Need help with solving this equation

Assume that during each second, a job arrives at a webserver with probability 0.03. Use the Poisson distribution to estimate the

Savatey [412]

Answer:

The probability that at lest one job will be missed in 57 second is $=1- e^{-1.71}$ =0.819134

Step-by-step explanation:

Poisson distribution:

A discrete random variable X having the enumerable set {0,1,2,....} as the spectrum, is said to be Poisson distribution.

$P(X=x)=\frac{e^{-\lambda t}({-\lambda t})^x}{x!}$ for x=0,1,2...

λ is the average per unit time

Given that, a job arrives at a web server with the probability 0.03.

Here λ=0.03, t=57 second.

The probability that at lest one job will be missed in 57 second is

=P(X≥1)

=1- P(X<1)

=1- P(X=0)

$=1-\frac{e^{-1.71}(1.71)^0}{0!}$

$=1- e^{-1.71}$

=0.819134

6 0

3 years ago

What is the difference between cross-section and intersection in geometry?

Triss [41]

A cross-section is the shape that would be exposed when cutting straight through a 3-D object. for example, cutting horizontally through a regular cone would expose a circle.

an intersection is where 2 or more shapes (does not matter how many dimensions) overlap. for example, the point at which two lines overlap is the intersection. or the area that two overlapping circles share is the intersection.

5 0

3 years ago

Please helpp!

Kitty [74]

Answer:

it may be B but I think it's C

Step-by-step explanation:

not sure if this is right, but I would say not congruent. it may be sas but in order for it to be sas then the Angle must be in-between the two congruent sides

5 0

2 years ago

How to know if a function is periodic without graphing it ?

zhenek [66]

A function $f(t)$ is periodic if there is some constant $k$ such that $f(t+k)=f(k)$ for all $t$ in the domain of $f(t)$ . Then $k$ is the "period" of $f(t)$ .

Example:

If $f(x)=\sin x$ , then we have $\sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x$ , and so $\sin x$ is periodic with period $2\pi$ .

It gets a bit more complicated for a function like yours. We're looking for $k$ such that

$\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5$

Expanding on the left, you have

$\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2$

and

$1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5$

It follows that the following must be satisfied:

$\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}$

The first two equations are satisfied whenever $k\in\{0,\pm4,\pm8,\ldots\}$ , or more generally, when $k=4n$ and $n\in\mathbb Z$ (i.e. any multiple of 4).

The second two are satisfied whenever $k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}$ , and more generally when $k=\dfrac{10n}7$ with $n\in\mathbb Z$ (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when $k$ is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

$\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2$

$\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5$

More generally, it can be shown that

$f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))$

is periodic with period $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .

4 0

3 years ago

Different shapes are drawn on cards and then the cards are placed in a bag. The number of cards for each shape is shown in the t

Anvisha [2.4K]

We can first add up the cards so we know how many we have in all:
16 + 16 + 18 = 50 cards

We can do this a little bit easier if we get the "16"-cards in one number total.

16 + 16 = 32

$\frac{32}{50}$ = 32 x 2 = $\frac{64}{100}$
50 x 2

$\frac{64}{100}$ = 64 : 2 = 32 %
100
We did just divide the % of two types cards on 2, so we get the %-chance of 1 type card.

I am not quite sure, but I think that 32 % is the correct answer.

6 0

3 years ago