20 Points Can anyone help me with these Geometry questions? Will give Brainliest.

Mathematics

1 answer:

Tom [10]3 years ago

4 0

Answer:

give brain list first

Step-by-step explanation:

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objective: central limit theorem assumptions. the factor(s) to be considered when assessing if the central theorem holds is/are

Eddi Din [679]

Answer:

Sample size

Step-by-step explanation:

Central Limit Theorem states that population with mean and standard deviation and if the sample size is large then the distribution of sample mean will be will be normally distributed. The central limit theorem holds assumptions that the factors to be considered when assessing central limit theorem is sample size.

8 0

3 years ago

Does anybody know how to do time with exponential decay

My name is Ann [436]

<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

$b_n=0.12\times b_{n-1}$

Why does this work? Initially, you start with 500 mL of air that you breathe in, so $b_1=500\text{ mL}$ . After the second breath, you have 12% of the original air left in your lungs, or $b_2=0.12\timesb_1=0.12\times500=60\text{ mL}$ . After the third breath, you have $b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}$ , and so on.

You can find the amount of original air left in your lungs after $n$ breaths by solving for $b_n$ explicitly. This isn't too hard:

$b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots$

and so on. The pattern is such that you arrive at

$b_n=0.12^{n-1}b_1$

and so the amount of air remaining after $50$ breaths is

$b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}$

which is a very small number close to zero.</span>

5 0

3 years ago

How much heat is released when 12. 0 grams of helium gas condense at 2. 17 K? The latent heat of vaporization of helium is 21 J/

aleksandr82 [10.1K]

The amount of heat released when 12.0g of helium gas condense at 2.17 K is; -250 J

The latent heat of vapourization of a substance is the amount of heat required to effect a change of state of the substance from liquid to gaseous state.

However, since we are required to determine heat released when the helium gas condenses.

The heat of condensation per gram is; -21 J/g.

Therefore, for 12grams, the heat of condensation released is; 12 × -21 = -252 J.

Approximately, -250J.