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borishaifa [10]
3 years ago
5

Function do lappet which faced vultures serve in a temperate grassland

Physics
2 answers:
vfiekz [6]3 years ago
7 0
Um this doesn't make since to me since you did not clearly state your awnser
Blababa [14]3 years ago
3 0

Answer:

free

Explanation:

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g (12 points) The time between incoming phone calls at a call center is a random variable with exponential density p(x) = 1 r e
rusak2 [61]

Answer:

(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0

Explanation:

A function f(x) is a Probability Density Function if it satisfies the following conditions:

(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0

Given the function:

p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}

(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in [0,\infty).

(2)

\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1

The function p(x) satisfies the conditions for a probability density function.

6 0
3 years ago
Does sound travel much faster than light ?
Kitty [74]

Answer:

nothing travels faster than light

Example:

You’ll always see lightning before you hear it, because typically lightning will be a mile away, two miles away.

4 0
3 years ago
A 1502.7 kg car is traveling at 33.1 m/s when
Bond [772]

By definition of average acceleration,

<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²

Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:

∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N

If you just want the magnitude, drop the negative sign.

5 0
3 years ago
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker
Aleksandr-060686 [28]
B4 the tackle: 

<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>

<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>


<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>

<span>The vector triangle is right angled: </span>

<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>

<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>

<span>v(f) = 5.6 m/s (to 2 sig figs) </span>


<span>direction of v(f) is the same as the direction of the final momentum </span>

<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>


<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>




<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
4 0
3 years ago
What force could be applied to a box to make the net force in the horizontal direction of zero​
Kazeer [188]

If there is no friction, the force that moves the box forward horizontally must be matched by the same force.

If there is friction, then the force moving it forward = frictional force + the additional force you need to add.

6 0
3 years ago
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