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avanturin [10]
3 years ago
6

The charge to rent a trailer is ​$2525 for up to 2 hours plus ​$99 per additional hour or portion of an hour. Find the cost to r

ent a trailer for 2.82.8 ​hours, 33 ​hours, and 8.78.7 hours. Then graph all ordered​ pairs, (hours,​ cost), for the function.

Mathematics
1 answer:
Ksju [112]3 years ago
4 0

The question is not written properly! Complete question along with answer and step by step explanation is provided below.

Question:

The charge to rent a trailer is ​$25 for up to 2 hours plus ​$9 per additional hour or portion of an hour.

Find the cost to rent a trailer for 2.8 ​hours, 3 ​hours, and 8.7 hours.

Then graph all ordered​ pairs, (hours,​ cost), for the function.

Answer:

ordered pair = (2.8, 34)

ordered pair = (3, 34)

ordered pair = (8.7, 88)

Step-by-step explanation:

Charge for 2.8 ​hours:

$25 for 2 hours

$9 for 0.8 hour

Total = $25 + $9

Total = $34

ordered pair = (2.8, 34)

Charge for 3 ​hours:

$25 for 2 hours

$9 for 1 hour

Total = $25 + $9

Total = $34

ordered pair = (3, 34)

Charge for 8.7 ​hours:

$25 for 2 hours

$9 for 1 hour

$9 for 1 hour

$9 for 1 hour

$9 for 1 hour

$9 for 1 hour

$9 for 1 hour

$9 for 0.7 hour

Total = $25 + $9 + $9 + $9 + $9 + $9 + $9 + $9

Total = $88

ordered pair = (8.7, 88)

The obtained ordered pairs are graphed, please refer to the attached graph.

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BigorU [14]

Answer:

29.5+/-1.11

= ( 28.39, 30.61)

Therefore, the 90% confidence interval (a,b) =( 28.39, 30.61)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 29.5

Standard deviation r = 5.2

Number of samples n = 59

Confidence interval = 90%

z-value (at 90% confidence) = 1.645

Substituting the values we have;

29.5+/-1.645(5.2/√59)

29.5+/-1.645(0.676982337100)

29.5+/-1.113635944529

29.5+/-1.11

= ( 28.39, 30.61)

Therefore, the 90% confidence interval (a,b) =( 28.39, 30.61)

4 0
3 years ago
Assume, for the sake of this question, that the data were collected through a well-designed, well-implemented random sampling me
amid [387]

Answer:

Since the pvalue of the test is 0.2743 > 0.1, the threshold probably was met.

Step-by-step explanation:

The widget manufacturing company had established a threshold of 60% preferring the proposed new widget to move forward with producing the new widgets.

This means that at the null hypothesis we test if the proportion is at least 60%, that is:

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And the alternate hypothesis is:

H_{a}: p < 0.6

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.6 is tested at the null hypothesis:

This means that:

\mu = 0.6

\sigma = \sqrt{0.6*0.4}

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This means that n = 575, X = \frac{338}{575} = 0.5878

Value of the test-statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.5878 - 0.6}{\frac{\sqrt{0.6*0.4}}{\sqrt{575}}}

z = -0.6

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We want to find the probability of a proportion of 0.5878 or lower, which is the pvalue of z = -0.6.

Looking at the z-table, z = -0.6 has a pvalue of 0.2743.

Since 0.2743 > 0.1, the threshold probably was met.

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stich3 [128]

Answer:

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8 0
3 years ago
a customer buys a diiferent book that has an original selling price of $38 the books is discounted 25% the customer must pay a 6
Nimfa-mama [501]
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7 0
3 years ago
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