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andreyandreev [35.5K]

3 years ago

13

Marthy has exactly five blue pens,six black pens,and four red pens in his knapsack. if he pulls out one pen at random from his k

napsack.what is the probability that the pen is either red or black?

1 answer:

Arte-miy333 [17]3 years ago

5 0

(red pens + black pens) / total pens
(4 + 6) / 15 = 10/15 = 2/3

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Lena [83]

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2 years ago

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First person to give answer with work will get brainliest!!

Ket [755]

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4 0

2 years ago

Use Cramer's Rule to solve the following system: –2x – 6y = –26 5x + 2y = 13

andriy [413]

$\bf \begin{cases}
-2x-6y&=-26\\
\quad 5x+2y&=13
\end{cases}\stackrel{\textit{determinant of the coefficients}}{D=
\begin{bmatrix}
-2&-6\\5&2
\end{bmatrix}}\implies (-4)-(-30)
\\\\\\
D=-4+30\implies \boxed{D=26}\\\\
-------------------------------\\\\$

$\bf x=\cfrac{D_x}{D}\implies x=\cfrac{
\begin{bmatrix}
\boxed{-26}&-6\\\\ \boxed{13}&2
\end{bmatrix}}{D}\implies x=\cfrac{(-52)-(-78)}{26}
\\\\\\
x=\cfrac{-52+78}{26}\implies x=\cfrac{26}{26}\implies \boxed{x=1}\\\\
-------------------------------\\\\
y=\cfrac{D_y}{D}\implies y=\cfrac{
\begin{bmatrix}
-2&\boxed{-26}\\\\ 5&\boxed{13}
\end{bmatrix}}{D}\implies y=\cfrac{(-26)-(-130)}{26}
\\\\\\
y=\cfrac{-26+130}{26}\implies y=\cfrac{104}{26}\implies \boxed{y=4}$

4 0

3 years ago

A printer takes 5mins to print 90 copies of a repor. At this rate, how many copies of the same report can it print in 1/2 hours

just olya [345]

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3 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago