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Zigmanuir [339]
3 years ago
11

pablo will rent a car for the weekend. he can choose one of two plans. the first plan has an initial fee of $55.96 with addition

al costs of $0.12 per mile driven. the second plane has an initial fee of $63.96 and costs an additional $0.08 per mile driven. how many miles does pablo drive to make the miles the same?
Mathematics
1 answer:
kow [346]3 years ago
8 0

Answer:

200 miles

Step-by-step explanation:

First, set up the equations.

plan 1: initial fee of $55.96, $0.12 per mile

  • y = 0.12x + 55.96

plan 2: initial fee of $63.96, $0.08 per mile

  • y = 0.08x + 63.96

We want to know at what distance will the cost be the same. So, set the equations equal to each other.

0.12x + 55.96 = 0.08x + 63.96

Combine the variables.

0.04x + 55.96 = 63.96

Combine the constants.

0.04x = 8

Divide by 0.04 to isolate x.

x = 200 miles

Check by plugging x back into each equation.

y = 0.12(200) + 55.96

y = 24 + 55.96

y = $79.96

y = 0.08(200) + 63.96

y = 16 + 63.96

y = $79.96

You are correct!

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. upper left chamber is enlarged, the risk of heart problems is increased. The paper "Left Atrial Size Increases with Body Mass
Sonbull [250]

Answer:

Part 1

(a) 0.28434

(b) 0.43441

(c) 29.9 mm

Part 2

(a) 0.97722

Step-by-step explanation:

There are two questions here. We'll break them into two.

Part 1.

This is a normal distribution problem healthy children having the size of their left atrial diameters normally distributed with

Mean = μ = 26.4 mm

Standard deviation = σ = 4.2 mm

a) proportion of healthy children have left atrial diameters less than 24 mm

P(x < 24)

We first normalize/standardize 24 mm

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (24 - 26.4)/4.2 = -0.57

The required probability

P(x < 24) = P(z < -0.57)

We'll use data from the normal probability table for these probabilities

P(x < 24) = P(z < -0.57) = 0.28434

b) proportion of healthy children have left atrial diameters between 25 and 30 mm

P(25 < x < 30)

We first normalize/standardize 25 mm and 30 mm

For 25 mm

z = (x - μ)/σ = (25 - 26.4)/4.2 = -0.33

For 30 mm

z = (x - μ)/σ = (30 - 26.4)/4.2 = 0.86

The required probability

P(25 < x < 30) = P(-0.33 < z < 0.86)

We'll use data from the normal probability table for these probabilities

P(25 < x < 30) = P(-0.33 < z < 0.86)

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c) For healthy children, what is the value for which only about 20% have a larger left atrial diameter.

Let the value be x' and its z-score be z'

P(x > x') = P(z > z') = 20% = 0.20

P(z > z') = 1 - P(z ≤ z') = 0.20

P(z ≤ z') = 0.80

Using normal distribution tables

z' = 0.842

z' = (x' - μ)/σ

0.842 = (x' - 26.4)/4.2

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Part 2

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Population Standard deviation = σ = 5 mm

The central limit theory explains that the sampling distribution extracted from this distribution will approximate a normal distribution with

Sample mean = Population mean

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σₓ = (5/√100) = 0.5 mm

So, probability that the sample mean distance ¯x for these 100 will be between 64 and 67 mm = P(64 < x < 67)

We first normalize/standardize 64 mm and 67 mm

For 64 mm

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For 67 mm

z = (x - μ)/σ = (67 - 65)/0.5 = 4.00

The required probability

P(64 < x < 67) = P(-2.00 < z < 4.00)

We'll use data from the normal probability table for these probabilities

P(64 < x < 67) = P(-2.00 < z < 4.00)

= P(z < 4.00) - P(z < -2.00)

= 0.99997 - 0.02275 = 0.97722

Hope this Helps!!!

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