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3 years ago

PLEASE HELP I DONT GET IT

Mathematics

1 answer:

irga5000 [103]3 years ago

5 0

I think it's b but i'm not sure!

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Matt Ali deposited $25,000 in a savings account. The account earns 5.5 percent interest compounded daily. Use the formula “ Amou

katen-ka-za [31]

Answer:

A) Amount = $ 116,601.31

B) Compound Interest = $ 91,601.31

Step-by-step explanation:

Compound interest Formula

A = P(1 + r/n)^nt

Where

P = Initial amount invested = $25,000

r = interest rate = 5.5 %

n = Compounding frequency = daily

t = time in years = 28 years

A). What amount will he have in his account 28 years later ?

First, convert R percent to r a decimal

r = R/100

r = 5.5%/100

r = 0.055 per year,

Then, solve our equation for A

A = P(1 + r/n)^nt

A = 25,000.00(1 + 0.000150685/365)^(365)(28)

A = $ 116,601.31

B.) How much will be compound interest ?

I = Amount - Principal

A(Amount ) = $ 116,601.31

P (principal) = $ 25,000.00

I = $ 116,601.31 - $ 25,000.00

I (interest) = $ 91,601.31

6 0

2 years ago

If the sum of the first 12 terms of a geometric series is 8190 and the common ratio is 2. Find the first term and the 20th term.

Reil [10]

The first term is 2 and the 20th term is 1048576 .

Step-by-step explanation:

Here we have , If the sum of the first 12 terms of a geometric series is 8190 and the common ratio is 2. We need to Find the first term and the 20th term. Let's find out:

We know that Sum of a GP is :

⇒ $S_n = \frac{a(r^n-1)}{r-1}$

So ,Sum of first 12 terms is :

⇒ $S_1_2 = \frac{a(2^{12}-1)}{2-1}$

⇒ $8190=a(2^{12}-1)$

⇒ $\frac{8190}{4095}=a$

⇒ $a=2$

Now , nth term of a GP is

⇒ $a_n = ar^{n-1}$

So , 20th term is :

⇒ $a_2_0 = 2(2)^{20-1}$

⇒ $a_2_0 = (2)^{20}$

⇒ $a_2_0 = 1048576$

Therefore , the first term is 2 and the 20th term is 1048576 .

8 0

3 years ago

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found: