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Artist 52 [7]
3 years ago
7

The monthly cost in dollars of a long distance phone plan is a linear function of the total calling time in minutes. The monthly

cost for 42 minutes of calls is $18.15 and the monthly cost for 90 minutes is $22.47. What is the monthly cost for 45 minutes of calls
Mathematics
1 answer:
Deffense [45]3 years ago
3 0

Answer:

The monthly cost for 45 minutes is $18.42

Step-by-step explanation:

Let

x -----> the total calling time in minutes

y ----> the monthly cost in dollars

we have the ordered pairs

(42,18.15) and (90,22.47)

step 1

Find the slope

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

substitute the values

m=\frac{22.47-18.15}{90-42}

m=\frac{4.32}{48}

m=\$0.09\ per\ minute

step 2

Find the linear equation in point slope form

y-y1=m(x-x1)

we have

m=0.09\\(x1,y1)=(42,18.15)

substitute

y-18.15=0.09(x-42)

step 3

Find the monthly cost for 45 minutes of calls

For x=45 min

substitute the value of x in the linear equation and solve for y

y-18.15=0.09(45-42)

y-18.15=0.09(3)

y-18.15=0.27

y=0.27+18.15

y=18.42

therefore

The monthly cost for 45 minutes is $18.42

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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
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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

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

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

= P(z < 0.86) - P(z < -0.33)

= 0.80511 - 0.37070 = 0.43441

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

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Population mean = μ = 65 mm

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

¯x = μₓ = μ = 65 mm

Standard deviation of the distribution of sample means = σₓ = (σ/√n)

where n = Sample size = 100

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

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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

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