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3 years ago

May u hellp me plz. which is correct

Mathematics

1 answer:

marissa [1.9K]3 years ago

8 0

The answer is x-3y=6 that is the correct answer

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Anybody know the right answer?

qwelly [4]

Quadrant I : cot(x) > 0 → -cot(x) < 0

Quadrant II : cot(x) < 0 → -cot(x) > 0

Quadrant III : cot(x) > 0 → -cot(x) < 0

Quadrant IV : cot(x) < 0 → -cot(x) > 0

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$a)\ \left(\dfrac{11\pi}{6},\ -\sqrt3\right)\\\\\dfrac{11\pi}{6}\in\text{Quadrant IV, then -cot(x)} > 0,\ \text{we have}\ -\cot\dfrac{11\pi}{6}=-\sqrt3 < 0$

Answer: a)

$b)\\\dfrac{2\pi}{3}\in\text{Quadrant II},\ -\cot(x) > 0,\ \text{we have}\ \dfrac{\sqrt3}{3} > 0\\\\c)\\\dfrac{5\pi}{4}\in\text{Quadrant III},\ -\cot(x) < 0,\ \text{we have}\ -1 < 0\\\\d)\\\dfrac{4\pi}{3}\in\text{Quadrant III},\ -\cot(x) < 0,\ \text{we have}\ -\dfrac{\sqrt3}{3} < 0$

We know that one answer is only true, so we did not check the correctness of the cotangent value, only to the number sign

8 0

3 years ago

Where can I learn basics of algebra

svetlana [45]

Answer:

khan academy

Step-by-step explanation:

khan academy its something that I usually use to study algebra and Geometry this page includes videos with a really good information.

6 0

3 years ago

Read 2 more answers

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 ).