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gregori [183]
3 years ago
7

Difference in the following polynomials (6x^3 - 2x^2 + 4) -(2x^3 + 4x^2 - 5)

Mathematics
1 answer:
Colt1911 [192]3 years ago
6 0

\bold{[ \ Answer \ ]}

\boxed{\bold{4x^3-6x^2+9}}

\bold{[ \ Explanation \ ]}

  • \bold{Find \ Difference: \ \left(6x^3\:-\:2x^2\:+\:4\right)-\left(2x^3\:+\:4x^2\:-\:5\right)}

\bold{-----------------------}

  • \bold{Remove \ Parenthesis}

\bold{6x^3-2x^2+4-\left(2x^3+4x^2-5\right)}

  • \bold{Remove \ Parenthesis \ / \ Simplify}

\bold{-\left(2x^3+4x^2-5\right): \ -2x^3-4x^2+5}

  • \bold{Rewrite}

\bold{6x^3-2x^2+4-2x^3-4x^2+5}

  • \bold{[Simplify \ 6x^3-2x^2+4-2x^3-4x^2+5] \ 4x^3-6x^2+9}

\bold{4x^3-6x^2+9}

\boxed{\bold{[] \ Eclipsed \ []}}

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Step-by-step explanation:

Step by step

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In words.

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3 years ago
The National Transportation Safety Board publishes statistics on the number of automobile crashes that people in various age gro
ra1l [238]

Answer:

a) Null hypothesis:p\leq 0.12  

Alternative hypothesis:p > 0.12  

b) z_{\alpha}=1.64

c) z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362  

d) p_v =P(z>1.362)=0.0866  

e) For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

Step-by-step explanation:

Information given

n=1000 represent the random sample selected

X=134 represent the number of young drivers ages 18 – 24 that had an accident

\hat p=\frac{134}{1000}=0.134 estimated proportion of young drivers ages 18 – 24 that had an accident

p_o=0.12 is the value that we want to verify

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic

p_v{/tex} represent the p valuePart aWe want to verify if the population proportion of young drivers, ages 18 – 24, having accidents is greater than 12%:  Null hypothesis:[tex]p\leq 0.12  

Alternative hypothesis:p > 0.12  

The statistic would be given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Part b

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the right and we got:

z_{\alpha}=1.64

Part c

For this case the statistic would be given by:

z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362  

Part d

The p value can be calculated with the following probability:

p_v =P(z>1.362)=0.0866  

Part e

For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

8 0
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