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Tcecarenko [31]
3 years ago
9

What is the interest earned on $3000 at a rate of 0.04 for three years? The formula is interest equals (principal) (rate) (time)

Substitute and multiply
Mathematics
1 answer:
bekas [8.4K]3 years ago
5 0

<u>Answer:</u>

The interest after 3 years is $360

<u>Explanation:</u>

Given the principal amount (P) = $3000

Rate of interest (R) = 0.04

Time period (T) is given as 3 years

The Simple Interest is calculated by the formula;

SI = Principal \times Rate of Interest \times Time

Substituting the values in the above formula,

SI = 3000 \times 0.04 \times 3

SI = $360

Therefore, the interest after 3 years is $360

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It's a whole number. It cannot be a mixed number. Anyways I would have to know the denominator.

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Which of the following best defines 3 to the power of 2 over 3 ? A.Square root of 9 B.Cube root of 9 C.Cube root of 3 D.Square r
blondinia [14]
The answer is b. Cube root of 9
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18x-(-3x+9)help!!! Please
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Answer:

21x - 9

Step-by-step explanation:

3 0
3 years ago
A cake manufacturer boxes Swiss chocolate and German chocolate cakes at one site. A box of Swiss chocolate cake contains 0.55 lb
Kobotan [32]

Answer:

2,500 German chocolate cake boxes.

1,500 Swiss chocolate cake boxes.

Step-by-step explanation:

Let 'S' be the number of Swiss chocolate cakes boxed and 'G' the number of German cholocate cakes boxed. If all of the available ingredients are used:

0.55*S +0.59*G = 2,300\\0.45*S+0.41*G = 1,700

Solving the linear system above:

(0.55 *S +0.59*G)- \frac{0.55}{0.45} (0.45*S+0.41*G)= 2,300-\frac{0.55}{0.45}*1,700*\\0.088888*G = 222.222\\G=2,500\\S =\frac{2,300-2,500*0.59}{0.55}\\S=1500

2,500 German chocolate cake boxes and 1,500 Swiss chocolate cake boxes can be made each day.

6 0
4 years ago
Sherman has two sequences . The first sequence is described by the explicit rule f(n) = 15n + 4 and the second sequence is descr
Viktor [21]

Answer:

Step-by-step explanation:

Given the explicit function as

f(n) = 15n+4

The first term of the sequence is at when n= 1

f(1) = 15(1)+4

f(1) = 19

a = 19

Common difference d = f(2)-f(1)

f(2) = 15(2)+4

f(2) = 34

d = 34-19

d = 15

Sum of nth term of an AP = n/2{2a+(n-1)d}

S20 = 20/2{2(19)+(20-1)15)

S20 = 10(38+19(15))

S20 = 10(38+285)

S20 = 10(323)

S20 = 3230.

Sum of the 20th term is 3230

For the explicit function

f(n) = 4n+15

f(1) = 4(1)+15

f(1) = 19

a = 19

Common difference d = f(2)-f(1)

f(2) = 4(2)+15

f(2) = 23

d = 23-19

d = 4

Sum of nth term of an AP = n/2{2a+(n-1)d}

S20 = 20/2{2(19)+(20-1)4)

S20 = 10(38+19(4))

S20 = 10(38+76)

S20 = 10(114)

S20 = 1140

Sum of the 20th terms is 1140

7 0
3 years ago
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