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MariettaO [177]
1 year ago
15

6 points Emily’s family needs to rent a moving truck to move their belongings to a different house. The rental cost for Trucks-A

-Lot is $42 per day plus $0.72 per mile. The rental cost for Move-It-Truckers is $70 per day plus $0.12 per mile. Let m represent Emily’s mileage. Write an equation to represent the cost of renting each truck, and determine what will be the number if miles for which the truck cost the same amount.
Mathematics
1 answer:
Alexxx [7]1 year ago
7 0

Answer/Step-by-step explanation:

Equation to represent the daily rental cost for each type of truck can be written as follows:

Daily rental cost for Trucks-A-Lot = 42 + 0.72m

Daily rental cost for Move-in-Truckers = 70 + 0.12m

Where, m = Emily's mileage

To determine the number of miles for which the truck cost the same amount, set both equations equal to each other and solve for m.

42 + 0.72m = 70 + 0.12m

Collect like terms

0.72m - 0.12m = 70 - 42

0.6m = 28

Divide both sides by 0.6

\frac{0.6m}{0.6} = \frac{28}{0.6}

m = 46.7

At approximately 47 miles, both trucks would cost the same amount.

Check:

Daily rental cost for Trucks-A-Lot = 42 + 0.72m

Plug in the value of x = 47

= 42 + 0.72(47) = $75.84 ≈ $76

Daily rental cost for Move-in-Truckers = 70 + 0.12m

Plug in the value of x = 47

= 70 + 0.12(47) = $75.64 ≈ $76

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Lenovo uses the​ zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as​ follows:
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Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}

The forcast for a period F_{t+1} is given by the formular

F_{t+1}=\alpha A_t+(1-\alpha)F_t

where A_t is the actual value for the preceding period and F_t is the forcast for the preceding period.

Part 1A:
Given <span>α ​= 0.1 and the initial forecast for october of ​$1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\  \\ =0.1(1.57)+(1-0.1)(1.83) \\  \\ =0.157+0.9(1.83)=0.157+1.647 \\  \\ =1.804

Therefore, the foreast for period 11 is $1.80


Part 1B:

</span>Given <span>α ​= 0.1 and the forecast for november of ​$1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

F_{12}=\alpha&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.1(1.62)+(1-0.1)(1.80) \\  \\ &#10;=0.162+0.9(1.80)=0.162+1.62 \\  \\ =1.782

Therefore, the foreast for period 12 is $1.78</span>



Part 2A:

Given <span>α ​= 0.3 and the initial forecast for october of ​$1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

F_{11}=\alpha&#10; A_{10}+(1-\alpha)F_{10} \\  \\ =0.3(1.57)+(1-0.3)(1.76) \\  \\ &#10;=0.471+0.7(1.76)=0.471+1.232 \\  \\ =1.703

Therefore, the foreast for period 11 is $1.70

</span>
<span><span>Part 2B:

</span>Given <span>α ​= 0.3 and the forecast for November of ​$1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

F_{12}=\alpha&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.3(1.62)+(1-0.3)(1.70) \\  \\ &#10;=0.486+0.7(1.70)=0.486+1.19 \\  \\ =1.676

Therefore, the foreast for period 12 is $1.68



</span></span>
<span>Part 3A:

Given <span>α ​= 0.5 and the initial forecast for october of ​$1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

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Therefore, the forecast for period 11 is $1.65

</span>
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Thus, the forecast for period 12 is given by:

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</span></span>
<span><span>Part 6:

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Therefore, the mean absolute deviation </span><span>using exponential smoothing where α ​= 0.5 of October, November and December is given by: 0.097</span></span>
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If 164 of the 400 eighth-graders take none of these courses, then 400-164=236 students take some courses.

Let x students take all three courses, then:

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3. 29-x take only both advanced computer and industrial technology.

Now let's count how many student take only one course:

1. 117-(x+34-x+70-x)=13+x take algebra,

2. 109-(x+29-x+70-x)=10+x take advanced computer,

3. 114-(x+29-x+34-x)=51+x take industrial technology.

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Answer:

The correct option is (A).

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Let <em>X</em> = number of orange  milk chocolate M&M’s.

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The event of an milk chocolate M&M being orange is independent of the other candies.

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The expected value of a Binomial random variable is:

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E(X)=np

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It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

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This probability is quite larger than 0.05.

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2 years ago
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andre [41]

Answer:

The company has to produce more than 92 guitars and sell them for making a profit.

Step-by-step explanation:

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⇒ x = 92

Therefore, the company has to produce more than 92 guitars and sell them for making a profit. (Answer)

7 0
2 years ago
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