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frutty [35]
3 years ago
15

Javier is considering two lawn care services. Premier Landscaping charges $15.00 for travel to the job site and $55.00 per hour

for maintaining his lawn. Ace Landscaping does not have a travel fee, but charges $65.00 per hour for lawn maintenance. At what amount of time do the two lawn companies cost the same?
Mathematics
1 answer:
Wittaler [7]3 years ago
6 0

Answer:

Time in which both companies charge same cost = 1.5 hour

Step-by-step explanation:

Given:

                                                        Fixed    Variable

Premier Landscaping charges          $15        $55

Ace Landscaping charges                               $65

Find:

Time in which both companies charge same cost:

Computation:

Assume in X time both companies charge same cost:

So,

Premier Landscaping total cost = Ace Landscaping total cost

⇒ $15 + $55(Time taken) = $65 (Time taken)

⇒ $15 = $65 (Time taken) - $55(Time taken)

⇒ $15 = $10 (Time taken)

⇒ Time taken = 1.5 hour

Time in which both companies charge same cost = 1.5 hour

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To the nearest tenth, find the perimeter of ABC with vertices A (-2,-2) B (0,5) and C (3,1)
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the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.


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5 0
3 years ago
Rewa Delta Union Rugby CEO has become concerned about the slow pace of the rugby games played in the current union rugby, fearin
ozzi

Answer:

We conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

Step-by-step explanation:

We are given that Before the meeting, the mean duration of the 15-sided rugby game time was 3 hours, 5 minutes, that is, 185 minutes.

A random sample of 36 of the 15-sided rugby games after the meeting showed a mean of 179 minutes with a standard deviation of 12 minutes.

Let \mu = <em><u>mean duration of 15-sided union rugby games after the meeting.</u></em>

So, Null Hypothesis, H_0 : \mu \geq 185 minutes      {means that the mean duration of 15-sided union rugby games has increased or remained same after the meeting}

Alternate Hypothesis, H_A : \mu < 185 minutes     {means that the mean duration of 15-sided union rugby games has decreased after the meeting}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                            T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean duration of 15-sided union rugby games = 179 min

            s = sample standard deviation = 12 minutes

            n = sample of 15-sided rugby games = 36

So, <u><em>the test statistics</em></u>  =  \frac{179-185}{\frac{12}{\sqrt{36} } }  ~ t_3_5

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The value of t test statistics is -3.

<u>Now, at 1% significance level the t table gives critical value of -2.437 at 35 degree of freedom for left-tailed test.</u>

Since our test statistic is less than the critical value of t as -3 < -2.437, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

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muminat

Answer:

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