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forsale [732]
3 years ago
6

F20 in-1 30 in. 13 in. 13 in.

Mathematics
1 answer:
scoray [572]3 years ago
7 0

Answer:

13

Step-by-step explanation:

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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)

= 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

z' = 0.842

z' = (x' - μ)/σ

0.842 = (x' - 26.4)/4.2

x' = 29.9364 = 29.9 mm

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

z = (x - μ)/σ = (64 - 65)/0.5 = -2.00

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!!!

7 0
3 years ago
Jerry started doing sit-ups every day. The first day he did 18 sit-ups. Every day after that he did 4 more sit-ups than he had d
USPshnik [31]

Answer:

18 + (4*5) = 38


because 4 * 5 is 20 then 20 + 18 is 38, and that's your answer.



have a nice day!   :)


7 0
3 years ago
Let omega be a complex number such that omega^3 = 1. Find all possible values of {1}{1 + \omega} + {1}{1 + \omega^2}. Enter all
Norma-Jean [14]

Answer:

1 is Answer.

Step-by-step explanation

\frac{1}{1+omega^{2} } + \frac{1}{1+omega }\\

= \frac{(-1)*(omega + 1)}{omega^{2}  }

As we know that ω²+ω+1=0

Thus putting in above equation, we get

= \frac{1}{(-1)*omega } + \frac{1}{(-1)*omega^{2}  }

Rearranging and simplifying:

= \frac{-1}{omega } + \frac{-1}{omega^{2}  }

= \frac{(-1)*(omega + 1)}{omega^{2}  }

= \frac{(-1)*(- omega^{2} )}{omega^{2}  }

= 1 Answer

8 0
3 years ago
How much is f(10) if f=x^2-2
dangina [55]

To find this value, simply substitute the number 10 in for x in your equation:

f(x) = x^2 - 2

f(10) = 10^2 - 2 = \boxed{98}

The answer is 98.

8 0
3 years ago
If the mean of 20, y, 10,5,15 is 22 find the value of y​
QveST [7]
Looks like 14.4 because you have to add them all together and divide by the total numbers
(20+22+10+5+15)/ 5
8 0
3 years ago
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