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CaHeK987 [17]
3 years ago
14

Find Y pls Thank you

Mathematics
2 answers:
Kaylis [27]3 years ago
5 0

Answer:

y=3

y=1

y=0

Step-by-step explanation:

For this problem, we are given the x-value. TO find the y-value, you plug in x and solve.

x=-1

y=-\frac{1}{2}(-1)+\frac{5}{2}              [multiply]

y=\frac{1}{2}+\frac{5}{2}                        [add]

y=\frac{6}{2}                               [divide]

y=3

---------------------------------------------------------------------------------------------------------------

x=3

y=-\frac{1}{2}(3)+\frac{5}{2}                [multiply]

y=-\frac{3}{2}+\frac{5}{2}                    [add]

y=\frac{2}{2}                              [divide]

y=1

---------------------------------------------------------------------------------------------------------------

x=5

y=-\frac{1}{2}(5)+\frac{5}{2}               [multiply]

y=-\frac{5}{2}+\frac{5}{2}                   [add]

y=0

ss7ja [257]3 years ago
3 0

Answer:

y=3

y=1

y=0

Step-by-step explanation:

y=-\frac{1}{2}x+ \frac{5}{2} \\\\y=\frac{5-x}{2} \\\\x=-1\\\\y=\frac{5-(-1)}{2}\\\\y=\frac{5+1}{2}\\\\y=\frac{6}{2}\\\\y=3\\\\x=3\\\\y=\frac{5-3}{2}\\\\y=\frac{2}{2}\\\\y=1\\\\y=\frac{5-5}{2}\\\\y=\frac{0}{2}\\\\y=0

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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

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)

= 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

P(z ≤ z') = 0.80

Using normal distribution tables

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Part 2

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

Hope this Helps!!!

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