If the sixth term of a sequence is 128 and the common ratio is 2, then what is the first term? 1, 2 or ,4

GREYUIT [131]

Pay attention here because I'm adding an extra letter to our circle to help keep track of the values in our formula. OUTSIDE of the intercepted arc I'm adding the point E. So the major arc is arc BEG and the minor arc is arc BG. The formula then for us is ∠ $BFG= \frac{1}{2}(arcBEG-arcBG)$ . We just don't have values for the arcs yet. If the measure of the central angle is 4x+238, then the measure of arcBG is also 4x+238. Around the outside of the circle is 360°. So we will use it in an expression. ArcBEG=360-(4x+238). Fitting that into our formula we have $2x+146= \frac{1}{2}[(360-4x-238)-(4x+238)]$ . Doing all the simplifying inside there we have $2x+146= \frac{1}{2}(122-4x-4x-238)$ and $2x+146= \frac{1}{2}(-8x-116)$ . Multiply both sides by 2 to get rid of the fraction: 4x+292=-8x-116. Combine like terms to get 12x = -408 and divide to solve for x. x = -34. Fourth choice down from the top.

3 0

3 years ago

Which equation is y= 2x ^2 - 8x +9 rewritten in vertex form.

tankabanditka [31]

y= 2x ^2 - 8x +9

y = a(x - h)2 + k, where (h, k) is the vertex of the parabola
so
y= 2x ^2 - 8x + 9
y= 2x ^2 - 8x + 8 + 1
y = 2(x^2 - 4x - 4) + 1
y = 2(x - 2)^2 + 1 ....<---------vertex form

7 0

3 years ago

30 divided by (21-6)•4-27+82

damaskus [11]

Rounded 62 .1
But it goes on for a while but its 62.58...

7 0

4 years ago

Which of these changes in today's world is a product of advances in biology?

zavuch27 [327]

"Increased crop yields" is the one change in today's world among the choices given in the question that is a product of advances in biology. The correct option among all the options that are given in the question is the first option. I hope that this is the answer that has come to your desired help.

5 0

3 years ago

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

a.

$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require

$\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}$

b. Find the marginal joint density of $X$ and $Y$ by integrating the joint density with respect to $z$ :

$f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz$

$\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}$

Then

$\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

$\approx\boxed{0.12886}$

c. This probability can be found by simply integrating the joint density:

$\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

$\approx\boxed{0.012262}$

7 0

3 years ago