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JulijaS [17]
3 years ago
7

Place the correct coefficients x²y³ + in the difference. xy² + y

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
8 0

The difference between the two polynomials is 11x²3y³ - 4xy².

<h3>What is the Difference between the Polynomials?</h3>

The difference between two polynomials is the result you get by subtracting one from the other.

Thus:

= (4x²2y³ + 2xy² – 2y) – (–7x²y³ + 6xy² – 2y)

= 4x²2y³ + 2xy² – 2y + 7x²y³ - 6xy² + 2y

= 11x²3y³ - 4xy²

Learn more about polynomials on:

brainly.com/question/2833285

#SPJ1

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Writing your answer as a power, the simplified expression is what
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8^{14}

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8^{10}  *  8^{4}        <u>add the powers 10 and 4</u>

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3 years ago
A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ t
Fiesta28 [93]

Answer:

100% probability that the sample mean scores will be between 87 and 124 points

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 105, \sigma = 20, n = 20, s = \frac{20}{\sqrt{20}} = 4.47

What is the probability that the sample mean scores will be between 87 and 124 points

This is the pvalue of Z when X = 124 subtracted by the pvalue of Z when X = 87. So

X = 124

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

By the Central Limit Theorem

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

Z = \frac{124 - 105}{4.47}

Z = 4.25

Z = 4.25 has a pvalue of 1

X = 87

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

Z = \frac{87 - 105}{4.47}

Z = -4.25

Z = -4.25 has a pvalue of 0

1 - 0 = 1

100% probability that the sample mean scores will be between 87 and 124 points

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4 years ago
Given the arithmetic sequence a, = -5 + 3(n - 1), what is the domain for n? (1 point)
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Answer:

n=a/3+8/3

Step-by-step explanation:

isolate the variable by dividing each side by factors that don't contain the variable

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