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3 years ago

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Reptile [31]3 years ago

7 0

Step-by-step explanation:

18% compounded monthly

charges 18% compounded semi-annually.

#1 is worse than #2 because...

Since the nominal rates are the same, the card with the more frequent compounding so it would have the higher effective rat.

Let's compare them!:

(1 + .18/12)^12 = 1.19562 ---> effective rate is 19.562 per annum

(1+.18/2)² = 1.1881

1.1881 has the effective rate of 18.88% per annum

(1 + /17/4)^4 = 1.18114.. --> effective rate is 18.114% per annum

18% monthly is .18 over 12 each month = 0.015/month

so every month multiply by 1.015

total cost = $607.99 * 1.01^number of months

18% semi annual = .18/2 every six months = .09 /half year

so every half year (6 months) multiply by 1.09

total cost = $607.99 * 1.09^ (number of months/6)

17% quarterly = .17/4 every 3 months = .0425 /quarter year

so every quarter year (3 months) multiply by 1.0425

total cost = $607.99 * 1.025^(number of months/3)

HOPE THIS HELPS YOU HAVE A GREAT GREAT GREAT FANTASTIC FRIDAY!

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Jobisdone [24]

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4 0

3 years ago

A car travels 2 1/3 miles in 3 1/2 minutes at a constant speed. Write an equation to represent the car travels in miles and minu

Korvikt [17]

Answer:

d = 0.666t , where d is in miles and t is in minutes.

d = 39.94h, where d is in miles and h is in hours.

Step-by-step explanation:

A car travels $2\frac{1}{3} = 2.33$ miles in $3\frac{1}{2} = 3.5$ minutes at a constant speed.

Let the relation between the distance (d) traveled by car in miles after traveling t minutes is d = kt ......... (1)

Now, putting d = 2.33 miles and t = 3.5 minutes in the above equation we get,

2.33 = 3.5k

⇒ k = 0.666 (Approx.)

So, the equation (1) becomes d = 0.666t. (Answer)

Let us assume that the relation between the distance (D) traveled by car in miles after traveling h hours is d = Kh ............ (2)

Now, putting D = 2.33 miles and T = 3.5/60 = 0.058 hours in the above equation, we get

2.33 = 0.058K

⇒ K = 39.94

So, the equation (2) we get, d = 39.94h (Answer)

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3 years ago

You earn a 90% on his science test you answer 18 questions correctly how many questions were on the test

Bogdan [553]

Hello there!

The correct answer is 20.

Hope This Helps You!
Good Luck :)

8 0

3 years ago

Read 2 more answers

Find the area of a regular hexagon inscribed in a circle with a radius of 6 feet. Round to the nearest tenth.

raketka [301]

Answer:

93.5ft

Step-by-step explanation:

7 0

3 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago