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Sedbober [7]
4 years ago
7

In problems 16-20, calculate the expected price in the year 2008 if you assume that there was a consistent 5% inflation rate and

use the given 1988 price. answers should be rounded to the nearest penny. 16. median salary, $27,225 17. gallon of gas, $1.08 18. dozen eggs, $0.89 19. movie admission, $3.50 20. mcdonalds hamburger, $0.62
Mathematics
1 answer:
Lena [83]4 years ago
5 0
16. 72236.03 
17. <span>2.87 
</span>18. 2.36
19. 9.29
20. 1.65
The exponential growth formula is as follows
y=a(1+r)^{x}:
a is initial value
(<span>$27,225 17. gallon of gas, $1.08 18. dozen eggs, $0.89 19. movie admission, $3.50 20. mcdonalds hamburger, $0.62)
r is rate of increase (0.05)
x is time (2008-1988 = 20 years)
to solve plug in values for each one.
</span>y=27225(1+0.05)^{20} = 72236.03

1.08(1+0.05)^{20} = 2.87

.89(1+0.05)^{20} = 2.36

3.50(1+0.05)^{20} = 9.29

.62(1+0.05)^{20} = 1.65



<span>

</span>
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The equation that is an example of inverse variation is B none

<h3>How to determine the equations that are direct variation?</h3>

There are several types of variations.

Some of them include:

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  • Inverse variation

Equations that are inverse variation are represented as:

y = k/x

Where k represents the constant of proportionality or variation

From the list of options, none of the equations has the form y = k/x

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2 years ago
A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the
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Answer:

a) \hat p=\frac{471}{1024}=0.460

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Data given and notation  

n=1024 represent the random sample taken    

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

\hat p=\frac{471}{1024}=0.460 estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering    

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305

Part b

If we replace the values obtained we got:

0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429

0.460+1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.491

The 99% confidence interval would be given by (0.429;0.491)

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t = number of years = 1

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=> A = 2000 (1 + 0.03/4)^4

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