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luda_lava [24]
3 years ago
6

Which option will be most cost effective?why?

Mathematics
1 answer:
anygoal [31]3 years ago
7 0
Option 1. If you have to pay $42 per person and you have 72 people to pay for you multiply 42X72= $3,024
It is more efficient to pay $500 per night
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HigherOrder thing name two rays with the same endpoint in the figure below. Do they form an angle? Explain.
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where is the figure?

Step-by-step explanation:

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Yareli and Rosa won a total of 125 tickets at an arcade. The number of tickets that Rosa won is 11 more than half the number of
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Step-by-step explanation:

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Solve for x: two thirds(x − 2) = 4x.
Sveta_85 [38]

Answer:

x =  - 0.4

Step-by-step explanation:

\frac{2}{3} (x - 2) = 4x

Expand with the distributive rule:

\frac{2}{3} x -  \frac{2}{3}  \times 2 = 4x

\frac{2}{3} x -  \frac{4}{3}  = 4x

Add 4/3 to both sides:

\frac{2}{3} x = 4x +  \frac{4}{3}

Subtract 4x from both sides

- \frac{10}{3} x =  \frac{4}{3}

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x =  - 0.4

5 0
3 years ago
An appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W. From pre
melamori03 [73]

Answer:

We conclude that a compact microwave oven consumes a mean of more than 250 W.

Step-by-step explanation:

We are given that an appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W with a population standard deviation of 15 W.

They take a sample of 20 microwave ovens and find that they consume a mean of 257.3 W.

Let \mu = <u><em>mean power consumption for microwave ovens.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 250 W     {means that a compact microwave oven consumes a mean of no more than 250 W}

Alternate Hypothesis, H_A : \mu > 250 W     {means that a compact microwave oven consumes a mean of more than 250 W}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                                T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean power consumption for ovens = 257.3 W

            σ = population standard deviation = 15 W

            n = sample of microwave ovens = 20

So, <em><u>the test statistics</u></em>  =  \frac{257.3-250}{\frac{15}{\sqrt{20} } }

                                      =  2.176

The value of z test statistics is 2.176.

<u>Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.</u>

Since our test statistic is more than the critical value of t as 2.176 > 1.645, 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 a compact microwave oven consumes a mean of more than 250 W.

7 0
3 years ago
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C !.!.!.!.!!.!.!.!.!!.!!.!.!.!.!!!.!!!.!.!.
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