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2 years ago

[4 -4 0 0]+[4 -2 2 -4]

Mathematics

1 answer:

jok3333 [9.3K]2 years ago

6 0

Answer:

[8 -6 2 -4].

Step-by-step explanation:

[4 -4 0 0]+[4 -2 2 -4]

= [4+4 -4+-2 0+2 0+-4]

= [8 -6 2 -4]

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A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true

kvv77 [185]

Answer:

99.74% probability that the sample proportion will be less than 0.1

Step-by-step explanation:

I am going to 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:

$n = 276, p = 0.06$

$\mu = E(X) = np = 276*0.06 = 16.56$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{276*0.06*0.94} = 3.9454$

What is the probability that the sample proportion will be less than 0.1

This is the pvalue of Z when X = 0.1*276 = 27.6. So

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

$Z = \frac{27.6 - 16.56}{3.9454}$

$Z = 2.8$

$Z = 2.8$ has a pvalue of 0.9974

99.74% probability that the sample proportion will be less than 0.1

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Suppose that your business is operating at the 4.5-Sigma quality level. If projects have an average improvement rate of 50% annu

Luda [366]

Answer:

11.75 years

Step-by-step explanation:

If we ignore the fact that "6-sigma" quality means the error rate corresponds to about -4.5σ (3.4 ppm) and simply go with ...

P(z ≤ -6) ≈ 9.86588×10^-10

and

P(z ≤ -4.5) ≈ 3.39767×10^-6

the ratio of these error rates is about 0.000290372. We're multiplying the error rate by 0.5 each year, so we want to find the power of 0.50 that gives this value:

0.50^t = 0.000290372

t·log(0.50) = log(0.00290372) . . . . take logarithms

t = log(0.000290372)/log(0.50) ≈ -3.537045/-0.301030

t ≈ 11.75

It will take about 11.75 years to achieve Six Sigma quality (0.99 ppb error rate).

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<em>Comment on Six Sigma</em>

A 3.4 ppm error rate is customarily associated with "Six Sigma" quality. It assumes that the process may have an offset from the mean of up to 1.5 sigma, so the "six sigma" error rate is P(z ≤ (1.5 -6)) = P(z ≤ -4.5) ≈ 3.4·10^-6.

Using that same criteria for the "4.5-Sigma" quality level, we find that error rate to be P(z ≤ (1.5 -4.5)) = P(z ≤ -3) ≈ 1.35·10^-3.

Then the improvement ratio needs to be only 0.00251699, and it will take only about ...

t ≈ log(0.00251699)/log(0.5) ≈ 8.6 . . . . years

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