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Mars2501 [29]
3 years ago
6

Explain how to subtract the polynomial (The first attachment) from (The second attachment)

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
5 0
(5x^3-3x-6)-(4x^3+7x+1)\\\\=5x^3-3x-6-4x^3-7x-1\\\\=(5x^3-4x^3)+(-3x-7x)+(-6-1)\\\\=x^3+(-10x)+(-7)\\\\=\boxed{x^3-10x-7}
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562 ÷ Choose... = 14 R2
Zolol [24]

Answer:

7868

Step-by-step explanation:

⇒ This can be written algebraically (with a variable <em>x</em>) as:

562 ÷ x = 14

⇒ Convert the division as a fraction:

\frac{562}{x} = 14

⇒ Multiply both sides by 562 to get rid of the fraction and to isolate the variable <em>x</em>:

562 · \frac{562}{x} = 14 · 562

⇒ Simplify:

x = 7868

<u>Answer:</u> 7868

<em></em>

<em>Hope this helps!</em> :)

4 0
3 years ago
What is the slope of the line
Naily [24]

Step-by-step explanation:

(1, -3). (3, 0)

(0 + 3)/(3 - 1) = 3/2 is the slope

6 0
3 years ago
If line BA is a midsegment of triangle XYZ, find x.
DIA [1.3K]
For midsegment theorem
And parallels lune are twice

So AB = YZ /2

AB = 10/2 =5

BUT AB = X–1 =5

X–1 = 5 = X = 1+5 =6

X =;6


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MARK AS Brainliest
6 0
3 years ago
ASAP HELP PLEASE I WILL GIVEE 100 just help me please and ASAP
Ksenya-84 [330]

Answer:

A

Step-by-step explanation:

6 0
3 years ago
Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o
timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

\lambda t=8\times 1=8

The probability distribution of a Poisson distribution is:

P(X=x)=\frac{e^{-8}(8)^{x}}{x!}

Compute the value of P (X = 6) as follows:

P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

P(X\geq 10)=1-P(X

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 2.5=20

Compute the value of P (X ≥ 20) as follows:

P(X\geq 20)=1-P(X

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0
3 years ago
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