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hjlf
3 years ago
12

Please help!!!! I don’t understand how to figure it out

Mathematics
1 answer:
lubasha [3.4K]3 years ago
4 0

Answer:

(sin x)^2*(sec x) is positive in QII

Step-by-step explanation:

(sin x)^2 is always 0 or positive.  Here x lies in QII.

sec x is positive when the adjacent side is positive and negative when the adjacent side is negative.  In QII the adjacent side is positive.

In summary, (sin x)^2*(sec x) is positive in QII

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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
4 years ago
Identify the area, show work
Wittaler [7]

Answer:

30

Step-by-step explanation:

To calculate the area of a triangle you will do it like a square or rectangle but you will divide your answer by 2. So...

10 * 6 = 60

60 / 2 = 30

7 0
3 years ago
A ribbon is 30cm long and it is cut into three pieces such that each piece is 2cm longer than the next. Represent this as an equ
Alisiya [41]

Answer:

Step-by-step explanation:

3 pieces such that each piece is 2 cm longer then the next...

x , x + 2, x + 4

x + x + 2 + x + 4 = 30

3x + 6 = 30

3x = 30 - 6

3x = 24

x = 24/3

x = 8

x + 2 = 8 + 2 = 10

x + 4 = 8 + 4 = 12

check...

8 + 10 + 12 = 30

18 + 12 = 30

30 = 30 (correct)...it checks out

the ribbon lengths are : 8 cm, 10 cm, and 12 cm

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B. $27,000

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3 years ago
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Which of the following rational numbers is equal to 3/5?
Vedmedyk [2.9K]
3/5 = 60%
20 ÷ 5 = 4
3 x 4 = 12

Answer is A) 12/20
4 0
3 years ago
Read 2 more answers
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