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3 years ago

If an outcome measure is normally distributed, this means that

1 answer:

5 0

A normally distributed result means that when the data is plotted on a graph, we can actually see a bell shaped curve. That highest frequency or peak is centered on the average value. Because it is shaped like this, we can say that:

most scores are around the average with some scores being lower or higher

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Solve x2 + 4x − 4 = 0

hoa [83]

Answer:

third option

Step-by-step explanation:

Given

x² + 4x - 4 = 0 ( add 4 to both sides )

x² + 4x = 4

Using the method of completing the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(2)x + 4 = 4 + 4, that is

(x + 2)² = 8 ( take the square root of both sides )

x + 2 = ± $\sqrt{8}$ = ± 2 $\sqrt{2}$ ( subtract 2 from both sides )

x = - 2 ± 2 $\sqrt{2}$

7 0

3 years ago

Read 2 more answers

Item 20<br> Factor 42x+28yusing the GCF.

Yanka [14]

So 7×6=42 and 7×4=28 GCF is 7 so 7x(6x+4)

4 0

2 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

2 years ago

0.60 gal to 0.12 gal as a fraction

adell [148]

Answer:

0.60 gal to 0.12 gal can be rewritten as:

$\frac{0.60}{0.12} =\frac{0.6}{0.6(0.2)} =\frac{1}{0.2}$