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Gnoma [55]
3 years ago
9

A new bank customer with $3,000 wants to open a money market account. The bank is offering a simple interest rate of 1.1%. How m

uch interest will the customer earn in 20 years? What will be the account balance after 20 years?
Mathematics
1 answer:
pickupchik [31]3 years ago
3 0
$3,660

The formula is P*R*T and that stands for principal * rate * time (in years). When you use the formula and plug in all of the numbers, you get the answer $3,660.

Hopefully this helps!
-Coconut;)
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How do I solve rate of change problems? (With picture) thanks!
denis-greek [22]

5)

\bf \begin{array}{|cc|cccc|ll} \cline{1-6} sodas&x&\underline{24}&28&\underline{32}&36\\ \cline{1-6} cost&y&\underline{18}&21&\underline{24}&27\\ \cline{1-6} \end{array}~\hspace{9em} (\stackrel{x_1}{24}~,~\stackrel{y_1}{18})\qquad (\stackrel{x_2}{32}~,~\stackrel{y_2}{24}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{24-18}{32-24}\implies \cfrac{6}{8}\implies \cfrac{3}{4}

6)

\bf \begin{array}{|cc|cc|ll} \cline{1-4} year&x&0&12\\ \cline{1-4} \$&y&720&1080\\ \cline{1-4} \end{array}~\hspace{10em} (\stackrel{x_1}{0}~,~\stackrel{y_1}{720})\qquad (\stackrel{x_2}{12}~,~\stackrel{y_2}{1080}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1080-720}{12-0}\implies \cfrac{360}{12}\implies \cfrac{30}{1}\implies 30

7)

slope as you should already know is rise/run, or how much something moves in relation something else, namely how much the y-axis go up as the x-axis moves sideways, one moves, the other follows, but the increments will be different, sometimes the same, but usually different.

the y-intercept means, when the graph of the equation touches or intercepts the y-axis, and when that happens x = 0, or the horizontal distance is at bay.

for the slope on  6), 30 or 30/1 means, for every 1 year(x) passed, the worth(y) increased by 30, or jumped by 30 units, so as the x-axis moved 1, the y-axis moved 30.  After 12 years 30 * 12 = 360, and we add the initial 720 and we end up with 1080.

the y-intercept, well, as aforementioned is when x = 0, is year 0.

6 0
3 years ago
Read 2 more answers
A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hit the ground. a.Find
Lyrx [107]
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2 
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
4 0
4 years ago
Use the present value formula to determine the amount to be invested now, or the present value needed.
Vladimir79 [104]

Answer:

The value needed now is $102,296.60

Step-by-step explanation:

Here, we want to calculate the present value needed.

We shall be using the formula for compound interest here

Mathematically;

A = I(1 + r/n)^nt

According to the question;

I is the initial amount which is what we want to calculate

A is the accumulated amount with interest = 140,000

r is the interest rate = 8% = 8/100 = 0.08

n is the number of times interest is compounded which is 4, since it is quarterly

t is the number of years = 2

Substituting these values, we have ;

140,000 = I(1 + 0.08/2)^(4 * 2)

140,000 = I(1.04)^8

I = 140,000/(1.04)^8

I = 102,296.62870027774

Which is I = $102,296.60 to the nearest cent

8 0
3 years ago
2. What is the equation of the line that passes through points (-8,3/2) and (-12, 5/2)
MissTica

Answer:

y=2x+17.5

Step-by-step explanation:

17.5=35/2

7 0
1 year ago
Which of the following names a line segment
Eva8 [605]

Answer:

The answer is C.

Step-by-step explanation:

A. name of ray

B. name of line

C. name of line segment

D. length of segment

3 0
3 years ago
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