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Feliz [49]
2 years ago
9

Susan purchased company ABC's stock and invested the balance in her money market account. ABC's stock yielded 13% last year and

her money market account yielded 5% last year. If Susan's initial investments amount to $15000, and the annual income is $1070, how much money is invested in stock and how much is invested in her money market account?
Mathematics
1 answer:
garri49 [273]2 years ago
7 0

$4,000 is invested in ABC's stock and $11,000 is invested in her market account

Step-by-step explanation:

Susan purchased company ABC's stock and invested the balance in her money market account

  • ABC's stock yielded 13% last year and her money market account yielded 5% last year
  • Susan's initial investments amount to $15000, and the annual income is $1070

We need to find how much money is invested in stock and how much is invested in her money market account

Assume that she invested $x in ABC's stock and $y in her money market account

∵ She invested $x in ABC's stock

∵ She invested $y in her market account

∵ Her initial investments amount is $15000

∴ x + y = 15000 ⇒ (1)

The interest formula is I = Prt, where P is the initial investment, r is the rate of interest in decimal and t is the time

In ABC's stock:

∵ She invested $x

∴ P = x

∵ The rate is 13%

∴ r = 13% = 13 ÷ 100 = 0.13

∵ t = 1

- Substitute all of these values in the formula of interest above

∴ I_{ABC} = x(0.13)(1)

∴ I_{ABC} = 0.13x

In her market account:

∵ She invested $y

∴ P = y

∵ The rate is 5%

∴ r = 5% = 5 ÷ 100 = 0.05

∵ t = 1

- Substitute all of these values in the formula of interest above

∴ I_{market} = y(0.05)(1)

∴ I_{market} = 0.05y

∵ Her annual income is $1070

- Add the interest of ABC's stock and the interest of her market

   account, equate the sum by 1070

∴ 0.13x + 0.05y = 1070 ⇒ (2)

Now we have a system of equations to solve it

Multiply equation (1) by -0.05 to eliminate y

∵ -0.05x - 0.05y = -750 ⇒ (3)

- Add equations (2) and (3)

∴ 0.08x = 320

- Divide both sides by 0.08

∴ x = 4000

- Substitute value of x in equation (1) to find y

∵ 4000 + y = 15000

- Subtract 4000 from both sides

∴ y = 11000

$4,000 is invested in ABC's stock and $11,000 is invested in her market account

Learn more:

You can learn more about the system of equations in brainly.com/question/2115716

#LearnwithBrainly

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(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)

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\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

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