25

watch was 25 before discount

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3x+2y=8 how to solve
GarryVolchara [31]
8-3x
Divide the whole thing by 2 so it would be y=4-1.5x
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3 years ago
How old am I if 250 reduced by 5 times my age is 50?
Korolek [52]
250 - 5(x) = 50

250 (-250) - 5x = 50 (-250)

-5x = -200

-5x/-5 = -200/-5

x = 40

hope this helps

you are 40 years old!
4 0
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Help Please.Use complete sentences to explain the steps required in converting 180 degrees to its equivalent radian measure. Inc
Phoenix [80]
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180 (π/180)
The result is π radians.
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3 years ago
Explain how finding 7×20 is similar to finding 7×2,000 . then Find each product
3241004551 [841]
7x20=140. 7x2000=14000. The only thing that changes the product of these numbers is the amount of zeros behind the 2. Since the only two numbers that affect the answer is the 7 and the 2. (7x2=14). The number of zeros behind the 2 affect how many zeros will be included in the product.
6 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
2 years ago
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