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1 year ago

What’s the correct answer answer asap for brainlist

Mathematics

2 answers:

svetlana [45]1 year ago

6 0

Answer:

Iron

Step-by-step explanation:

Just learned about this

Musya8 [376]1 year ago

5 0

Answer:

C. Hydrogen

Step-by-step explanation:

Carbon, Hydrogen, Oxygen, and Nitrogen are most commonly found in lipids.

if this helps you, pls leave a heart. if it dosent, im so sorry. :(

You might be interested in

Solve 12.7 ≥ s − 5.3 Graph the solution.

Julli [10]

Hello!

I dont a graph to show you but I will solve the answer.

12.7 + 5.3 ≥ s

18 ≥ s

s ≤ 18
That is your answer.

Enjoy.
~Isabella

8 0

3 years ago

Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????

LenaWriter [7]

Answer:

$E[X^2]= \frac{2!}{2^1 1!}= 1$

$Var(X^2)= 3-(1)^2 =2$

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

$\phi(t) = E[e^{tX}]$

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

$\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx$

And we have that the moment generating function can be write like this:

$\phi(t) = e^{\frac{t^2}{2}$

And we can write this as an infinite series like this:

$\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...$

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

$E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]$

$E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...$

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

$\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}$

And then we have this:

$E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...$

And then we can find the $E[X^2]$

$E[X^2]= \frac{2!}{2^1 1!}= 1$

And we can find the variance like this :

$Var(X^2) = E[X^4]-[E(X^2)]^2$

And first we find:

$E[X^4]= \frac{4!}{2^2 2!}= 3$

And then the variance is given by:

$Var(X^2)= 3-(1)^2 =2$

7 0

3 years ago

What's the length of QR?

tigry1 [53]

Answer:

using the pithagerum therom

a² + b² = c²

this formula is to find the side of your triangle with 2 sides already known

the answer is A) 17.7

8 0

2 years ago

How can you write the statement "Obtuse angles have a greater measure than acute angles" so that it is a good definition?

Ilia_Sergeevich [38]

Answer:

Obtuse angles have a measure that is greater than 90 but less than 180 degrees, while acute angles have a measure that is less than 90 but greater than 0.

Step-by-step explanation:

Obtuse angles have a measure that is greater than 90 but less than 180 degrees because 180 degree angle is a straight angle. Any angle that is greater than 181 but less than 360 degrees is known as a reflex angle. Acute angles have a measure that is less than 90 but greater than 0 because anything with a higher of lower degree of measure would be consider a different angle.

6 0

3 years ago

Randy divides (2x4 – 3x3 – 3x2 + 7x – 3) by (x2 – 2x + 1) as shown below. What error does Randy make?

jarptica [38.1K]

error is in constant term of quotient

4 0

4 years ago

Read 2 more answers