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3 years ago

What is the percent of change? initial value:30 final value:180

Mathematics

2 answers:

sp2606 [1]3 years ago

5 0

600% I looked it up and multiplied 30 by 600 and it equals 180.

Alexxandr [17]3 years ago

3 0

600% 30 is 1/6 of 180

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Corrado has 27 books in his room that he has already read. He wants to give them away to his 3 younger siblings. How many books

Fantom [35]

Answer:

3b = 27

Step-by-step explanation:

They will each receive 9 books

7 0

3 years ago

I need help with this problem Someone explain this to me

Anton [14]

Answer:

16.5

Step-by-step explanation:

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3 years ago

The following table shows the percent increase of donations made on behalf of a non-profit organization for the period of 1984 t

pashok25 [27]

Giving the table below which shows the percent increase of donations made on behalf of a non-profit organization for the period of 1984 to 2003.

Year: 1984 1989 1993 1997 2001 2003

Percent: 7.8 16.3 26.2 38.9 49.2 62.1

The scatter plot of the data is attached with the x-axis representing the number of years after 1980 and the y-axis representing the percent increase of donations made on behalf of a non-profit organization.

To find the equation for the line of regression where the x-axis representing the number of years after 1980 and the y-axis representing the percent increase of donations made on behalf of a non-profit organization.
$\begin{center}
\begin{tabular}{ c| c| c| c| }
 x & y & x^2 & xy \\ [1ex] 
 4 & 7.8 & 16 & 31.2 \\ 
 9 & 16.3 & 81 & 146.7 \\ 
13 & 26.2 & 169 & 340.6 \\ 
17 & 38.9 & 289 & 661.3 \\ 
21 & 49.2 & 441 & 1,033.2 \\ 
23 & 62.1 & 529 & 1,428.3 \\ [1ex]
\Sigma x=87 & \Sigma y=200.5 & \Sigma x^2=1,525 & \Sigma xy=3,641.3 
\end{tabular}
\end{center}$

Recall that the equation of the regression line is given by
$y=a+bx$
where
$a= \frac{(\Sigma y)(\Sigma x^2)-(\Sigma x)(\Sigma xy)}{n(\Sigma x^2)-(\Sigma x)^2} = \frac{200.5(1,525)-87(3,641.3)}{6(1,525)-(87)^2} \\ \\ = \frac{305,762.5-316793.1}{9,150-7,569} = \frac{-11,030.6}{1,581} =-6.977$
and
$b= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{n(\Sigma x^2)-(\Sigma x)^2} = \frac{6(3,641.3)-(87)(200.5)}{6(1,525)-(87)^2} \\ \\ = \frac{21,847.8-17,443.5}{9,150-7,569} = \frac{4,404.3}{1,581} =2.7858$

Thus, the equation of the regresson line is given by
$y=-6.977+2.7858x$

The graph of the regression line is attached.

Using the equation, we can predict the percent donated in the year 2015. Recall that 2015 is 35 years after 1980. Thus x = 35.

The percent donated in the year 2015 is given by
$-6.977+2.7858(35)=-6.977+97.503=90.526$

Therefore, the percent donated in the year 2015 is predicted to be 90.5

7 0

3 years ago

Help me plz thanks......................

katovenus [111]

The answer is a linear function. For y=8/11x+22

8 0

3 years ago

Read 2 more answers

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

3 years ago