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3 years ago

If the APR of a savings account is 3.6% and interest is compounded monthly, what is the approximate APY of the account?

Mathematics

2 answers:

Mila [183]3 years ago

8 0

The Answer is ''3.66%''

Nadusha1986 [10]3 years ago

4 0

Answer:

3.66% ( approx )

Step-by-step explanation:

Since, the formula of annual percentage yield is,

$APY = (1+\frac{r}{n})^n-1$

Where,

r = stated annual interest rate,

n = number of compounding periods,

Here, r = 3.6% = 0.036,

n = 12 ( ∵ 1 year = 12 months )

Hence, the annual percentage yield is,

$APY=(1+\frac{0.036}{12})^{12}-1=1.03659 - 1 = 0.036599\approx 0.0366 = 3.66\%$

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Are cells living? or not? 3 times 3 minus 6 (3x3-6) = ?

Arlecino [84]

Answer:

Yes, cells are living, because they make up living things, and can do what most living things are able to, like reproduce.

(3 x 3 - 6) = ?

First, multiply 3 x 3, which = 9

(9 - 6)

Then subtract 9 by 6, giving you 3.

So, :

(3 x 3 - 6) = 3

3 0

4 years ago

Two lamps marked 100 W - 110 V and 100 W - 220 V are connected i

ELEN [110]

<h2>Answer:</h2>

The power consumed in the lamp marked 100W - 110V is 15.68W

The power consumed in the lamp marked 100W - 220V is 62.73W

<h2>Step-by-step explanation:</h2>

Given:

First lamp rating

Power (P) = 100W

Voltage (V) = 110V

Second lamp rating

Power (P) = 100W

Voltage (V) = 220V

Source

Voltage = 220V

i. Get the resistance of each lamp.

Remember that power (P) of each of the lamps is given by the quotient of the square of their voltage ratings (V) and their resistances (R). i.e

P = $\frac{V^2}{R}$

Make R subject of the formula

⇒ R = $\frac{V^2}{P}$ ------------------(i)

For first lamp, let the resistance be R₁. Now substitute R = R₁, V = 110V and P = 100W into equation (i)

R₁ = $\frac{110^2}{100}$

R₁ = 121Ω

For second lamp, let the resistance be R₂. Now substitute R = R₂, V = 220V and P = 100W into equation (i)

R₂ = $\frac{220^2}{100}$

R₂ = 484Ω

ii. Get the equivalent resistance of the resistances of the lamps.

Since the lamps are connected in series, their equivalent resistance (R) is the sum of their individual resistances. i.e

R = R₁ + R₂

R = 121 + 484

R = 605Ω

iii. Get the current flowing through each of the lamps.

Since the lamps are connected in series, then the same current flows through them. This current (I) is produced by the source voltage (V = 220V) of the line and their equivalent resistance (R = 605Ω). i.e

V = IR [From Ohm's law]

I = $\frac{V}{R}$

I = $\frac{220}{605}$

I = 0.36A

iv. Get the power consumed by each lamp.

From Ohm's law, the power consumed is given by;

P = I²R

Where;

I = current flowing through the lamp

R = resistance of the lamp.

For the first lamp, power consumed is given by;

P = I²R [Where I = 0.36 and R = 121Ω]

P = (0.36)² x 121

P = 15.68W

For the second lamp, power consumed is given by;

P = I²R [Where I = 0.36 and R = 484Ω]

P = (0.36)² x 484

P = 62.73W

Therefore;

The power consumed in the lamp marked 100W - 110V is 15.68W

The power consumed in the lamp marked 100W - 220V is 62.73W

8 0

3 years ago

Luke threw balls into these buckets at the carnival. The number on the bucket gives the number of points for each throw . what i

Arada [10]

I am not sure that there is enough information to answer this question.

3 0

4 years ago

Read 2 more answers

Join our among us Code: KFTJFF

Pavlova-9 [17]

Ok bet I tryyyyy!!!!!

8 0

3 years ago

Read 2 more answers

Find the area of the surface generated by revolving about the x-axis the portion of the astroid x Superscript two thirds Baseli

allochka39001 [22]

The correct question is:

Find the area of the surface generated by revolving about the x-axis, the portion of the astroid

$x^\frac{3}{2} + y^\frac{2}{3} = 1$

Answer: The Surface Area is

$\frac{6}{5}\pi$

Step-by-step explanation:

First, we rewrite the expression in terms of x, because we are revolving about the x-axis, we want to integrate in terms of x. Doing that, we have

$y = \left(1 - x^\frac{2}{3}\right)^\frac{3}{2}$

Next, we differentiate y with respect to x

$\frac{dy}{dx} = \frac{3}{2}\left(1 - x^\frac{2}{3}\right)^\frac{1}{2}\left(-\frac{2}{3}\right)x^\frac{-1}{3}}\\ \\= -\frac{\sqrt{\left(1 - x\right)^\frac{2}{3}}}{x^\frac{1}{3}}$

Thus,

$\left(\frac{dy}{dx}\right)^2 = \frac{\left(1 - x\right)^\frac{2}{3}}{x^\frac{2}{3}}}$

and so

$1 + \left(\frac{dy}{dx}\right) ^2 \\ \\ = 1 +\frac{(1 - x)^\frac{2}{3}}{x^ \frac{2}{3}} \\ \\=\frac{1}{x^\frac{2}{3}}$

Therefore, the Surface Area is given as:

$\int_{0}^{1}2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx \\ \\= \int_{0}^{1}2\pi\left(1 - x^\frac{2}{3}\right)^\frac{3}{2}\sqrt{\frac{1}{x^\frac{2}{3}}}dx \\ \\=\int_{0}^{1}2\pi\left(1-x^\frac{2}{3}\right)^\frac{3}{2}x^\frac{-3}{2}dx. \\ \\$

If we let

$u = 1-x^\frac{2}{3}$

then

$du = -\frac{2}{3}x^{-\frac{1}{3}}dx,$

so we see that

$= -\frac{3}{2}\int_{0}^{1}2\pi\left(1-x^\frac{2}{3}\right)^\frac{3}{2} - \frac{2}{3}x^{-\frac{1}{3}} dx \\ \\= -3\pi \int_{0}^{1}u^\frac{3}{2}du \\ \\= 3\pi\frac{2}{5}u^\frac{5}{2}\left \{ {{u=1} \atop {u=0}} \right. \\ \\= \frac{6}{5}\pi$

3 0

3 years ago