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galina1969 [7]
3 years ago
14

All of the following ratios are equivalent except _____. ANSWER QUICK FIRST CORRECT ANSWER GETS BRAINLIEST

Mathematics
2 answers:
Alisiya [41]3 years ago
8 0
The answer Is going to be 15/10
jok3333 [9.3K]3 years ago
3 0

Answer:

  • 15/10  

Step-by-step explanation:

<u>Ratios given:</u>

  • 15/10  = 3:2  this is different from the others
  • 8 to 12  = 8:12 = 2:3
  • 2/3  = 2:3
  • 6:9 = 2:3
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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
Line w passes through the points (-2, 3) and (3, 13) . What<br> is the y-intercept of the line w?
Virty [35]
0.83 is they answer
7 0
3 years ago
Please help ASAP !!! Thank you !
snow_tiger [21]
<h2>(1)</h2><h2> =(a+b)(3c-d)</h2><h2> =a(3c-d)+b(3c-d)</h2><h2> =3ac-ad+3bc-bd</h2>

<h2>(2)</h2><h2> =(a-b)(c+2d)</h2><h2> =a(c+2d)-b(c+2d)</h2><h2> =ac+2ad-bc-2bd</h2>

<h2>(3)</h2><h2> =(a-b)(c-2d)</h2><h2> =a(c-2d)-b(c-2d)</h2><h2> =ac-2ad-bc+2bd</h2>

<h2>(4)</h2><h2> =(2a+b)(c-3d)</h2><h2> =2a(c-3d)+b(c-3d)</h2><h2> =2ac-6ad+bc-3bd</h2>

8 0
3 years ago
Whoever answers this correctly gets brainliest
ELEN [110]

Answer: 11. Is A

Step-by-step explanation:

5 0
3 years ago
To make cheesecake you need about 21/2pounds of creamcheese how many pounds of cream cheese will you need to make 2 cheesecakes
Elina [12.6K]
2 1/2 pounds of cream cheese for a cheesecake
to find the amount needed to make 2 cheesecakes multiply 2 1/2 times 2
2 1/2 times 2 = 5
you need 5 pounds of cream cheese to make 2 cheesecakes
6 0
3 years ago
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