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Butoxors [25]
2 years ago
12

Lisa paid $43.95 for 16.1 gallons of gasoline. What was the cost per gallon, rounded to the nearest hundredth?

Mathematics
2 answers:
Marizza181 [45]2 years ago
5 0

Answer:

$2.37

Step-by-step explanation:

43.95 ÷ 16.1 = 2.37

                  =$2.37

DerKrebs [107]2 years ago
4 0
$2.73, 42.95 divided by 16.1 = <span>2.7298136646, which rounded to the nearest hundredth is 2.73, then add the dollar sign, and you have $2.73</span>
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Mia's work to find the slope of a trend line through the points (3, 10) and (35, 91) is shown below. Mia's Work Step 1 StartFrac
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Step-by-step explanation:

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Marisol is making a rectangular wooden frame. She wants the length of the frame to be no more than 12 inches. She has less than
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4 0
2 years ago
in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru
melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is  

n = 200

the proportion is

p = 154/200 = 0.77

<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>

<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant  difference. </em>

The approximation would be to a Normal curve with this parameters:

<em>Mean </em>

p = 0.77

<em>Standard deviation </em>

\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298

The 90% confidence interval for the proportion would be then  

\bf  [0.77-z^*0.0298, 0.77+z^*0.0298]

where \bf z^* is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval \bf [-z^*,z^*] is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

<em>NORMSINV(0.95) and we get a value of 1.645 </em>

The 90% confidence interval for the proportion is then

\bf  [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

5 0
2 years ago
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