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IrinaK [193]
3 years ago
6

Hei has $1,500 in a retirement account earning 4% interest compounded annually. Each year after the first, she makes additional

deposits of $1,500. After 5 years, what was her account balance if she did not make any withdrawals? Round each year's interest to the nearest cent if necessary.
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

$150,000

Step-by-step explanation:

so its about rounding to the nearest if you round $1,500 you will get $150,000

Hope this helps! (And I'm not wrong...)

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Help on #9 I don't get it!
Fantom [35]

Your trying to rearrange the equation with v were the r was.

2( v - h ) / k = r

2( v - h ) k / k = rk

2( v - h ) = rk

2( v - h ) / 2 = rk / 2

v - h = rk / 2

v - h + h = rk / 2 + h

v = rk / 2 + h

8 0
4 years ago
Help please for both questions !
uysha [10]

Answer:

bro hounestly I'm just doing this to setup my profile I don't know the awnser sorry

3 0
2 years ago
Ts Which conclusions can you draw from the data shown<br> in the table? Select all that apply.
timama [110]

Answer:

Twelve tickets cost $30.00

Step-by-step explanation:

7 0
3 years ago
Use the given information to find the exact value of the trigonometric function
eimsori [14]
\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

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7 0
1 year ago
Consider an experiment whose sample space consists of a countably infinite number of points. Show that not all points can be equ
ella [17]

Answer: if you have infinity points, which i will asume are the events, they cant have the same probability because then the probability will not be normalized, because in graph of prob vs variable, you will se infinite area under the curve if the probability is constant.

And yes, can all points have positive probability of occurring, but besides you medium value (the bell for example) you will see an asintotic decrease to the zero.

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