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oksian1 [2.3K]
3 years ago
5

What is 37% out of 87%?

Mathematics
1 answer:
vichka [17]3 years ago
4 0
37% out of 87% equals to 42.5
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The left and right page numbers of an open book are two consecutive integers whose sum is 87.
Lisa [10]
Left page numbers  are always even  and right page will have a number 1 greater that the left. so  its not possible to have a sum of 87.
If even page  is 2n then right page = 2n + 1:-

2n + 2n + 1 = 87

4n = 86

n = 86 / 4 = 21.5  and 2n = 43  but left page is even  so its not possible to have a sum of left + right = 87.


3 0
3 years ago
Lets find 1/3 - 1/12<br><br> 1/3 - 1/12 = ___/12 - 1/12 =___/___<br><br> Fill in the blanks
kicyunya [14]
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4 0
3 years ago
A 20-year loan of 1000 is repaid with payments at the end of each year. Each of the first ten payments equals 150% of the amount
Alja [10]

Answer:

x = 97

Step-by-step explanation:

Given

t = 20 --- time (years)

A =1000 --- amount

r = 10\% --- rate of interest

Required

The last 10 payments (x)

First, calculate the end of year 1 payment

y_1(end) = 10\% * 1000 * 150\%

y_1(end) = 150

Amount at end of year 1

A_1=A - y_1(end) - r * A

A_1=1000 - (150 - 10\% * 1000)

A_1 =1000 - (150- 100)

A_1 =950

Rewrite as:

A_1 = 0.95 * 1000^1

Next, calculate the end of year 1 payment

y_2(end) = 10\% * 950 * 150\%

y_2(end) = 142.5

Amount at end of year 2

A_2=A_1 - (y_2(end) - r * A_1)

A_2=950 - (142.5 - 10\%*950)

A_2 = 902.5

Rewrite as:

A_2 = 0.95 * 1000^2

We have been able to create a pattern:

A_1 = 1000 * 0.95^1 = 950

A_2 = 1000 * 0.95^2 = 902.5

So, the payment till the end of the 10th year is:

A_{10} = 1000*0.95^{10}

A_{10} = 598.74

To calculate X (the last 10 payments), we make use of the following geometric series:

Amount = \sum\limits^{9}_{n=0} x * (1 + r)^n

Amount = \sum\limits^{9}_{n=0} x * (1 + 10\%)^n

Amount = \sum\limits^{9}_{n=0} x * (1 + 0.10)^n

Amount = \sum\limits^{9}_{n=0} x * (1.10)^n

The amount to be paid is:

Amount = A_{10}*(1 + r)^{10} --- i.e. amount at the end of the 10th year * rate of 10 years

Amount = 1000 * 0.95^{10} * (1+r)^{10}

So, we have:

Amount = \sum\limits^{9}_{n=0} x * (1.10)^n

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+r)^{10}

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+10\%)^{10}

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+0.10)^{10}

\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1.10)^{10}

The geometric sum can be rewritten using the following formula:

S_n = \sum\limits^{9}_{n=0} x * (1.10)^n

S_n =\frac{a(r^n - 1)}{r -1}

In this case:

a = x

r = 1.10

n =10

So, we have:

\frac{x(r^{10} - 1)}{r -1} = \sum\limits^{9}_{n=0} x * (1.10)^n

\frac{x((1.10)^{10} - 1)}{1.10 -1} = \sum\limits^{9}_{n=0} x * (1.10)^n

\frac{x((1.10)^{10} - 1)}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n

x * \frac{1.10^{10} - 1}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n

So, the equation becomes:

x * \frac{1.10^{10} - 1}{0.10} = 1000 * 0.95^{10} * (1.10)^{10}

Solve for x

x = \frac{1000 * 0.95^{10} * 1.10^{10} * 0.10}{1.10^{10} - 1}

x = 97.44

Approximate

x = 97

4 0
4 years ago
If q is the smallest composite number greater than 2 and p is the smallest prime number less than 10, what is p/q? a. 2 b. 0.5 c
Mrac [35]

Answer:

b. 0.5

Step-by-step explanation:

knowing that a prime number is an integer with exactly two divisors and a composite number is an integer with more than two divisors, we have that if q is a composite number greater than 2 these can be 4, 6, 8, .. But how the smallest of them should be is 4, and  p is a prime number can be 2,3,5,7,....

but  is the smallest prime number less than 10 then p=2, also p/q=2/4 = 0.5

6 0
4 years ago
}
Juliette [100K]

Answer:

the items new price is $60

Step-by-step explanation:

4 0
3 years ago
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