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Orlov [11]
3 years ago
11

Find the compound interest and the total amount after 4 years and 6 months if the interest is compounded annually

Mathematics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

Compound Interest,I- I=P[(1+I)^{4.5}-1]

Total Amount, A- A=P(1+i)^{4.5}

Step-by-step explanation:

Compound interest is the difference between the initial amount invested at time t=0and the final amount of the investment at time t.

-Let the initial amount invested be P, the annual rate be i and A be the final amount of the investment.

-Let I be the compound interest earned.

-Given that the investment term n=4yrs 6 months, the compound interest is calculated as;

A=P(1+i)^n\\\\I=A-P\\\\\# n=4.5,P=P, i=i\\\\\therefore I=P(1+i)^{4.5}-P\\\\=P[(1+I)^{4.5}-1]\\\\I=P[(1+I)^{4.5}-1]

Hence, the compound interest after 4 1/2 years is I=P[(1+I)^{4.5}-1]

#The total amount after the same period is A=P(1+i)^{4.5}

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Consider the vector field f(x,y,z)=(2z+3y)i+(3z+3x)j+(3y+2x)kf(x,y,z)=(2z+3y)i+(3z+3x)j+(3y+2x)k.
natima [27]
\dfrac{\partial f}{\partial x}=2z+3y\implies f(x,y,z)=2xz+3xy+g(y,z)

\dfrac{\partial f}{\partial y}=3x+\dfrac{\partial g}{\partial y}=3z+3x
\dfrac{\partial g}{\partial y}=3z\implies g(y,z)=3yz+h(z)
\implies f(x,y,z)=3xz+3xy+3yz+h(z)


\dfrac{\partial f}{\partial z}=3x+3y+\dfrac{\mathrm dh}{\mathrm dz}=3y+2x
\dfrac{\mathrm dh}{\mathrm dz}=-x

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3 years ago
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Anvisha [2.4K]

Answer:

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7 0
2 years ago
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The scale of a map is 1in.:75km. Determine the distance between two towns that are 5.6 in. apart on the map.
sergeinik [125]

Answer:

420 km

Step-by-step explanation:

75 times 5 = 375

75/10 = 7.5 times 6 = 45

375 + 45 = 420

6 0
3 years ago
The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

Z = \frac{X - \mu}{s}

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

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Answer:

One Solution

Step-by-step explanation:

Trust me

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