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Alborosie
3 years ago
11

Plz help me so I don’t fail

Mathematics
2 answers:
Thepotemich [5.8K]3 years ago
7 0

Answer:

B

Step-by-step explanation:

below 0 is negative

so -15

Leviafan [203]3 years ago
5 0

<em>In</em><em> </em><em>My</em><em> </em><em>Knowledge</em><em> </em>

<em>Option</em><em> </em><em>B</em><em>.</em><em> </em><em>-15</em><em>°</em>

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Step-by-step explanation:

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I'll give brainliest!!! please help!
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Puritans believed that people (children in particular) were inherently bad (sinners) and had to learn to behave. Salvation lay in learning to check one's natural instincts and behave as an adult.

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at the drive in it cost 5 dollars per vehicle plus 2 dollars per person in the vehicle construct an equations from this informat
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Step-by-step explanation:

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3 years ago
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
Giving away 33 please help i suck at math
Harman [31]

D. (I used 3.14 as pi so my answer was a little off so I just went with the one closest to my answer)

8 0
3 years ago
Read 2 more answers
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