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zubka84 [21]
3 years ago
8

An amount of 19,000 is borrowed for 7 years at 7.75% interest, compounded annually. If the loan is paid in full at the end of th

at period, how much be paid back?
Mathematics
1 answer:
Kruka [31]3 years ago
7 0

Answer:

The Amount paid back after 7 years is Rs 32,038.674  .

Step-by-step explanation:

Given as :

The loan amount borrowed = p = Rs 19,000

The rate of interest applied = r = 7.75% compounded annually

The time period of loan = t = 7 years

Let The Amount paid back after 7 years = Rs A

<u>Now,From Compound Interest method</u>

Amount = Principal × (1+\dfrac{\texrm rate}{100})^{\textrm time}

Or, A = p × (1+\dfrac{\texrm r}{100})^{\textrm t}

Or, A = Rs 19,000 × (1+\dfrac{\texrm 7.75}{100})^{\textrm 7}

Or, A = Rs 19,000 × (1.0775)^{7}

Or, A = Rs 19,000 × 1.686246

∴   A = Rs 32,038.674

So, The Amount paid back after 7 years = A = Rs 32,038.674

Hence, The Amount paid back after 7 years is Rs 32,038.674  . Answer

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sergiy2304 [10]

Answer:

p(x)= x^3-2x^2+x-2

Step-by-step explanation:

Here we are given that a polynomial has zeros as 2 , i and -i . We need to find out the cubic polynomial . In general we know that if \alpha , \ \beta \ \& \ \gamma are the zeros of the cubic polynomial , then ,

\sf \longrightarrow p(x)= (x -\alpha )(x-\beta)(x-\gamma)

Here in place of the Greek letters , substitute 2,i and -i , we get ,

\sf\longrightarrow p(x)= (x -2 )(x-i)(x+i)

Now multiply (x-i) and (x+i ) using the identity (a+b)(a-b)=a² - b² , we have ,

\sf  \longrightarrow p(x)= (x-2)\{ x^2 - (i)^2\}

Simplify using i = √-1 ,

\sf \longrightarrow p(x)= (x-2)( x^2 + 1 )

Multiply by distribution ,

\sf \longrightarrow p(x)= x(x^2+1) -2(x^2+1)

Simplify by opening the brackets ,

\sf\longrightarrow p(x)= x^3+x-2x^2-2

Rearrange ,

\sf\longrightarrow \underline{\boxed{\blue{\sf p(x)= x^3-2x^2+x-2}}}

4 0
2 years ago
Please Help Best answer gets brainliest first answer gets five stars
Brrunno [24]

Hi,

(1/4)^−2−(5^0)(2)(1^−1)

=16−(5^0)(2)(1^−1)

=16−(1)(2)(1^−1)

=16−2(1^−1)

=16−(2)(1)

=16−2

=14

Have a great day!

4 0
3 years ago
Read 2 more answers
The whole number 140 is divisible by 2, 3, 4, 5, 7, and 10.<br><br><br> True or False
Mariulka [41]
False. It is not divisible by 3.
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3 years ago
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Determine the actual cost, including interest, of a $3,450 used car with a down payment of $200 and 48 monthly payments of 75$ e
Fantom [35]

Answer:

$3,800

Step-by-step explanation:

$200 (down payment) + 48 x $75 (monthly payments)


$200 + $3600 = $3800


Since the original car was $3450, there was $350 in interest ($3800 - $3450 = $350)



5 0
3 years ago
A survey said that 3 out of 5 students enrolled in higher education took at least one online course last fall. Explain your calc
marysya [2.9K]

Answer:

a) 60% probability that student took at least one online course

b) 40% probability that student did not take an online course

c) 12.96% probability that all 4 students selected took online courses.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they took at least one online course last fall, or they did not. The probability of a student taking an online course is independent of other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

3 out of 5 students enrolled in higher education took at least one online course last fall.

This means that p = \frac{3}{5} = 0.6

a) If you were to pick at random 1 student enrolled in higher education, what is the probability that student took at least one online course?

This is P(X = 1) when n = 1. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{1,1}.(0.6)^{1}.(0.4)^{0} = 0.6

60% probability that student took at least one online course.

b) If you were to pick at random 1 student enrolled in higher education, what is the probability that student did not take an online course?

This is P(X = 0) when n = 1.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{1,1}.(0.6)^{0}.(0.4)^{1} = 0.4

40% probability that student did not take an online course

c) Now, consider the scenario that you are going to select random select 4 students enrolled in higher education. Find the probability that all 4 students selected took online courses

This is P(X = 4) when n = 4.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{4,4}.(0.6)^{4}.(0.4)^{0} = 0.1296

12.96% probability that all 4 students selected took online courses.

3 0
3 years ago
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