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3 years ago

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When I count as a principal of $1000 and earns 4% simple interest per year and other account as a principal $1000 and earns 4% i

nterest compounded annually which account has the greater balance at the end of four years

1 answer:

OLga [1]3 years ago

3 0

Answer: the account that earned compound interest has the greater balance at the end of four years.

Step-by-step explanation:

The formula for determining simple interest is expressed as

I = PRT/100

Where

I represents interest paid on the amount invested.

P represents the principal or amount invested.

R represents interest rate

T represents the duration of the investment in years.

From the information given,

P = 1000

R = 4%

T = 4 years

I = (1000 × 4 × 4)/100 = 160

Total amount earned is

1000 + 160 = $1160

The formula for determining compound interest is expressed as

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

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 1000

r = 4% = 4/100 = 0.04

n = 1 because it was compounded once in a year.

t = 4 years

Therefore,.

A = 1000(1+0.04/1)^1 × 4

A = 1000(1.04)^4

A = $1170

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3 years ago

Algebra 2, please help! thank you

Hunter-Best [27]

Answer:

-2pi/3

Step-by-step explanation:

y = 2 cos 3(x + 2π∕3) +1

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7 0

3 years ago

Read 2 more answers

If JKLM is a rhombus, MK = 30, NL = 13, and mZMKL = 41°, find each measure.

oksian1 [2.3K]

Answer:

$NK = 15$

$JL = 26$

$KL = 19.85$

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

Step-by-step explanation:

Given

$MK = 30$

$NL = 13$

$\angle MKL = 41$

Solving (a): NK

MK is a diagonal and NK is half of the diagonal. So:

$NK = \frac{1}{2} * MK$

$NK = \frac{1}{2} * 30$

$NK = 15$

Solving (b): JL

JL is a diagonal, and it is twice of NL.

$JL = 2 * NL$

$JL = 2 * 13$

$JL = 26$

Solving (c): KL

To solve for KL, we consider triangle KNL where:

$\angle KNL = 90$

and

$KL^2 = NL^2 + NK^2$

$KL^2 = 13^2 + 15^2$

$KL^2 = 394$

$KL = \sqrt{394$

$KL = 19.85$

Solving (d - h):

To do this, we consider triangle JKN

$\angle KNL = \angle LNM = \angle MNJ = \angle JNK = 90$ -- diagonals bisect one another at right angle

Alternate interior angles are equal. So:

$\angle MKL = \angle KMJ = \angle KJL = \angle JLM = 41$

Similarly:

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 90 - 41$

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 49$

So:

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = \angle MLJ + \angle JLK$

$\angle MLK = 49 + 41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

5 0

3 years ago

Given g(x) = -x – 4, find g(-5).

riadik2000 [5.3K]

The value of g(-5) from the given. function; g(x) = -x – 4 is; g(-5) = 1

<h3>Evaluation of functions</h3>

The given function is;

g(x) = -x – 4

Hence, the value of g(-5) can be obtained by substituting -5 for x in the function as follows;

g(-5) = -(-5) -4

g(-5) = 5 -4

g(-5) = 1.

Read more on functions;

brainly.com/question/10439235

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3 years ago

Read 2 more answers

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.

8 0

3 years ago