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4 years ago

Problems 7-12 Simplify each expression,

Mathematics

1 answer:

vredina [299]4 years ago

6 0

Answer:

7. 81

8. 6t³-27t²+2t-9

1667

9. -14y/\4+21y²+63y

10. 6x-1

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A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa

UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

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 interval.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so $\mu = \frac{14}{60} = 0.2333$

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

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

We want $P(X \geq 1)$ . So

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

In which

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

$P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081$

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0

4 years ago

Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the Laplace transform

sveticcg [70]

Answer:

L(f(t)) = $2 \frac{e^{-s} }{s} - \frac{1}{s}$

Step-by-step explanation:

let f be a function defined for t ≥ 0

we can write the function f(t) in terms of unit function as follows

f(t) = 2 u,(t) - 1 where

0≤ t < 1

f(t) = (2 * 0) -1 = -1

when t ≥ 1

f(t) = (2*1 )- 1 = 1

Now the Laplace transform L(F(T)) = 2L( u, (t) ) - L(1) --------equation 1

this is because L(u,(t)) = $\frac{e^{-cs} }{s}$

c = 1 hence L(1) = 1/s

back to equation 1

L(f(t)) = 2 $\frac{e^{-cs} }{s}$ - 1/s laplace transform

also L(u(t) ) = $\frac{e^{-s} }{s}$

3 0

3 years ago

What is the value of x that makes l1||l2

Sholpan [36]

Answer: 10

Step-by-step explanation:

By the alternate interior angles theorem,

$4x=2x+20\\2x=20\\x=10$

3 0

2 years ago

2pq-2rs+2pr+pq-2rs+3pr what is the answer??

sergejj [24]

The answer is 1pq +1pr

3 0

4 years ago

What’s the an and the a10

malfutka [58]

Answer:

an = -6 + n

a10 = -6+ 10 =4

Step-by-step explanation:

an should start at one behind the first number so it should be -6. Then we add whatever the sequence is increasing by, which is 1 in this case, multiplied by n.

an = -6 + n

a10 = -6+ 10 =4

6 0

3 years ago