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3 years ago

5

You deposit $250 in a bank account that pays an annual interest rate of 2%. How much simple interest will you earn after two yea

rs?

1 answer:

Elanso [62]3 years ago

7 0

I believe the answer is $260.1 and an interest of $10.1

250(1.02)=255 --> one year
255(1.02)=260.1 --> two years

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A gallon of gas cost $2.16. Rob buys 13.5 gallons of gas. How much did he pay?

pashok25 [27]

Answer:

$29.16

Step-by-step explanation:

Each gallon of gas is $2.16.

13.5 gallons of gas is 2.16 (13.5) = $29.16.

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3 years ago

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There are 10 yellow, 6 green, 9 orange, and 5 red cards in a stack of cards turned facedown. Once a card is selected, it is not

Lilit [14]

Answer:

9/29

Step-by-step explanation:

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3 years ago

Which is the coefficient in the expression 7x + 14?

guajiro [1.7K]

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3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Korolek [52]

Answer:

A sample of 499 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 given by:

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

In this question, we have that:

$\pi = 0.21$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 90% confidence level with an error of at most 0.03

We need a sample of n, which is found when $M = 0.03$ . So

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

$0.03 = 1.645\sqrt{\frac{0.21*0.79}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.21*0.79}$

$\sqrt{n} = \frac{1.645\sqrt{0.21*0.79}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.21*0.79}}{0.03})^2$

$n = 498.81$

Rounding up

A sample of 499 is needed.

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2 years ago

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3 years ago