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3 years ago

what function is represented below? graph begins in the first quadrant and decreases quickly while crossing the ordered pair

Mathematics

2 answers:

posledela3 years ago

8 0

Hi my name is bubs I need sine money plz help

olchik [2.2K]3 years ago

8 0

Answer:

बूप बीप बूप मैं एक रोबोट हूं जो आपकी सभ्यता को संभालने के लिए आता है! मैं भी पार्ट मार्टियन हूं।

Step-by-Step Explanation:

भाई मेरे पास वास्तव में कोई स्पष्टीकरण नहीं है, लेकिन मॉड शायद इस का अनुवाद नहीं।

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let X represent the amount of time till the next student will arriv ein the library partking lot at the university. If we know t

AlekseyPX

Answer:

0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean of 4 minutes

This means that $m = 4, \mu = \frac{1}{4} = 0.25$

Find the probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot:

This is:

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2)$

In which

$P(X \leq 132) = 1 - e^{-0.25*132} = 1$

$P(X \leq 2) = 1 - e^{-0.25*2} = 0.393469$

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2) = 1 - 0.393469 = 0.606531$

0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.

4 0

2 years ago

What is the answer please ??

dmitriy555 [2]

To solve this, you plug in the first value (3) for x in the equations and and the second value (-6) for y in the equations! if the two equations then equal each other, it is true! if not, it’s false

6 0

2 years ago

Read 2 more answers

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = 1.5 m/s downwards

(b) The rate the area formed by the ladder is changing is approximately -75.29 ft.²/sec

Reasons:

The given parameter are;

Length of the ladder, l = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let y represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = 1.5 m/s downwards

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ -6.86 m/s downwards

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ -75.29 ft.²/sec

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ 0.286 rad/sec

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0

2 years ago

The hypotenuse of a triangle is 30 cm long and the length of leg A is 15 cm. How long is leg B? (round the the nearest tenth) PL

makvit [3.9K]

Answer:about 26

Step-by-step explanation:

30^2-15^2

Square root it 25.98 about 26

3 0

3 years ago

In circle O below, what is the measure of arc ED?

Serga [27]

Triangle FOD is an isosceles with base FD.
FD = 2 · 4 = 8
FD = CB
Arc CB = 56°
Arc ED = 1/2 Arc FD = 1/2 Arc CB
Arc ED = 56° : 2
Answer:
Arc ED = 28°

5 0

2 years ago