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RUDIKE [14]
2 years ago
15

X^4 -(3m+2)x^2 +3m+1 +0 tìm m để phương trình có 4 nghiệm phân biệt nhỏ hơn 2

Mathematics
1 answer:
Kobotan [32]2 years ago
8 0

Answer:

3

4

4

3

4

4

3

5

5

3

4

5

5

4848

Step-by-step explanation:

hope I helped

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Point K is on line segment JL. Given JK = 2x – 2, KL
Scorpion4ik [409]

Answer:

KL = 10

Step-by-step explanation:

JK + KL = JL

2x – 2 + x – 9 =  2x + 8

Combine like terms

3x -11 = 2x+8

Subtract 2x from each side

3x-2x -11 = 2x+8-2x

x-11 = 8

Add 11 to each side

x-11+11 = 8+11

x = 19

KL = x – 9

     = 19-9

     = 10

8 0
3 years ago
−6.2<−4.5, yet |−6.2|>|−4.5|. Use a number line to show why.
romanna [79]

Answer:

-27.9

Step-by-step explanation:

3 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
Fiona, Peter and Angad share some sweets in the ratio 2:5:1. Fiona gets 26 sweets. How many did Angad get?
Ksju [112]

Answer:

13 sweets

Step-by-step explanation:

Fiona : Angad

2 : 1

26 : X

26/2 = X/1

X = 13

7 0
3 years ago
What is multiplication
Ne4ueva [31]

Answer:

Multiplication is one of the four elementary mathematical operations of arithmetic, with the others being addition, subtraction and division. Multiplication is marked as an "x." A mathematical operation performed on a pair of numbers in order to derive a third number called a product. For positive integers, multiplication consists of adding a number (the multiplicand) to itself a specified number of times. Thus multiplying 6 by 3 means adding 6 to itself three times.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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