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MaRussiya [10]
3 years ago
14

What is the greatest common factor of the polynomial's terms?

Mathematics
1 answer:
Alinara [238K]3 years ago
4 0

Step-by-step explanation:

Write the prime factorization of each term:

9r⁵s = 3² × r⁵ × s

6r⁴s² = 2 × 3 × r⁴ × s²

12r²s = 2² × 3 × r² × s

The greatest common factor will have all the common factors raised to their lowest exponent.

So all three terms have 3, r, and s as factors.  The lowest exponent of 3 is 1.  The lowest exponent of r is 2.  The lowest exponent of s is 1.

GCF = 3 × r² × s

GCF = 3r²s

Factor out the GCF:

9r⁵s + 6r⁴s² − 12r²s

3r²s (3r³ + 2r²s − 4)

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The waverly company issued 4,000 shares of common stock worth 200,000.00 total. What is the par value of each share?
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In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
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Answer:

The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus 11 - 1 = 10.

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

  • The area to the left of the lower end of the interval shall be 1/2 \cdot (1 - 0.90)= 0.05.
  • The area to the left of the upper end of the interval shall be 0.05 + 0.90 = 0.95.

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

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t \approx 1.812.

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

\displaystyle t \cdot \frac{s_{n-1}}{\sqrt{n}},

where

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  • s_{n-1} is the unbiased estimate for the standard deviation, and
  • n is the sample size.

For this confidence interval:

  • t \approx 1.812,
  • s_{n-1} = 14, and
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Hence the width of the 90% confidence interval is

\displaystyle 1.812 \times \frac{14}{\sqrt{10}} \approx 7.65.

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

(38.3, 53.7).

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