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3 years ago

What is 12 divided by 1230 in long division

2 answers:

5 0

It is 102.5. You made it kind of confusing at first by putting the divisor first in your question then putting your dividend, but I understand what you meant.

algol133 years ago

3 0

What is 12 divided by 1230 in long division? Answer:

0 1 0 2 5

1 2 / 1 2 3 0 0

- 1 2

0 3 0

- 2 4

6 0

- 6 0

Since 12 x 102.5 does equal 1230, our solution checks out.

hope it helps

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The graph of which trigonometric function(s) include the point (0,1)? Select all that apply.

iogann1982 [59]

Im pretty sure
Y=Sin(2x)+1
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3 0

3 years ago

The compound interest on a sum of money in

Yakvenalex [24]

Answer:

The difference between the principal and the compound interest in three years is Rs 17,994

Step-by-step explanation:

The compound interest is given according to the following formula;

$C.I. = P \cdot \left ( 1 + \dfrac{r}{n} \right ) ^{n\cdot t} - P$

The given amount of the compound after 2 years = Rs 5,460

The given amount of the compound after 4 years = Rs 12,066.60

Therefore, we have;

$5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P$ ...(1)

$12,066.60 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{4} - P$ ...(2)

Dividing equation (2) by (1), we have;

$\dfrac{12,066.60}{5,460} = \dfrac{P \cdot \left ( \left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1\right )}{P \cdot \left (\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 \right ) } =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 }$

$Let \ \left ( 1 + \dfrac{r}{100} \right ) ^{2} = x, we \ get;$

$\dfrac{12,066.60}{5,460} =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 } = \dfrac{x^2 - 1}{x - 1}$

∴ 12,066.60 × (x - 1) = 5,460 × (x² - 1) = 5,460 × (x - 1) ×(x + 1)

∴ 12,066.60 × (x - 1)/(x - 1) = 5,460 × (x + 1)

12,066.60/5,460 = x + 1

x = 12,066.60/5,460 - 1 = 1.21 = 121/100

x = 121/100

$\left ( 1 + \dfrac{r}{100} \right ) ^{2} = x = \dfrac{121}{100}$

$1 + \dfrac{r}{100} =\sqrt{ \dfrac{121}{100}} = \dfrac{11}{10}$

We get

$\dfrac{12,066.60}{5,460} =\dfrac{221}{100}$

$\therefore \dfrac{12,066.60}{5,460} =\dfrac{221}{100} = \left ( 1 + \dfrac{r}{100} \right ) ^{2}$

$1 + \dfrac{r}{100} = \sqrt{ \dfrac{221}{100} } = \dfrac{\sqrt{221} }{10}$

$\dfrac{r}{100} = \dfrac{\sqrt{221} }{10} - 1$

$\dfrac{r}{100} = \dfrac{11}{10} - 1 = \dfrac{1}{10} = 0.1$

r = 100 × 0.1 = 10%

r = 10%

Therefore, we have;

$5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P = P \times \left ( 1 + 0.1\right ) ^{2} - P$

$5,460 = P \times \left ( 1 + 0.1\right ) ^{2} - P = P \times \left (\left ( 1 + 0.1\right ) ^{2} - 1\right) = P \times \dfrac{21}{100}$

$P = \dfrac{100}{21} \times 5,460 = 26,000$

The principal = Rs. 26,000

The compound interest in 3 years is therefore;

$CI_3 = 26,000 \times \left ( 1 + \dfrac{10}{100} \right ) ^{3} - 26,000= 8606$

The difference, 'd', between the principal and the compound interest in three years, is given as follows;

d = P - CI₃

d = 26,600 - 8606 = 17994

The difference between the principal and the compound interest in three years, d = Rs 17,994.

5 0

3 years ago

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and tes

EleoNora [17]

Answer:

a) The probability that this whole shipment will be accepted is 30%.

b) Many of the shipments with this rate of defective aspirin tablets will be rejected.

Step-by-step explanation:

We have a shipment of 3000 aspirin tablets, with a 5% rate of defects.

We select a sample of size 48 and test for defectives.

If more than one aspirin is defective, the batch is rejected.

The amount of defective aspirin tablets X can be modeled as a binomial distribution random variable, with p=0.55 and n=48

We have to calculate the probabilities that X is equal or less than 1: P(X≤1).

$P(X\leq1)=P(X=0)+P(X=1)\\\\\\P(0)=\binom{48}{0}(0.05)^0(0.95)^{48}=1*1*0.0853=0.0853\\\\\\P(1)=\binom{48}{1}(0.05)^1(0.95)^{47}=48*0.05*0.0897=0.2154\\\\\\P(X\eq1)=0.0853+0.2154=0.3007$

3 0

4 years ago

The alphabet game costs $.25 to play. Before the game, 26 slips of paper with a different letter of the alphabet on it are put i

iragen [17]

Answer: 25/676

Step-by-step explanation:

Number of possible outcomes = 26

In other to win, one must draw must be either (A, E, I, O or U)

Therefore required drws to win = 5

First draw:

P(win) = Total required outcome / Total possible outcome

P(win) = 5/26

Second draw:

P(win) = Total required outcome / Total possible outcome

P(win) = 5/26

Therefore,

P(winning twice) = (5/26) × (5/26) = 25/676

3 0

4 years ago

Hannah does push ups in sets of 3. She does 10 push ups today. Then her coach asked her to do 15 more push ups. How many push up

Dennis_Churaev [7]

If they meant 10 sets of push-ups, then she did 45. if they just meant 10 push-ups, it is 25.

5 0

3 years ago