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3 years ago

Suppose you invest $1400 at interest rate of 4.3% compounded continuously. How much will you have in the account after 7 years?

Round to the nearest dollar.

Mathematics

1 answer:

STatiana [176]3 years ago

3 0

$\bf ~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$1400\\ r=rate\to 4.3\%\to \frac{4.3}{100}\dotfill &0.043\\ t=years\dotfill &7 \end{cases} \\\\\\ A=1400e^{0.043\cdot 7}\implies A=1400e^{0.301}\implies A\approx 1891.69\implies \stackrel{\textit{rounded up}}{A=1892}$

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PLEASE HELP, GOOD ANSWERS GET BRAINLIEST. +40 POINTS WRONG ANSWERS GET REPORTED

MA_775_DIABLO [31]

1. Ans:(A) 123

Given function: $f(x) = 8x^2 + 11x$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)$
=> $\frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11$
=> $\frac{d}{dx} f(x) = 16x + 11$

Now at x = 7:
$\frac{d}{dx} f(7) = 16(7) + 11$

=> $\frac{d}{dx} f(7) = 123$

2. Ans:(B) 3

Given function: $f(x) =3x + 8$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)$
=> $\frac{d}{dx} f(x) = 3*1 + 0$
=> $\frac{d}{dx} f(x) = 3$

Now at x = 4:
$\frac{d}{dx} f(4) = 3$ (as constant)

=>Ans: $\frac{d}{dx} f(4) =$ 3

3. Ans:(D) -5

Given function: $f(x) = \frac{5}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})$
=> $\frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = -5x^{-2}$

Now at x = -1:
$\frac{d}{dx} f(-1) = -5(-1)^{-2}$

=> $\frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}$
=> Ans: $\frac{d}{dx} f(-1) = -5$

4. Ans:(C) 7 divided by 9

Given function: $f(x) = \frac{-7}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})$
=> $\frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = 7x^{-2}$

Now at x = -3:
$\frac{d}{dx} f(-3) = 7(-3)^{-2}$

=> $\frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}$
=> Ans: $\frac{d}{dx} f(-3) = \frac{7}{9}$

5. Ans:(C) -8

Given function:
$f(x) = x^2 - 8$

Now if we apply limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)$

=> $\lim_{x \to 0} f(x) = (0)^2 - 8$
=> Ans: $\lim_{x \to 0} f(x) = - 8$

6. Ans:(C) 9

Given function:
$f(x) = x^2 + 3x - 1$

Now if we apply limit:
$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)$

=> $\lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1$
=> Ans: $\lim_{x \to 2} f(x) = 4 + 6 - 1 = 9$

7. Ans:(D) doesn't exist.

Given function: $f(x) = -6 + \frac{x}{x^4}$
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
$f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same$

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: $f(x) = 9 + \frac{x}{x^3}$
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: $\lim_{x \to 9} f(x)$ where,
$f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.$

If both sides are equal on applying limit then limit does exist.

Let check:
If x $\textless$ 9: answer would be 9+9 = 18
If x $\geq$ 9: answer would be 9-9 = 0

Since both are not equal, as $18 \neq 0$ , hence limit doesn't exist.

12. Ans:(B) Limit doesn't exist.

Find out: $\lim_{x \to 1} f(x)$ where,

$f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1$

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 $\neq$ 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x f(x)=1/(x-9)

----------------------------------------

8.9 -10

8.99 -100

8.999 -1000

8.9999 -10000

9.0 -∞

Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =9 (correct)

14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = $\frac{ds(t)}{dt}$

Therefore,

$\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6$

At t=2,

Inst. velocity = -6

15. Ans: +∞, x =7

f(x) = 1/(x-7)^2.

Table:

x f(x)= 1/(x-7)^2

--------------------------

6.9 +100

6.99 +10000

6.999 +1000000

6.9999 +100000000

7.0 +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞, x =7 (correct)

-i

7 0

3 years ago

Read 2 more answers

Use an addition or subtraction formula to find the exact value of the expression, as demonstrated in example 1. tan(105°)

strojnjashka [21]

Use an addition or subtraction formula to find the exact value of the expression, as demonstrated in example 1. tan(105°)

7 0

3 years ago

What is the solution to –2(8x – 4) < 2x + 5?

zhuklara [117]

the answer is x > 1/6

7 0

3 years ago

A sample survey interviews SRSs of 500 female college students and 550 male college students. Researchers want to determine whet

stira [4]

Answer:

Option (d) and (e)

Step-by-step explanation:

We are given that a sample survey interviews SRS of 500 female college students and 550 male college students. And Researchers want to determine whether there is a difference in the proportion of male and female college students who worked for pay last summer.

In all, 410 of the females and 484 of the males say they worked for pay last summer.

From this, Null Hypothesis, $H_0$ : $p_1 = p_2$ {means proportion of male and female college students who worked for pay last summer are same}

Alternate Hypothesis, $H_1$ : $p_1\neq p_2$ {means there is a difference in the proportion of male and female college students who worked for pay last summer}

Now, since we know that results were statistically significant at the 1% level.

So, if p-value is less than the significance level ⇒ we will reject $H_0$

if p-value is more than the significance level ⇒ we will not reject $H_0$

From the options given it is sure that P-value is less than 1%,i.e.;

<em>P-value is less than the significance level, so we will reject null hypothesis and conclude that there is convincing evidence that the proportion of all male college students in the study who worked for pay last summer is different from the proportion of all female college students in the study who worked for pay last summer. </em>

Also, option (d) and (e) are same in my opinion.

4 0

3 years ago

Any help please? For extra credit

Law Incorporation [45]

Answer:

see explanation

Step-by-step explanation:

(a)

A recursive formula allows any term in the sequence to be found by adding the common difference d to the previous term.

Here d = - 4 , then recursive formula is

$a_{n}$ = $a_{n-1}$ - 4 with a₁ = 2

(b)

The explicit formula for an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 2 and d = - 4, thus

$a_{n}$ = 2 - 4(n - 1) = 2 - 4n + 4 = 6 - 4n ← explicit formula

(c)

Using the recursive formula

a₁ = 2

a₂ = 2 - 4 = - 2

a₃ = - 2 - 4 = - 6

Using the explicit formula

a₅ = 6 - 4(5) = 6 - 20 = - 14

a₁₀ = 6 - 4(10) = 6 - 40 = - 34

a₁₀₀ = 6 - 4(100) = 6 - 400 = - 394

4 0

3 years ago