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10 months ago

At what rates percent compound interest will Rs2000 amount to Rs 2420 in 2 years.

1 answer:

4 0

The interest rates required to get a total amount of $2,420 from compound interest on a principal of $2,000 compounded 1 times per year over 2 years is 10% per year.

<h3>What is compound interest?</h3>

The interest on savings that is calculated on both the initial principal and the interest accrued over time is known as compound interest.

The concept of compound interest, also known as "interest on interest," is thought to have first appeared in Italy in the 17th century. It will accelerate the growth of a sum more quickly than simple interest, which is calculated only on the principal sum.

Money is multiplied more quickly through compounding, and the more times it is compounded, the higher the compound interest will be.

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

Solving for rate r as a decimal

r = n[(A/P)^(1/nt) - 1]

r = 1 × [(2,420/2,000)^{1/(1)(2)} - 1]

r = 0.1

Then convert r to R as a percentage

R = r × 100

R = 0.1 × 100

R = 10%/year

Learn more about compound interest

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The mayor of a town has proposed a plan for the construction of an adjoining bridge. A political study took a sample of 13001300

Ket [755]

Answer:

No, there is not enough evidence at the 0.02 level to support the strategist's claim.

Step-by-step explanation:

We are given that a political study took a sample of 1300 voters in the town and found that 57% of the residents favored construction.

And, a political strategist wants to test the claim that the percentage of residents who favor construction is more than 54%, i.e;

Null Hypothesis, $H_0$ : p = 0.54 {means that the percentage of residents who favor construction is 54%}

Alternate Hypothesis, $H_1$ : p > 0.54 {means that the percentage of residents who favor construction is more than 54%}

The test statistics we will use here is;

T.S. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }$ ~ N(0,1)

where, p = actual percentage of residents who favor construction = 0.54

$\hat p$ = percentage of residents who favor construction in a sample of

1300 voters = 0.57

n = sample of voters = 1300

So, Test statistics = $\frac{0.57 -0.54}{\sqrt{\frac{0.57(1- 0.57)}{1300} } }$

= 2.185

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the percentage of residents who favor construction is 54% and the strategist's claim is not supported.

4 0

3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$