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aleksandrvk [35]
2 years ago
11

If you put $1000 in a savings account that yields an 7% annual rate of interest that is compounded weekly, how much would that b

e worth in 25 weeks
Mathematics
1 answer:
Tanzania [10]2 years ago
7 0

Answer:

A = $1034

Step-by-step explanation:

The relevant Amount function is A = P(1 + r/n)^(n*t), where r is the annual interest rate as a decimal fraction, n is the number of compounding periods per year, and t is the number of years.

There are 52 weeks in a year.  Hence, n = 52, and 25 weeks = 25/52 year, or 0.481 year.  

The accumulation (value of the savings) would be

A = ($1000)(1 + 0.07/52)^(52*[25/52]), or approximately

A = ($1000)(1.0013)^25, or approximately A = $1034

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What is the value of X?<br> Help please!
anastassius [24]

Answer:

(11√3)/3

Step-by-step explanation:

In order to solve for the variable, you will need to use a trig function. In this case, you will need to use the trig function tangent.

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⇒ Tan 30° = x/11

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4 0
3 years ago
Solve for p. <br><br> i = prt<br><br> p = i-rt<br> p = i/(rt)<br> p = (rt)/i
Debora [2.8K]

Answer:

<h2>p = (rt)/i</h2>

Step-by-step explanation:

i=prt\to prt=i\qquad\text{divide both sides by}\ rt\neq0\\\\\dfrac{prt}{rt}=\dfrac{i}{rt}\\\\p=\dfrac{i}{rt}

5 0
3 years ago
A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021
egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

  • \displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}

Therefore, for the last term we have;

  • \displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

  • A ≈ <u>2020.022</u>

Learn more about the sum of a series here:

brainly.com/question/190295

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2 years ago
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Y = -1/3x + 4 . 4 is the intercept -1/3x is the slope
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