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1 year ago

14

Jim is building a rectangular deck and wants the length to be 1 ft greater than the width. what will be the dimensions of the de

ck if the perimeter is to be 66 ft?

1 answer:

Travka [436]1 year ago

3 0

Let x = width
x+1 is then the length

2x+2(x+1)=66
2x+2x+2=66
4x=64
x=16
deck will be 16x17, nice for a BBQ. :)

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A hot air balloon is flying at a constant speed of 20 mi/h at a bearing of N 36° E. There is a 10mi/h crosswind blowing due east

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Answer:

Step-by-step explanation:

v = {[(20sin36°)i + (20cos36°)j] + 10i} mi/h

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2 years ago

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Answer:

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During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

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Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

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2 years ago

Every year the United States Department of Transportation publishes reports on the number of alcohol related and non-alcohol rel

Damm [24]

Complete question:

The line graph relating to the question was not attached. However, the line graph has can be found in the attachment below.

Answer:

17,209

Step-by-step explanation:

The line graph provides information about alcohol-related highway fatalities between year 2001 to 2010.

Determine the average number of alcohol-related fatalities from 2001 to 2006. Round to the nearest whole number.

The average number of alcohol related fatalities between 2001 - 2006 can be calculated thus :

From the graph:

Year - - - - - - - - - - Number of fatalities

2001 - - - - - - - - - - 17401

2002 - - - - - - - - - 17525

2003 - - - - - - - - - 17013

2004 - - - - - - - - - 16694

2005 - - - - - - - - - 16885

2006 - - - - - - - - - 17738

To get the average :

Sum of fatalities / number of years

(17401 + 17525 + 17013 + 16694 + 16885 + 17738) / 6

= 103256 / 6

= 17209.333

Average number of alcohol related fatalities is 17,209 (to the nearest whole number)

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2 years ago