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Westkost [7]
3 years ago
7

Fatore a equação 2x²-3xy-2y²-2x-11y-12

Mathematics
1 answer:
Keith_Richards [23]3 years ago
3 0

2x²-3xy-2y²-2x-11y-12

Write -3xy as a difference

2x²+xy-4xy-2y²-2x-11y-12

write -2x as a difference

2x²+xy-4xy-2y²+4x-6x-11y-12

Write -11y as a difference

2x²+xy-4xy-2y²+4x-6x-8y-3y-12

Factor out 2x from the expression

2x×(x-2y-3)+xy-2y²+4x-8y-3y-12

Factor out y from the expression

2x×(x-2y-3)+y×(x-2y-3)+4x-8y-12

Factor out 4 from the expression

2x×(x-2y-3)+y×(x-2y-3)+4(x-2y-3)

Factor out x-2y-3 from the expression

Answer: (x-2y-3)x(2x+y+4)

PLEASE MARK ME AS BRAINLIEST!!

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Say that taking a quarter of student out of the group will add a quarter present of days for food so by take 20 students out you’ll get 60 days (that’s what I got hope this helps)
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A sample with mean of 85 and SD of 12 is transformed into z-scores. After the transformation, what are the values for the mean a
MrRa [10]

Answer:

The value remain unchanged.

Step-by-step explanation:

By Z score, we mean:

Z = \frac{x-\mu}{\sigma}, if we substitute the given values of mean and standard deviation, we obtain the z score value. That is,

Z =  \frac{85 - 0}{12} = 85/12 = 7.083333.

NB: We assume that mean and standard deviation are in the same units.

5 0
3 years ago
A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu
telo118 [61]

Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

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 = 110, p = 0.2

So

\mu = E(X) = np = 110*0.2 = 22

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

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

Z = \frac{25 - 22}{4.1952}

Z = 0.715

Z = 0.715 has a pvalue of 0.7626.

X = 17

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

Z = \frac{17 - 22}{4.1952}

Z = -1.19

Z = -1.19 has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

4 0
3 years ago
Carolyn is using the table to find 360% of 15. What values do X and Y represent in her table?
Ganezh [65]

★  100% of 15 = X

\implies \mathsf{100\%\;of\;15 = \left(\dfrac{100}{100} \times 15\right) = 15}

\implies \mathsf{X = 15}

★  20% of 15 = Y

\implies \mathsf{20\%\;of\;15 = \left(\dfrac{20}{100} \times 15\right)}

\implies \mathsf{20\%\;of\;15 = 3}

\implies \mathsf{Y = 3}

<u>Answers</u> : X = 15 and Y = 3

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