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alekssr [168]
3 years ago
12

Mitchell deposits $1,200 in an account that pays 1.5% simple interest. He keeps the money in the account for three years without

any deposits or withdrawals. How much is in the account after three years?
Mathematics
2 answers:
rosijanka [135]3 years ago
7 0

The amount that will be in the account after three years is $1,254.00.

Using this formula

B = P+ ( P x R x T)

Where:

B=Amount ?

P=Principal=$1,200

r=Rate=1.5% or 0.015

T=Time=3 years

Let plug in the formula

B=$1,200+($1,200×0.015×3)

B=$1,200+$54

B= $1,254.00

Inconclusion the amount that will be in the account after three years is $1,254.00.

Learn more here:

brainly.com/question/20348538

maks197457 [2]3 years ago
3 0

The simple interest is calculated as the product of the <em>principal, rate and time</em>. Hence, Mitchell will have $1254 in her account.

Given the Parameters :

  • Principal, P = 1200
  • Rate, r = 1.5%
  • Time, t = 3 years

Amount in account, A = P(1 + rt)

Amount in account = 1200(1 + 0.015 × 3)

Amount in account = 1200(1.045) = 1254

Therefore, Mitchell will have $1254 after 3 years

Learn more :brainly.com/question/25281429

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Answer:

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Step-by-step explanation:

20+12-E=2

32-E=2

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3 years ago
If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.
nydimaria [60]

Answer:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

Step-by-step explanation:

Given

\cos(\theta) = -\frac{2}{3}

\theta \to Quadrant III

Required

Determine \tan(\theta) \cdot \cot(\theta) + \csc(\theta)

We have:

\cos(\theta) = -\frac{2}{3}

We know that:

\sin^2(\theta) + \cos^2(\theta) = 1

This gives:

\sin^2(\theta) + (-\frac{2}{3})^2 = 1

\sin^2(\theta) + (\frac{4}{9}) = 1

Collect like terms

\sin^2(\theta)  = 1 - \frac{4}{9}

Take LCM and solve

\sin^2(\theta)  = \frac{9 -4}{9}

\sin^2(\theta)  = \frac{5}{9}

Take the square roots of both sides

\sin(\theta)  = \±\frac{\sqrt 5}{3}

Sin is negative in quadrant III. So:

\sin(\theta)  = -\frac{\sqrt 5}{3}

Calculate \csc(\theta)

\csc(\theta) = \frac{1}{\sin(\theta)}

We have: \sin(\theta)  = -\frac{\sqrt 5}{3}

So:

\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}

\csc(\theta) = \frac{-3}{\sqrt 5}

Rationalize

\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}

\csc(\theta) = \frac{-3\sqrt 5}{5}

So, we have:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)

Substitute: \csc(\theta) = \frac{-3\sqrt 5}{5}

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}

Take LCM

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

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7. The probability that a randomly chosen student will be left-handed is .09: a) In a class of 108, find the probability that th
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Solution :

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Using the normal approximation to binomial distribution,

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Using z table,

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Using z table,

= P(z< 0.94)-P(z< 0.60)

= 0.8294 - 0.7257

= 0.1006

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