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3 years ago

How to factorize -3pa-9b-18pc

1 answer:

7 0

-3pa - 9b - 18pc

For starters, you know that you can't take away any of the letters from the equation because none of the three have the exact same letter; however, they all have something they can be divided by and that is -3.

It would be -3 instead of just 3 because the equation itself starts with a negative. (you could probably do it anyways with just 3, but you'd end up taking out a negative one, anyways, so what's the point?)

Now, you would have: -3 ( pa + 3b + 6pc )

That is truly all you can take out. So, that would end up being your answer since all you can really do is simplify.

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92x12 rounder both factors

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Answer:

90 and 10

Step-by-step explanation:

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3 years ago

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Suppose statement p is true but statement q is false. Which of these compound statements are true? Check all that apply. p ∧ q,

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P -> p, p ^ q

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Water is pumped out of a holding tank at a rate of 6-6e^-0.13t liters/minute, where t is in minutes since the pump is started. I

astraxan [27]

Procedure:

1) Integrate the function, from t =0 to t = 60 minutues to obtain the number of liters pumped out in the entire interval, and

2) Substract the result from the initial content of the tank (1000 liters).

Hands on:

Integral of (6 - 6e^-0.13t) dt ]from t =0 to t = 60 min =

= 6t + 6 e^-0.13t / 0.13 = 6t + 46.1538 e^-0.13t ] from t =0 to t = 60 min =

6*60 + 46.1538 e^(-0.13*60) - 0 - 46.1538 = 360 + 0.01891 - 46.1538 = 313.865 liters

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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

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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.

(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.

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2 years ago

Make v the subject of the formula

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$T =\dfrac 5{V+1}\\\\\\\implies VT+T = 5\\\\\implies VT = 5-T\\\\\implies V =\dfrac{5-T}T$

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