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Fed [463]
3 years ago
13

Find the simple interest and amount when principal = Rs. 6000, rate = 6% per annum and time = 4 years

Mathematics
1 answer:
Alinara [238K]3 years ago
7 0

Given:

Principal = Rs. 6000

Rate of simple interest = 6% per annum.

Time = 4 years

To find:

The simple interest and amount.

Solution:

Formula for simple interest:

I=\dfrac{P\times r\times t}{100}

Where, P is principal, r is the rate of interest and t is the number of years.

Putting P=6000, r=6 and t=4, we get

I=\dfrac{6000\times 6\times 4}{100}

I=60\times 6\times 4

I=1440

Now,

Amount=Principal+Interest

Amount=6000+1440

Amount=7440

Therefore, the simple interest is Rs. 7440 and the amount is Rs 7440.

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arsen [322]

Answer:

13.76

Step-by-step explanation:

We want year 10; this means we substitute 10 in place of x:

y=-0.34(10²)+4.43(10)+3.46

y = -0.34(100)+4.43(10)+3.46

y = -34+44.3+3.46 = 13.76

3 0
3 years ago
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Would it be
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3 years ago
A random sample of 7 fields of durum wheat has a mean yield of 29.8 bushels per acre and standard deviation of 3.62 bushels per
meriva

Answer:

CI = 29.8 ± 3.53

Critical value is z = 2.58

Step-by-step explanation:

First of all let's find margin of error. It is given by the formula;

ME = zσ/√n

We are given;

Standard deviation; σ = 3.62

Sample size; n = 7

Mean; x¯ = 29.8

Now, z-value for 99% Confidence level is 2.58

Thus;

ME = (2.58 × 3.62)/√7

ME = 3.53

CI is written as;

CI = x¯ ± ME

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5 0
3 years ago
PLEASE SHOW ALL THE STEPS THAT YOU USE TO SOLVE THIS PROBLEM
Mademuasel [1]

Answer:

{x = 1 , y=1, z=0

Step-by-step explanation:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

{-2 x - y + z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x - 3 y - 2 z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y+0 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer:  {x = 1 , y=1, z=0

6 0
3 years ago
The yearbook staff receive 75 submissions for yearbook articles. Paul will accept 2/5
antiseptic1488 [7]

Answer:

2/5 of 75 is 30 thats all i have for because i am haveing trouble with the same question

Step-by-step explanation:

7 0
2 years ago
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