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weqwewe [10]
3 years ago
12

Recently, the texas junior college teachers association annual conference was held in austin. at the time a taxi ride in austin

was $ 1.75 $1.75 for the first 1 5 15 of a mile and $ 0.25 $0.25 for each additional 1 5 15 of a mile. if the distance from one of the convention hotels to the airport is 5.8 5.8 miles, how much will it cost to take a taxi from that hotel to the airport?
Mathematics
1 answer:
erma4kov [3.2K]3 years ago
4 0
The correct question is
<span>Recently, the texas junior college teachers association annual conference was held in austin. at the time a taxi ride in austin was $ 1.75 for the first 1/5 of a mile and $ 0.25 for each additional 1/5 of a mile. if the distance from one of the convention hotels to the airport is 5.8 miles, how much will it cost to take a taxi from that hotel to the airport?

we know that
(1/5) miles is equal to 0.20 miles
so

for the first 0.20 miles----------> $1.75
</span>Remaining Distance = 5.8 - 0.2 miles = 5.6 miles
<span>5.6/(0.2) =28 (this is the number of (1/5)th of a mile after the 1st (1/5)th
</span><span>Each of these costs 0.25 dollars
</span><span>28*0.25 = 7 dollars

total cost=$1.75+$7------> $8.75

the answer is
$8.75</span>
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⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden
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Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

        P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } ~ N(0,1)

where, \hat p = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = \frac{354}{1219}

           n = sample of US adults  = 1291

           p = population proportion of adults

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                                    significance level are -1.96 & 1.96}

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ]

                         = [ \frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } , \frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } ]

                         = [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

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what is the measure of the vertex angle of an isosceles triangle if one of its base angles measures 48 degrees?
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180-2\cdot48=180-96=84
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