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lesantik [10]
2 years ago
9

Can someone solve this

Mathematics
1 answer:
marysya [2.9K]2 years ago
4 0

From the Arithmetic Information given, Pr = 0.049375 ≠ -8.050625 ≠ 0.0325  See explanation below.

<h3>

What are the step by steps solution to the questions above?</h3>

First, lets us restate the question properly. We have

Pr = 1 - ((6*12+6)/80)² = 1 - ((9*12²)/1600) - ((3*12*6)/1600) - (6²/6400) = 1 - 0.005625 * 12² - 0.001875 * 12 * 6 - 0.00015625 * 6²

Note that there are three equals signs. So lets divide the problem according and solve for the different parts.

Lets   1 - ((6*12+6)/80)²  ............A

1 - ((9*12²)/1600) - ((3*12*6)/1600) - (6²/6400) .............B; and

1 - 0.005625 * 12² - 0.001875 * 12 * 6 - 0.00015625 * 6² ........C

Solving for A we have

1 - (78/80)²

= 1 - (0.975)²

= 1 - 0.950625

A = 0.049375


Solving for B we have

1 - ((9*12²)/1600) - ((3*12*6)/1600) - (6²/6400)

= 1- (14,256/1600) - (216/1600) - (36/6400)
= 1 - 8.91 - 0.135 - 0.005625

B = -8.050625

Solving for C we have

1 - 0.005625 * 12² - 0.001875 * 12 * 6 - 0.00015625 * 6²

= 1 - 0.005625 * 144 - 0.001875 * 12 * 6 - 0.00015625 * 144

= 0.0325


In summary we can state that:

A = 0.049375

B =  -8.050625

C = 0.0325

Given that there were no abstract quantities, we can state that

Pr = A ≠ B ≠C or

Pr = 0.049375 ≠ -8.050625 ≠ 0.0325

Learn more about equations with equal signs at

brainly.com/question/18924248
#SPJ1

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Hello

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So as we look at 8x, it needs to equal 40

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Compound interest In Exercise,$3000 is invested in an account at interest rate r,compounded continuously.Find the time required
xenn [34]

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(a) 8.15

(b) 12.92

Step-by-step explanation:

Given: P = $3000, r = 0.085

                A = Pe^{rt}

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

               A = 2 × $3000

               A = $6000

               6000 = 3000e^{0.085t}

               \frac{6000}{3000} = e^{0.085t}

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Take log_{e} of both sides

               log_{e}2 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln2 = 0.085t

               t = \frac{ln2}{0.085}

               t = \frac{0.693}{0.085}

               t = 8.15

(b) For the amount to double, A = 3 × P

               A = 3 × $3000

               A = $9000

               9000 = 3000e^{0.085t}

               \frac{9000}{3000} = e^{0.085t}

               3 = e^{0.085t}

Take log_{e} of both sides

               log_{e}3 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln3 = 0.085t

               t = \frac{ln3}{0.085}

               t = \frac{1.0986}{0.085}

               t = 12.92

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