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2 years ago

3. What is the measure of angle HEK in this figure?95 55 85 150

Mathematics

1 answer:

DIA [1.3K]2 years ago

7 0

Answer:

95 is the correct answer

Step-by-step explanation:

The initial or beginning side is H. Look on the protractor and you will see the top number is 55 degrees. That is where we start. Now, go to line E, that is where we end. The top number is 150. To find <HEK, we need to subtract 55 from 150 to get 95.

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What is 4,7,12,19,28 in quadratic form

Naddik [55]

Asked and answered elsewhere.
brainly.com/question/9247314

You obviously don't mind using "technology" (Brainly) to answer these questions. A graphing calculator can do quadratic regression on the sequence and tell you its formula.

If you want to do it by hand, you can write the equation
.. y = ax^2 +bx +c
and substitute three of the given points. Then solve the resulting three linear equations for a, b, and c.
.. 4 = a +b +c
.. 7 = 4a +2b +c
.. 12 = 9a +3b +c

Subtracting the first equation from the other two reduces this to
.. 3 = 3a +b
.. 8 = 8a +2b
The latter can be divided by 2, so reduces to
.. 4 = 4a +b
Subtracting the first of the reduced equations from this, you have
.. 1 = a
so
.. 3 = 3*1 +b
.. 0 = b
and
.. 4 = a + b + c = 1 + 0 + c
.. 3 = c

And your equation is
.. y = x^2 +3 . . . . . . as shown previously

7 0

3 years ago

Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :

$E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1$

so we end up with

$E[Y^n]=e^{\frac12n^2}$

3 0

3 years ago

Misha wrote the quadratic equation 0=-x2+4x-7 in standard form. what is the value of c in her equation

Masteriza [31]

Answer:

Final answer is c=-7.

Step-by-step explanation:

Given equation is $0=-x^2+4x-7$ .

Now question says that Misha wrote the quadratic equation 0=-x2+4x-7 in standard form. Now we need to find about what is the value of c in her equation.

We know that standard form of quadratic equation is given by $ax^2+bx+c=0$ .

compare given equation with the standard form, we find that -7 is written in place of +c

so that means +c=-7

or c=-7

Hence final answer is c=-7.

8 0

3 years ago

Solve 5w^2+25=115, Where w is a real number.

MArishka [77]

Answer:

w=+/-4.24 or whatever could be either since its squared

Step-by-step explanation:

subtract 25 so 5x^2=90. divide by 5 so w^2=18 then square root to get rid of the ^2 so w=+/-4.24

4 0

2 years ago

The rate at which a plant grows is constant. At the 3rd week it is 12 inches tall and at the 5th week it is 20 inches tall. If t

Likurg_2 [28]

It would be (12,20) or (20,12) I don't know which one

5 0

3 years ago

3. What is the measure of angle HEK in this figure?95 55 85 150​

3. What is the measure of angle HEK in this figure?95 55 85 150