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adelina 88 [10]
3 years ago
14

Amy will rent a car for the weekend. She can choose one of two plans. The first plan has an initial fee of $38 and costs an addi

tional $0.12 per mile driven. The second plan has an initial fee of $51 and costs an additional $0.08 per mile driven.
Mathematics
1 answer:
mars1129 [50]3 years ago
6 0

Answer:

The first plan would be cheaper

Step-by-step explanation:

Lets say she drives 12 miles. Write an equation y=0.12x+38 then substitute the x with 12:

y=0.12(12)+38

y=1.44+38

y=39.44

Now you have to write a new equation for the second plan y=0.08x+51 and substitute x with 12:

y=0.08(12)+51

y=.96+51

y=51.96

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A certain car rental location charges a daily fee as well as a mileage charge. On one trip a businessman rented a car for the da
nikklg [1K]

Answer:

x\approx 267\ miles

Step-by-step explanation:

<u>Linear Modeling</u>

Some events can be modeled as linear functions. If we are in a situation where a linear model is suitable, then we need two sample points to make the model and predict unknown behaviors.

The linear function can be expressed in the slope-intercept format:

f(x)=mx+b

For the problem at hand, we must pick the adequate variables according to the data provided.

The question states the charge for renting a car is a function of the mileage. It also provides two points from which we can build our model. Let's set the following variables:

c = the charge for renting a car in dollars

x = the distance driven by the businessman in miles

Representing the ordered pair as (x,c), we have the points: (150,79) and (65,63.70). Our model will be expressed as:

c = mx+b

We must find the values of m and b with the data provided. Substituting the first point:

79 = 150m+b

Substituting the second point:

63.70 = 65m+b

Both equations form the following system:

\left\{\begin{matrix}150m+b=79\\ 65m+b=63.70 \end{matrix}\right.

Subtracting both equations:

150m-65m=79-63.70

Note the variable b was canceled out in the operation, leaving only the variable m to solve. Joining like terms:

85m=15.3

Solving:

m=15.3/85=0.18

From the first equation

79 = 150m+b

Solving for b:

b=79-150m=79-150(0.18) = 52.

The model for the problem is:

c=0.18x+52

Now we need to calculate how many miles (x) could be driven for c=$100. From the equation above, substitute c=100

100=0.18x+52

Solve for x:

0.18x+52=100

0.18x=100-52=48

x=48/0.18=266.67

Rounding to the closest integer:

\boxed{x\approx 267\ miles}

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Answer:

78.81%

Step-by-step explanation:

We are given;

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Sample mean; x¯ = 147.8

Sample size; n = 88

standard deviation; σ = 14

Z-score is;

z = (x¯ - μ)/(σ/√n)

Plugging in the relevant values;

z = (147.8 - 149)/(14/√88)

z = -0.804

From z-distribution table attached, we have; p = 0.21186

P(X > 147.8) = 1 - 0.21186 = 0.78814

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Answer:

12

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(10/y + 13) -3

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6 1/4 for 25/8

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