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2 years ago

Find the simple interest and the total amount after three years.

Mathematics

1 answer:

ZanzabumX [31]2 years ago

5 0

Answer:

$\boxed{ \sf{Total \:Interest = \underline{ 2223} \: rupees}}$

$\boxed{ \sf{Total \: Amount = \underline{10023} \: rupees}}$

Step-by-step explanation:

<h3>♨ Given :</h3>

Principal ( P ) = 7800 rupees
Annual rate of interest ( R ) = 9.5 %
Time ( T ) = 3 years

Total interest ( I ) = ?
Total amount ( A ) = ?

<h3>✎ Finding the total interest :</h3>

$\boxed{ \sf{Simple \: interest \: ( \: I\: )\: = \frac{Principal \: ( \: P \: ) \: \times \: Time \: ( \: T \: ) \: \times \: Rate \: ( \: R \: )}{100}}}$

$\longrightarrow{ \sf{ \frac{7800 \times 3 \times 9.5}{100}}}$

$\longrightarrow{ \sf{ \frac{222300}{100}}}$

$\longrightarrow{ \sf{2223 \: rupees}}$

The total interest = <u>2</u><u>2</u><u>2</u><u>3</u> rupees.

<h3>✎ Finding the total Amount :</h3>

$\boxed{ \sf{Amount \: (A) = Principal \: ( \: P \: ) \: + \: Interest \: ( \: I \: )}}$

$\longrightarrow{ \sf{7800 \: rupees + 2223 \: rupees}}$

$\longrightarrow{ \sf{10023 \: rupees}}$

The total amount = <u>1002</u><u>3</u> rupees.

--------------------------------------------------------------

<h3>☞ Additional Info :</h3>

Principal : The money which is borrowed or deposited is called principal ( P ).
Interest : The additional amount of money which is paid by borrower to the lender is called Interest ( I ) .
Time : The duration of time for which principal is deposited or borrowed is termed as time period ( T ).
Rate : The condition under which the interest is charged is called rate ( R ).
Amount : The sum of principal and interest is called an amount ( A ).

<h3>✑ Important Formulaes :</h3>

$\sf{Interest \: = \frac{Principal \times Time \times Rate}{100}}$

$\sf{Principal = \frac{Interest \: \times \: 100}{Time \times Rate}}$

$\sf{Time = \frac{Interest \: \times \: 100}{Principal \times Rate}}$

$\sf{Rate = \frac{Interest \: \times \: 100}{Principal \times Time}}$

$\sf{Amount = Principal + Interest}$

$\sf{Principal = Amount - Interest}$

$\sf{Interest = Amount - Principal}$

Hope I helped!

Have a wonderful time ツ

~TheAnimeGirl

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Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

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∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

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So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$