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mafiozo [28]
3 years ago
8

Is 0.790 greater than or equal to 0.79?

Mathematics
1 answer:
dedylja [7]3 years ago
8 0

It's the same vallue. If they were different, i could argue that 2.000 is greater than 2.00. Impossible, because 2 is equal to 2.

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Simplify the expression X^-1
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Answer:

Step-by-step explanation:

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2 years ago
Choose the appropriate pattern and use it to find the product: (p4−q4)(p4+q4).
sp2606 [1]

The expression can be solved by expanding the bracket and multiplying out the terms

(p^4-q^4)(p^4+q^4)\begin{gathered} =p^4(p^4+q^4)-q^4(p^4+q^4) \\ =p^8+p^4q^4-p^4q^4-q^8 \\ =p^8-q^8 \end{gathered}

Therefore, the expression can be simplified as;

p^8-q^8

Alternatively, using the theorem of difference of two squares, which is

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1 year ago
Lorena and Sebastian are both five years old. Every year they each get a cash present from their ncighbor.
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14,  she will be 14 when she has more

8 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
2 years ago
15) Billy charges $15 to mow a yard. He needs at least $200 for the new bicycle that he wants. Write and solve an inequality to
andriy [413]
15x >= 200
x>=14
does this help?

5 0
3 years ago
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