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Stels [109]
3 years ago
7

A principal of $2200 is invested at 6.25% interest, compounded annually. How much will the investment be worth after 13 years

Mathematics
2 answers:
trasher [3.6K]3 years ago
7 0
The investment will be worth $4,838.37 after 13 years.
s2008m [1.1K]3 years ago
6 0

Answer:

Let S = Sum after 13 years

So = amount invested

t = time in years

i = annual interest rate = .0325

The S = So(1+i)t = $2,200(1.0325)13 = $3,334.21

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artcher [175]

you are driving onajor highway 94 mp

4 0
3 years ago
This is a math question find the complement of 50 degree​
AlexFokin [52]

Given:

The measure of an angle is 50 degrees.

To find:

The complement of the given angle.

Solution:

We know that the sum of an angle and its complement is 90 degrees.

Let x be the complement angle of 50 degrees.

x+50^\circ=90^\circ

x=90^\circ-50^\circ

x=40^\circ

Therefore, the measure of complement of the given angle is 40 degrees.

6 0
3 years ago
Twice the difference of a number and four is less than the sum of the number and five.
Whitepunk [10]
Let the number be x.

2*(x-4) < (x+5).

2x -8 < x +5
2x-x < 5 +8
x<13.

The number is less than 13.

That's it. Cheers.
7 0
3 years ago
The perimeter of a rectangle is 32 meters. The length is 5 less than twice the width. Find the length and the width of the recta
viktelen [127]
Let x and y be the length and width of the rectangle respectively
2(x+y)=32...(1)
x=2y-5...(2)

From (1),we have
2x+2y=32...(3)

By sub.(2) into (3),we have
2(2y-5)+2y=32
4y-10+2y=32
6y=42
y=7

By sub.y=7 into (2),we have
x=2*7-5
x=14-5
x=9

Length =9m
Width =7m
5 0
3 years ago
4) Sandeep, Hee, Sara, and Mohammad play euchre with a standard deck consisting of 24 cards (A, K, Q, J, 10, and 9 from each of
Keith_Richards [23]

Answer:

1.19685 × 10¹⁶

Step-by-step explanation:

From the given information.

Since each players receives 5 cards; then = 5 * 4 = 20

Now, the process to go about this is to first select 5 cards out of 24 that goes to player 1. Then from the remaining 19, we can select another 5 cards that goes to player 2, etc. until 4 cards are left, out of which one is remained faced up. Thus, the 4 sets of 5 cards selected can be shuffled in 4! ways.

∴

No fo ways = ^{24}C_5 \times ^{19}C_5 \times ^{14}C_5 \times ^{9}C_5 \times 4! \times ^{4}C_1

= \dfrac{24!}{5!(24-5)!}\times  \dfrac{19!}{5!(19-5)!}\times  \dfrac{14!}{5!(14-5)!}\times  \dfrac{9!}{5!4!}\times 4! \times \ \dfrac{4!}{1!(4-1)!}

= \dfrac{24!}{5!^4}\times 4

= 1.19685 × 10¹⁶

6 0
3 years ago
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