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Car A is traveling east at a steady speed of 40 miles per hour. After 2 hours, it is 95 miles east of Johnstown.

yarga [219]

Answer:

what graph?

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

4.

tresset_1 [31]

Answer:

The answer is C. 28.8 in

Step-by-step explanation:

Given a=16 and b=24,

c = 28.84441 = 8√13

∠α = 33.69° = 33°41'24" = 0.588 rad

∠β = 56.31° = 56°18'36" = 0.98279 rad

h = 13.3128

area = 192

perimeter = 68.84441

inradius = 5.57779

circumradius = 14.42221 = 4√13

5 0

3 years ago

Please help! (Picture Below!)

mixas84 [53]

Answer:

Step-by-step explanation:

Triangle 1,

Sum of the squares of the shorter lengths = 13² + 17²

= 169 + 289

= 458

Square of the longest length = 19² = 361

Since, sum of longest side is less than sum of squares of the smaller sides,

13²+ 17² > 19²

Therefore, it's acute triangle.

Triangle 2,

Sum of squares of the shorter lengths = 10² + 24²

= 576

Square of the largest length = 26² = 576

Since, sum of squares of the shorter lengths = square of the largest length

Therefore, it's a right triangle.

6 0

3 years ago

If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of

astraxan [27]

$M_X(t)=\mathbb E(e^{Xt})$
$M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)$
$M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots$
$M_X(t)=1+t+t^2+t^3+\cdots$
$M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}$

provided that $|t|$ .

Similarly,

$\varphi_X(t)=\mathbb E(e^{iXt})$
$\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots$
$\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}$

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt$

The integral can be rewritten as

$\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt$

so that

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt$

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to $x$ , which gives

$\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt$
$=\displaystyle\frac{\pi(1-s)}{2s(1+s)}$

and taking the inverse transform returns

$F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}$

which describes an exponential distribution with parameter $\lambda=1$ .

6 0

3 years ago

An airplane takes off from an airport. When the

Ivanshal [37]

The angle of elevation is = 70° and the distance the airplane travelled in the air is = 17,557ft

<h3>Calculation of the distance travelled</h3>

To calculate the angle of elevation of the airplane

tan x° = opposite/adjacent

where opposite = 16,500 feet

adjacent = 6,000 ft

tan x° = 16,500 / 6,000

tan x° = 2.75

X = arctan ( 2.75)

X = 70°

To calculate the distance the airplane travelled in the air Pythagorean Theorem is used.

C² = a² + b²

C² = 16,500² + 6,000²

C² = 272250000 + 36000000

C² = 308250000

C= √308250000

C= 17,557ft

Learn more about Pythagorean Theorem here:

brainly.com/question/343682

#SPJ1

3 0

3 years ago