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3 years ago

Two likes that do not intersect are parallel Always? Sometimes? Or Maybe?

Mathematics

1 answer:

earnstyle [38]3 years ago

3 0

Answer:

sometimes

Step-by-step explanation: They may not intersect at first but they may be put on a slight angle to where they may intersect in the longrun.

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Help ASAP!!!!!!!!!!!!!!!!!!!!!! Only if you know the answer

pentagon [3]

Step-by-step explanation:

try brainly app it will give you the correct answer for frr

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2 years ago

The number of requests for assistance received by a towing service is a Poisson process with rate θ = 4 per hour.a. Compute the

aliya0001 [1]

Answer:

a) 9.93% probability that exactly ten requests are received during a particular 2-hour period

b) 13.53% probability that they do not miss any calls for assistance

c) 2

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

Poisson process with rate θ = 4 per hour.

This means that $\mu = 4n$ , in which n is the number of hours.

a. Compute the probability that exactly ten requests are received during a particular 2-hour period.

n = 2, so $\mu = 4*2 = 8$

This is P(X = 10). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 10) = \frac{e^{-8}*8^{10}}{(10)!} = 0.0993$

9.93% probability that exactly ten requests are received during a particular 2-hour period

b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?

n = 0.5, so $\mu = 4*0.5 = 2$

This is P(X = 0). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

13.53% probability that they do not miss any calls for assistance

c. How many calls would you expect during their break?

n = 0.5, so $\mu = 4*0.5 = 2$

4 0

3 years ago

Find a pair of polar coordinates for the point with rectangular coordinates (5, –5).

Elanso [62]

Answer:

(5*sqrt(2), 5pi/4)

Step-by-step explanation:

In Polar coordinates, tan(theta)=y/x and r=sqrt(x^2+y^2)

tan(theta)=-5/5=-1. Theta=5pi/4

r=sqrt(5^2+5^2)=5*sqrt(2)

Hence the Polar coordinate is (5*sqrt(2), 5pi/4)

7 0

3 years ago

How many swimmers are holding exaclty 3 toy whales?

Setler79 [48]

About 40 swimmers maybe more

6 0

3 years ago

Need help finding the area of the shaded region. Round to the nearest tenth. Need help ASAP

Sergio [31]

Answer: 294.4 m²

<u>Step-by-step explanation:</u>

Separate the shaded region into two parts:

The section containing the central angle of 230° (360° - 130°)
The triangle with sides 11.1, 11.1 & 20.12 (use Law of Cosines)

$1.\ Area(A)=\pi\ r^2\ \bigg(\dfrac{\theta}{360}\bigg)\\\\\\.\qquad \qquad =\pi(11.1)^2\bigg(\dfrac{230}{360}\bigg)\\\\\\.\qquad \qquad =247.3$

$2.\ \text{Use Law of cosines to find the length of the third side.}\\\text{ Then use Heron's formula to find the Area of the triangle.}\\\\s=\dfrac{11.1+11.1+20.12}{2}=21.16\\\\\\A=\sqrt{s(s-a)(s-b)(s-c)}\\\\.\ =\sqrt{21.16(21.16-11.1)(21.16-11.1)(21.16-20.12)}\\\\.\ =\sqrt{2227}\\\\.\ =47.1$

Area of shaded region = Area of (1) + Area of (2)

= 247.3 + 47.1

= 294.4

8 0

3 years ago

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