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2 years ago

Use the following compound interest formula to complete the problem.

2 answers:

5 0

Card P's balance increased by $3.43 more than Card Q's balance. The accumulated total on Card P over the 4 years is $1080.70 and the accumulated total of Card Q is $1,206.28. Based on the principal outlay however, Card P would have netted a higher interest over Card Q when the principal is subtracted from the accumulated value. (For eg. Card P accumulated value $1080.70 less Principal $726.19 equals $354.51).The interests over the 4 years period would be $354.51 and $351.08 respectively, hence Card P having an increase in balance of $3.43 over Card Q.

scoundrel [369]2 years ago

3 0

Answer:

Your answer is C. i rather give the letter its easier.

Step-by-step explanation:

took test

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Capucine played a game where she earned 179 points from building 11 museums and 4 libraries. She earns 4 more points per museum

Korolek [52]

Capucine played a game where she earned 179 points from building 11 museums and 4 libraries.

Lets assume x points she earn per library

She earns 4 more points for museum than library.

4 more points per museum per x . so the points earned per museum is x+4

she earned 179 points from 11 museums and 4 libraries.

So the equation becomes

$11 (x+4) + 4x = 179$

$11x +44 + 4x = 179$

$15x + 44 = 179$ ( subtract 44 from both sides)

$15x= 135$ (divide by 15 from both sides)

x = 9

She earns 9 points per library

We know points per museum = x+4 = 9+4 = 13

She earns 13 points per museum .

5 0

2 years ago

Pamela has4/5 pound of sunflower seeds if she gives 2/3 pound of sunflower seeds to the squirrels in her backyard what fraction

Assoli18 [71]

4/5-2/3= 12/15-10/15 (common denominator is 15)=2/15 pounds of seeds left

8 0

2 years ago

Conservation of Species A certain species of turtle faces extinction because dealers collect truckloads of turtle eggs to be sol

JulijaS [17]

Answer:

The turtle population's rate of growth will be 32 turtles per year after 2 years and 248 per year after 6 years.

Ten years after the conservation measures are implemented the population will be 3260 turtles.

Step-by-step explanation:

To find the rate of growth of the turtle population at any time t you need to find $N'(t)$

$\frac{d}{dt}N(t)=\frac{d}{dt}(2t^3+3t^2-4t+1000)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dt}(2t^3)+\frac{d}{dt}(3t^2)-\frac{d}{dt}(4t)+\frac{d}{dt}(1000)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\N'(t)=6t^2+6t-4$

In particular, when t = 2 and t = 6, we have

$N'(2)=6(2)^2+6(2)-4=32\\\\N'(6)=6(6)^2+6(6)-4=248$

so the turtle population's rate of growth will be 32 turtles per year after 2 years and 248 per year after 6 years.

The turtle population at the end of the tenth year will be

$N(10)=2(10)^3+3(10)^2-4(10)+1000\\N(10)=3260 \:turtles$

3 0

2 years ago

1. Simplify: 2.6 x 8.4-5.4 A. 2.32 B. 7.8 C. 16.44 D. 21.3

zysi [14]

Answer:

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

An insurance company collected data from a class of high school sophomores on whether or not they have a cell phone and whether

NeX [460]

Answer:

C. Those who have a car tend to have a cell phone.

Step-by-step explanation:

The relative frequency for a data set n is calculated by dividing each frequency $x_i$ by n.

We have a table that relates the use of cars and cell phones.

First, we take from the table the population that has a car.

$n = 18$

Then, of the students who have a car, 12 have a cell phone and 6 do not have a cell phone.

Then we calculate the relative frequencies $f_{x_i}$ for those who have a cell phone and those who do not.

______________________________________________

Relative frequency Students with car n = 18.

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

Cell phone No cell phone Total n

12/18 6/18 18

_______________________________________________

Relative frequency Students with car n = 18.

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

Cell phone No cell phone Total n

0.666 0.333 18

________________________________________________

Most students who have a car also have a cell phone.

Now we calculate the relative frequency for students who do not have a car.

_______________________________________________

Relative frequency Students without a car n = 7

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

Cell phone No cell phone Total n

2/7 5/7 7

_______________________________________________

Relative frequency Students without a car n = 7

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

Cell phone No cell phone Total n

0.286 0.714 7

_______________________________________________

Most students who do not have a car do not have a cell phone either.

Then these data suggest that. Those who have a car tend to have a cell phone.

Option C

6 0

2 years ago