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Anna007 [38]
2 years ago
11

There are 12

Mathematics
1 answer:
notka56 [123]2 years ago
6 0

Answer:

just split them into six groups and two students on each groups

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MARK BRAINLIST !!!<br><br> What is the value of the expression below when z = 10 and w = 7
zepelin [54]

Answer:

163

Step-by-step explanation:

10(10)+9(7) <em>Stubstitute the variables for the values given</em>

100+63 <em>Multiply </em>

163 <em>Add</em>

5 0
3 years ago
April wants to paint three walls of her bedroom. The walls are each 90 square feet. Paint is only sold in whole quarts, and each
Ganezh [65]

Answer:

she should buy 3 quarts and keep the remainder which can possibly cover 30 square feet.

3 0
2 years ago
Pls send the answer ​
luda_lava [24]

Answer:

a = \dfrac{46}{17}

b = 0

Step-by-step explanation:

\dfrac{2\sqrt{5} + \sqrt{3}}{2\sqrt{5} - \sqrt{3}} + \dfrac{2\sqrt{5} - \sqrt{3}}{2\sqrt{5} + \sqrt{3}} =

= \dfrac{2\sqrt{5} + \sqrt{3}}{2\sqrt{5} - \sqrt{3}} \times \dfrac{2\sqrt{5} + \sqrt{3}}{2\sqrt{5} + \sqrt{3}} + \dfrac{2\sqrt{5} - \sqrt{3}}{2\sqrt{5} + \sqrt{3}} \times \dfrac{2\sqrt{5} - \sqrt{3}}{2\sqrt{5} - \sqrt{3}}

= \dfrac{23 + 4\sqrt{15}}{17} + \dfrac{23 - 4\sqrt{15}}{17}

= \dfrac{46}{17}

a = \dfrac{46}{17}

b = 0

4 0
2 years ago
. 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
What would the domain and range of this graph be? In interval form, if it can be.
skad [1K]

Answer:

Step-by-step explanation:

Domain is the x values.  The interval of x-values that the function encompasses are from 0 inclusive to 7 exclusive.  In interval notation that is [0, 7).  The range is the y values.  The interval of y-values that the function encompasses are from what looks like -2 to 4.  In interval notation that is [-2, 4].  The domain goes from the lowest x-value to the highest; the range goes from the lowest y-value to the highest.

6 0
3 years ago
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