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3 years ago

Divide 36x3 – 96x2 + 48x by 12x.

Mathematics

2 answers:

Pavel [41]3 years ago

8 0

( 36 x³ - 96 x² + 48 x ) : ( 12 x ) = 3 x² - 8 x + 4
Answer: A ) 3 x² - 8 x + 4
Thank you.

baherus [9]3 years ago

5 0

Answer:

<u>The answer is the option A</u>

$=3x^{2} -8x+4$

Step-by-step explanation:

we have

$\frac{36x^{3}-96x^{2}+48x}{12x}$

we know that

$\frac{36x^{3}-96x^{2}+48x}{12x}=\frac{36x^{3}}{12x}-\frac{96x^{2}}{12x}+\frac{48x}{12x}$

Simplify

$=3x^{2} -8x+4$

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Somebody help and explain!!!

Maslowich

Answer: The midpoint of segment PQ is the number 2.5

note: 2.5 as a fraction is 5/2; as a mixed number 2.5 converts to 2&1/2

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Explanation:

Apply the midpoint formula to get the midpoint of -8 and 6

We simply add up the values and divide by 2 and we get (-8+6)/2 = -2/2 = -1

So point Q is at -1 on the number line, which is exactly halfway from R to P

Focus on just points P and Q now. Apply the midpoint formula again

Q = -1

P = 6

(Q+P)/2 = (-1+6)/2 = 5/2 = 2.5

So the midpoint of segment PQ is 2.5

The decimal 2.5 can be written as the mixed number 2&1/2, showing that this new point is exactly halfway between 2 and 3.

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3 years ago

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hammer [34]

If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up (+1).

If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down (no change).

$207,689\approx210,000\\\\507,378\approx1,000,000\\\\3,422,087\approx3,422,000\\\\19,499,890\approx19,000000$

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3 years ago

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sergeinik [125]

Answer:

The answer is 72

Step-by-step explanation:

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3 years ago

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Nonamiya [84]

Answer:

Step-by-step explanation:

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3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).

$P(Y=0) + P(Y \geq 1) = 1 \implies P(Y \geq 1) = 1 -P(Y=0) = 1 - 0.0821 = 0.918$

3 0

3 years ago