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3 years ago

Pe

Mathematics

2 answers:

asambeis [7]3 years ago

7 0

A and B is the answer

sdas [7]3 years ago

3 0

Answer:

A: 150+(45•p)

B: 125+(50•p)

Step-by-step explanation:

You multiply the monthly fee by a letter representing an unknown number because we dont know how many months you are staying with each company and we add the one time fee because you are only paying once, so in conclusion you add the one time fee to the result of the monthly fee by how many months.

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A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t

julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of successes

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with $\mu = \lambda, \sigma = \sqrt{\lambda}$ , if $\lambda>10$ .

Poisson variable with the mean 3

This means that $\lambda= 3$ .

(a) At least 3 in a week.

This is $P(X \geq 3)$ . So

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

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

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768$

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

In which

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

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

$P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680$

$P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008$

$P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504$

$P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216$

Then

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988$