What is the measure of y? 2 X у 4 9 y = [?] Give your answer in simplest form. Enter

Mrs. Tyner drove about 150 km east from Pearland, to Houston, Texas. She

frozen [14]

Answer:

d = 167.7 m

Step-by-step explanation:

We have,

Mrs. Tyner drove about 150 km east from Pearland, to Houston, Texas. She drove about another 75 km north to Dallas.

It is required to find the approximate air distance from Pearland to Dallas, Texas.

The resultant vector will give the approximate air distance. So,

$d=\sqrt{150^2+75^2} \\\\d=167.7\ m$

So, the approximate air distance from Pearland to Dallas, Texas is 167.7 m.

3 0

2 years ago

Let f(x) = cxe−x2 if x ≥ 0 and f(x) = 0 if x < 0.

sergij07 [2.7K]

(a) If f(x) is to be a proper density function, then its integral over the given support must evaulate to 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx = \int_0^\infty cxe^{-x^2}\,\mathrm dx=1$

For the integral, substitute u = x ² and du = 2x dx. Then as x → 0, u → 0; as x → ∞, u → ∞:

$\displaystyle\frac12\int_0^\infty ce^{-u}\,\mathrm du=\frac c2\left(\lim_{u\to\infty}(-e^{-u})-(-1)\right)=1$

which reduces to

c / 2 (0 + 1) = 1 → c = 2

(b) Find the probability P(1 < X < 3) by integrating the density function over [1, 3] (I'll omit the steps because it's the same process as in (a)):

$\displaystyle\int_1^3 2xe^{-x^2}\,\mathrm dx = \boxed{\frac{e^8-1}{e^9}} \approx 0.3678$

5 0

2 years ago

Find the center of the circle that can be circumscribed about EFG with E(2,2),F(2,-2), and G(6,-2)

RSB [31]

Hmmm I'd say the center of the circle will end up at the center of the triangle, namely the Centroid of the triangle circumscribed in the circle.

$\bf \textit{Centroid of a Triangle}
\begin{array}{llll}
\stackrel{E(2,2)~~F(2,-2)~~G(6,-2)}{\left(\cfrac{2+2+6}{3}\quad ,\cfrac{2-2-2}{3}\quad \right)}
\end{array}$

6 0

3 years ago

Brian invests £1800 into his bank account.

SVEN [57.7K]

Step-by-step explanation:

$\underline{ \underline{ \text{Given}}} :$