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FrozenT [24]
3 years ago
10

Which expressions are completely factored? Select each correct answer. 30a6−24a2=3a2(10a4−8) 16a5−20a3=4a3(4a2−5) 12a3+8a=4(3a3+

2a) 24a4+18=6(4a4+3)
Mathematics
2 answers:
Sedaia [141]3 years ago
8 0
B and D are the correct ones!
irinina [24]3 years ago
6 0

When factoring an expression, we need to identify the Greatest Common Factor of the expression (GCF). By definition, GCF is the product of prime factors involved with its Lowest Exponent.

We need to divide each term by the GCF to get the expression inside the parenthesis.

30a^6-24a^2\\ GCF\; is \; 6a^2\\ \\ 30a^6-24a^2=6a^2(\frac{3a^6}{6a^2} -\frac{24a^2}{6a^2} )=6a^2(5a^4-4)\\ ---------------------\\ \\ 16a^5-20a^3\\ GCF \; is \; 4a^3\\ \\ 16a^5-20a^3=4a^3(\frac{16a^5}{4a^3} -\frac{20a^3}{4a^3})=4a^3(4a^2-5) \\ ---------------------\\ \\ 12a^3+8a\\ GCF \; is \; 4a\\ \\ 12a^3+8a=4a(\frac{12a^3}{4a}+\frac{8a}{4a})=4a(3a^2+2)\\   ---------------------\\ 24a^4+18\\GCF \; is \; 6\\\\24a^4+18=6(\frac{24a^4}{6} +\frac{18}{6})=6(4a^4+3) \\    ---------------------

Conclusion:

From above we can conclude that the below expressions are factored completely

16a^5-20a^3=4a^3(4a^2-5)\\ \\ 24a^4+18=6(4a^4+3)

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A certain recipe ask for 5/9 flour and 1/3 flour. how much flour is needed ?
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Time spent using​ e-mail per session is normally​ distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that
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Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know 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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 11, \sigma = 3

a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 25, s = \frac{3}{\sqrt{25}} = 0.6

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

Z = \frac{11.2 - 11}{0.6}

Z = 0.33

Z = 0.33 has a pvalue of 0.6293.

X = 10.8

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

Z = \frac{10.8 - 11}{0.6}

Z = -0.33

Z = -0.33 has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.5 and 11 ​minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

Z = \frac{11 - 11}{0.6}

Z = 0

Z = 0 has a pvalue of 0.5.

X = 10.5

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

Z = \frac{10.5 - 11}{0.6}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 100, s = \frac{3}{\sqrt{100}} = 0.3

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

Z = \frac{11.2 - 11}{0.3}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486.

X = 10.8

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

Z = \frac{10.8 - 11}{0.3}

Z = -0.67

Z = -0.67 has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0
3 years ago
What is -a-2 if a = -5?
Vladimir79 [104]

Answer:

3

Step-by-step explanation:

-(-5) -2

5 - 2

3

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