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2 years ago

Zoey invested $230 in an account paying an interest rate of 6.3% compounded daily.

Mathematics

2 answers:

Lady bird [3.3K]2 years ago

4 0

Answer:

500

Step-by-step explanation:

alukav5142 [94]2 years ago

3 0

Answer:

A ≈ $500

General Formulas and Concepts:

Pre-Alg

Order of Operations: BPEMDAS

Algebra I

Compounded Interest Rate: A = P(1 + r/n)ⁿˣ

A is final amount
P is initial (principle) amount
r is rate
n is number of compounds
x is number of years

Step-by-step explanation:

Step 1: Define

P = 230

r = 0.063

n = 365

x = 12

Step 2: Solve for A

Substitute: A = 230(1 + 0.063/365)³⁶⁵⁽¹²⁾
Divide: A = 230(1 + 0.000173)³⁶⁵⁽¹²⁾
Multiply: A = 230(1 + 0.000173)⁴³⁸⁰
Add: A = 230(1.00017)⁴³⁸⁰
Exponents: A = 230(2.1296)
Multiply: A = 489.808

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2 years ago

A rectangle has a length that is three feet more than twice it’s width. If the area of the rectangle is 90 square feet, then alg

Mkey [24]

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$Length=27feet\\Width=12 feet$

Step-by-step explanation:

Let, the length of the rectangle be 'L'

and, width of the rectangle be 'W'

length=2*Width+3

$L=(2W+3) feet\\\\$

Area of the rectangle= 90 square feet

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Using Factorization Method to solve quadratic equation:

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3 years ago

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro

Gnom [1K]

Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of $\pi$ , so we use the worst case scenario, which is $\pi = 0.5$

Then

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}$

$n = 1067.11$

Rounding up

A sample of 1068 is needed.

8 0

2 years ago