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2 years ago

What expression is equivalent to 3(6x-2)+2(x+5)

2 answers:

6 0

3(6x-2)+ 2(x+5)
= 3*(6x)-3*2 +2x +2*5 (distributive property)
= 18x -6 +2x +10
= (18x +2x) +(10-6)
= 20x +4

The final answer is 20x+4.

Hope this helps~

inessss [21]2 years ago

4 0

3(6x-2)+2(x+5)=18x-6+2x+10=20x+4=4(5x+1)

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a) Null hypothesis: $p\leq 0.12$

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b) $z_{\alpha}=1.64$

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

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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

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