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3 years ago

Find the total cost. $ 1,299 T.V.; 7% tax Math

Mathematics

1 answer:

Leto [7]3 years ago

4 0

The total to this would be $1,389.93 because $1,299 + 7% = $1,389.93

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Question 4 of 25

natali 33 [55]

Answer:

Step-by-step explanation:

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3 years ago

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

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3 years ago

Expanded notation for 55,032178

fenix001 [56]

74746394738284729274638292

8 0

3 years ago

Who Goes To Markham Middle School? If so what teachers do you have??? :3

lesya692 [45]

Answer:

Vaush

Step-by-step explanation:

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2 years ago

An Airliner has a capacity for 300 passengers. If the company overbook a flight with 320 passengers, What is the probability tha

TEA [102]

Answer:

The probability is $P(X >300 ) = 0.97219$

Step-by-step explanation:

From the question we are told that

The capacity of an Airliner is k = 300 passengers

The sample size n = 320 passengers

The probability the a randomly selected passenger shows up on to the airport

$p = 0.96$

Generally the mean is mathematically represented as

$\mu = n* p$

=> $\mu = 320 * 0.96$

=> $\mu = 307.2$

Generally the standard deviation is

$\sigma = \sqrt{n * p * (1 -p ) }$

=> $\sigma = \sqrt{320 * 0.96 * (1 -0.96 ) }$

=> $\sigma =3.50$

Applying Normal approximation of binomial distribution

Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

$P(X > k ) = P( \frac{ X -\mu }{\sigma } > \frac{k - \mu}{\sigma } )$

Here $\frac{ X -\mu }{\sigma } =Z (The \ standardized \ value \ of \ X )$

=> $P(X >300 ) = P(Z > \frac{300 - 307.2}{3.50} )$

Now applying continuity correction we have

$P(X >300 ) = P(Z > \frac{[300+0.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > \frac{[300.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > -1.914 )$

From the z-table

$P(Z > -1.914 ) = 0.97219$

$P(X >300 ) = 0.97219$

8 0

2 years ago