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Mamont248 [21]
3 years ago
5

Which of the following represent the number of units

Mathematics
1 answer:
aev [14]3 years ago
7 0

Answer:

l think that the answer is,-5-(-2)

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Simplify (4-3i)(5+2i)
Anna [14]
Hey I'm probably too late but your answer would be 26-7i

Hope this helps :)
8 0
3 years ago
Write the ordered pair that describes a point 12 units down from and 7 units to the right of the origin.
emmainna [20.7K]
(-7,-12) s it both down and to the right of the origin
8 0
3 years ago
Read 2 more answers
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
4 years ago
An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos
Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when X = 0.8*675 = 540. So

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

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

8 0
4 years ago
Anyone know how to do this?
Finger [1]

Answer:

X = 2

Step-by-step explanation:

3 0
3 years ago
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