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crimeas [40]
2 years ago
15

Annie gets a loan from her bank .She agrees yo borrow 8000 pounds at a fixed annual simple interest rate of 6%. She also agrees

to pay the loan back over a 10 year period. How much money in total will she have paid back at the end of 10 years
Mathematics
1 answer:
mojhsa [17]2 years ago
6 0

Annie paid 12800 pounds at end of 10 years

<em><u>Solution:</u></em>

Given that Annie borrows 8000 pounds at simple interest rate of 6 %

She also agrees to pay the loan back over a 10 year period

To find: total money paid back at the end of 10 years

The total amount paid is given as:

Total amount = simple interest + principal amount borrowed

The simple interest is given as:

S.I = \frac{pnr}{100}

Where, "p" is the principal sum

"n" is the number of years

"r" is the rate of interest

Here, p = 8000 ; r = 6 % ; n = 10 years

S.I = \frac{8000 \times 10 \times 6}{100}\\\\S.I = 4800

Therefore total amount paid at end of 10 years is:

Total amount = 4800 + 8000 = 12800

Thus Annie paid 12800 pounds at end of 10 years

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Answer:

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  • and the larger one is x = 6

Step-by-step explanation:

A "critical number" is a function argument where the function has zero or undefined slope. A polynomial function never has undefined slope, so we're looking for the x-values where f'(x) = 0.

A graphing calculator easily shows the points that have zero slope. They are found at x = 3 and x = 6.

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Setting the derivative to zero is another way to find the points with zero slope:

  f'(x) = 6x^2 -54x +108 = 6(x^2 -9x +18) = 6(x-3)(x-6)

The derivative will be zero where the factors are zero, at x=3 and x=6.

The critical numbers are x=3 and x=6.

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2 years ago
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3 years ago
Read 2 more answers
Find the value of k and x2<br> x^2+ 13x + k = 0, x1=-9
Anarel [89]

Given:

The quadratic equation is:

x^2+13x+k=0

x_1=-9

To find:

The value of k and x_1.

Solution:

We have,

x^2+13x+k=0                ...(i)

Putting x=-9, we get

(-9)^2+13(-9)+k=0

81-117+k=0

-36+k=0

k=36

Putting k=36 in (i), we get

x^2+13x+36=0

Splitting the middle term, we get

x^2+9x+4x+36=0

x(x+9)+4(x+9)=0

(x+9)(x+4)=0

x=-9,-4

Here, x_1=-9 and x_2=-4.

Therefore, the required values are k=36 and x_2=-4.

3 0
2 years ago
Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul
ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that p = 0.5

12 young adults were randomly selected

This means that n = 12

(a) What is the probability that more than 7 preferred McDonald's?

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)

In which

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

P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121

P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054

P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016

P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003

P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)

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

P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054

P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121

P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193

P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226

P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193

P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since p = 1-p = 0.5, this is the same as b) above.

So

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

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2 years ago
Please help hurry i will give brain
mars1129 [50]

Answer:

A

Step-by-step explanation I counted the numbers.

7 0
2 years ago
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