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Paha777 [63]
2 years ago
13

You are hiring a limousine. One company charges a $60 reservation fee plus another $15 per hour. A second company charges a $25

reservation fee plus another $20 per hour. Write and solve a linear equation to find the number of hours at which the total cost is the same for both companies.
Mathematics
1 answer:
anzhelika [568]2 years ago
5 0

Answer:

7 hours

Step-by-step explanation:

Step one:

given data

let the total charges be y and the number of hours be x

Company A charges

$60 reservation fee plus another $15 per hour.

so

y=60+15x----------1

Company B charges

$25 reservation fee plus another $20 per hour

y=25+20x------------2

Step two:

equate the 2 expressions above we have

60+15x=25+20x

collect like terms

60-25=20x-15x

35=5x

divide both sides by 5

x= 35/5

x= 7 hours

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57%:43% of 3.09 million
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Answer:

Step-by-step explanation:

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2 years ago
Multiply.
sergij07 [2.7K]

Answer:

D) 2x² + 4x - 30

Step-by-step explanation:

Follow the FOIL method:

FOIL =

First

Outside

Inside

Last

...and is the order you multiply to solve:

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Combine like terms. Simplify:

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5 0
3 years ago
A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard
bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44

In point a)  

\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}

  = \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2

value= (1- \frac{1}{k^2}) \times 100 \% =75\%

In point b)

\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma  \\\\= |(x-\mu)|\leq 2.5 \times 4.44  \\\\  = |(x-\mu)|\leq 1.11 \\\\ =  (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)

In point c)

\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\

In point d)

using standard normal variate

x=20.17\\\\  z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17

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2 years ago
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1st is parallel

2nd is perpendicular

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