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3. You want to have $4000 in your savings account after 2 years. Find the amount you should deposit for each of the situations d

pychu [463]

Answer:

Part A)

About $3767.34.

Part B)

About $3692.47.

Step-by-step explanation:

Part A)

Recall that compound interest is given by the formula:
$\displaystyle A = P\left(1+\frac{r}{n}\right)^{nt}$

Where A is the final amount, P is the initial amount, r is the interest rate, n is the number of times compounded per year, and t is the number of years.

To obtain $4000 after two years, let A = 4000 and t = 2.

Because the account pays 3% interest compounded monthly, r = 0.03 and n = 12.

Substitute and solve for P:

$\displaystyle \begin{aligned} (4000) & = P\left(1+\frac{(0.03)}{(12)}\right)^{(12)(2)} \\ \\ P & = \frac{4000}{\left(1+\dfrac{(0.03)}{(12)}\right)^{(12)(2)}} \\ \\ & \approx \$3767.34\end{aligned}$

In concluion, about $3767.34 should be deposited.

Part B)

Recall the formula for continuous compound:

$\displaystyle A = Pe^{rt}$

Where e is Euler's number.

Hence, let A = 4000, r = 0.04 and t = 2. Substitute and solve for P:

$\displaystyle \begin{aligned}(4000) & = Pe^{(0.04)(2)} \\ \\ P & = \frac{4000}{e^{(0.02)(4)}} \\ \\ & \approx \$3692.47 \end{aligned}$

In conclusion, about $3692.47 should be deposited.

8 0

2 years ago

You can't compare apples and orange

pychu [463]

This is a positive statement

3 0

3 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

3 years ago

a. 0.98 – 0.053 b. 0.67 – 0.4 c. 0.3 – 0.002 d. 3.2 – .789 e. 6.53 – 4.298 f. 6 – 4.32 g. 7 – 3.574 h. 4.83 – 1.8 i. 3.7 – 1.8 j

Pavlova-9 [17]

a. 0.927

We have:

0.98 – 0.053

We can re-write it as:

0. 9 8 0 -

0. 0 5 3

Moving digits to the right:

0. 9 7 10 -

0. 0 5 3

Digit-per-digit subtraction:

0. 9 2 7

b. 0.27

We have:

0.67 – 0.4

We can re-write it as:

0. 6 7 -

0. 4 0

Digit-per-digit subtraction:

0. 2 7

c. 0.298

We have:

0.3 – 0.002

We can re-write it as:

0. 3 0 0 -

0. 0 0 2

Moving digits to the right:

0. 2 10 0 -

0. 0 0 2

Again:

0. 2 9 10 -

0. 0 0 2

Digit-per-digit subtraction:

0. 2 9 8

d. 2.411

We have:

3.2 – .789

We can re-write it as:

3. 2 0 0 -

0. 7 8 9

We need to rewrite the first term by moving digits to the right several times:

3. 2 0 0 = 2. 12 0 0 = 2. 11 10 0 = 2. 11 9 10

So now we have:

2. 11 9 10 -

0. 7 8 9

Digit-per-digit subtraction:

2. 4 1 1

e. 2.232

We have:

6.53 – 4.298

We can re-write it as:

6. 5 3 0 -

4. 2 9 8

We need to rewrite the first term by moving digits to the right several times:

6. 5 3 0 = 6. 5 2 10 = 6. 4 12 10

So now we have:

6. 4 12 10 -

4. 2 9 8

Digit-per-digit subtraction:

2. 2 3 2

f. 1.68

We have:

6 – 4.32

We can re-write it as:

6. 0 0 -

4. 3 2

We need to rewrite the first term by moving digits to the right several times:

6. 0 0 = 5. 10 0 = 5. 9 10

So now we have:

5. 9 10 -

4. 3 2

Digit-per-digit subtraction:

1. 6 8

g. 4.426

We have:

7 – 3.574

We can re-write it as:

7. 0 0 0 -

3. 5 7 4

We need to rewrite the first term by moving digits to the right several times:

7. 0 0 0 = 6. 10 0 0 = 6. 9 10 0 = 6. 9 9 10

So now we have:

6. 9 9 10 -

3. 5 7 4 =

Digit-per-digit subtraction:

3. 4 2 6

h. 3.03

We have:

4.83 – 1.8

We can re-write it as:

4. 8 3 -

1. 8 0

We can immediately do the digit-per-digit subtraction:

3. 0 3

i. 2.9

We have:

3.7 – 1.8

We can re-write it as:

3. 7 -

1. 8

We need to rewrite the first term by moving digits to the right:

3. 7 = 2. 17

So now we have:

2. 17 -

1. 8 =

Digit-per-digit subtraction:

2. 9

j. 4.538

We have:

16.17 – 11.632

We can re-write it as:

1 6 . 1 7 0 -

1 1 . 6 3 2

We need to rewrite the first term by moving digits to the right:

1 6. 1 7 0 = 1 6. 1 6 10 = 1 5. 11 6 10

So now we have:

1 5. 11 6 10 -

1 1. 6 3 2 =

Digit-per-digit subtraction:

0 4. 5 3 8

3 0

3 years ago

A pie tin has an area of 81 pi square inches. Which choice below is closest to its

Brrunno [24]

Answer:

can i have the choices

Step-by-step explanation:

7 0

2 years ago