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valina [46]
3 years ago
9

What polynomial identity should be used to prove that 19 = 27 - 8? Difference of Cubes. a. Difference of Squares. b. Square of B

inomial. c. Sum of Cubes
Mathematics
2 answers:
Daniel [21]3 years ago
4 0
The answer is <span>Difference of Cubes.

Difference of cubes is </span>a³ - b³.<span>
27 can be written as 3</span>³ (= 3 × 3 × 3 = 27). So, a = 3.
8 can be written as 2³ (= 2 × 2 × 2 = 8). So, b = 2.

Difference of cubes can be expressed as:
       a³ - b³ = (a - b)(a² + ab + b²)
⇒   3³ - 2³ = (3 - 2)(3² + 3×2 + 2²) = 1 × (9 + 6 + 4) = 1 × 19 = 19
⇒  27 -  8  = 19
ExtremeBDS [4]3 years ago
3 0

Answer:

Difference of Cubes

Step-by-step explanation:

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The shape of the distribution of the time required to get an oil change at a 20 minute oil change facility. However, records ind
Katyanochek1 [597]

Answer:

For a mean oil change time of 20.51 minutes there would be a​ 10% chance of being at or​ below

Step-by-step explanation:

To solve this question, we need 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 21.3, \sigma = 3.9, n = 40, s = \frac{3.9}{\sqrt{40}} = 0.6166

Treating this as a random​ sample, at what mean​ oil-change time would there be a​ 10% chance of being at or​ below?

This is the 10th percentile, which is X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

By the Central Limit Theorem

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

-1.28 = \frac{X - 21.3}{0.6166}

X - 21.3 = -1.28*0.6166

X = 20.51

For a mean oil change time of 20.51 minutes there would be a​ 10% chance of being at or​ below

6 0
4 years ago
que condiciones debe cumplir las medidas de un triangulo para que puedas asegurar que es un triangulo rectangulo
nirvana33 [79]

Answer:

ENGLISH

One of the angles must equal 90 degrees and the other two angles must add up to 90 degrees.

SPANISH TRANSLATION

Uno de los ángulos debe ser igual a 90 grados y los otros dos ángulos deben sumar 90 grados.

3 0
3 years ago
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