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lisov135 [29]
3 years ago
13

Simplify (-8-1/2X + 2-1/2) + (-2-2/5x-5-2/3)

Mathematics
1 answer:
Citrus2011 [14]3 years ago
8 0
The answer to this problem is =<span><span><span><span>−9/</span>10</span>x</span>+<span><span>−85/</span><span>6</span></span></span>
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Please help!!! This is WAY overdue and I need to submit it by tonight or it won't be accepted!!!!
Masteriza [31]

Answer:

CCCCCCCCC

Step-by-step explanation:

7 0
2 years ago
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if the vertex of a parabola is (-4,6) and another point on the curve is (-3,14), what is the coefficient of the squared expressi
Softa [21]

Answer:

8

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (- 4, 6 ), thus

y = a(x + 4)² + 6

To find a substitute (- 3, 14) into the equation

14 = a(- 3 + 4)² + 6 ( subtract 6 from both sides )

8 = a

Thus the coefficient of the x² term is a = 8

3 0
2 years ago
Which of the following represents the solution to 3x^2+35=7x^2-4x<br> need help ASAP
mina [271]

Answer:

F

Step-by-step explanation:

4x^2 - 4x - 35 = 0

factor

(2x + 5)(2x - 7) = 0

x = -5/2, x = 7/2

that's F

4 0
3 years ago
Determine the measure of angle A, to the nearest degree.
Effectus [21]

Answer:

(a) 17°

(b) 78°

Step-by-step explanation:

(a) sin A= 0.2896

to calculate for "A" check for sin inverse of 0.2896

A= sin^-1(0.2896) = 16.83

to the nearest degree A= 17°

(b) tan A = 4.7046

A= tan^-1(4.7046)

A = 77.99

to the nearest degree A = 78°

4 0
1 year 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
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