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3 years ago

In how many ways may one A, three B’s, two C’s, and one F be distributed among seven

Mathematics

1 answer:

astraxan [27]3 years ago

5 0

7 different ways is the answer

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Assume that there is a 8% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit

romanna [79]

Answer:

0.9936 = 99.36% probability that during a year, you can avoid catastrophe with at least one working drive

Step-by-step explanation:

For each disk, there are only two possible outcomes. Either it works, or it does not. The probability of a disk working is independent of any other disk, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Assume that there is a 8% rate of disk drive failure in a year.

So 100 - 8 = 92% probability of working, which means that $p = 0.92$

Two disks are used:

This means that $n = 2$

What is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.92)^{0}.(0.08)^{2} = 0.0064$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0064 = 0.9936$

0.9936 = 99.36% probability that during a year, you can avoid catastrophe with at least one working drive

8 0

3 years ago

A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

Immediately after launch, the rocket is subject to gravity, so its acceleration is

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$