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Monica [59]
3 years ago
9

Can an answer to an equation be a decimal?

Mathematics
2 answers:
sesenic [268]3 years ago
6 0

Answer:

Yes

Step-by-step explanation:

An answer to an equation can be a decimal!

-Please mark as brainliest!-

-Hope this helps!-

-Good luck!-

Scorpion4ik [409]3 years ago
3 0

Answer:

Step-by-step explanation:

Yes, the answer to the equation can be a decimal.

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The answer would be B
8 0
3 years ago
*20 points*<br> how do you get the weighted average from this table?
sertanlavr [38]

Answer:

Weighted average is the average of a set of numbers, each with different associated “weights” or values. To find a weighted average, multiply each number by its weight, then add the results.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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
PLEASE HELP ME. ONLY HAVE COUPLE MORE MINS LEFT
ludmilkaskok [199]

Answer:

11

Step-by-step explanation:

i dont really know but i think it is 11 because its all two numbers distance

6 0
2 years ago
Given<br> f(x) = -4(x+5),<br> what is the value of<br> f(-3)?<br> 
sp2606 [1]

Answer:

f(-3)=-8

Step-by-step explanation:

plug -3 in as x

-4(-3+5)

12-20

-8

4 0
2 years ago
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