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2 years ago

14

If $1.00 at 5.5% interest compounded daily has a value of $1.00513 after 34 days, how much interest would Bonnie’s account have

earned on $4,000 after 34 days? (8 points)

1 answer:

ivolga24 [154]2 years ago

6 0

Answer:

$4020.52

Step-by-step explanation:

The future value is the principal multiplied by the amount $1.00 is worth in the future.

$4,000 × 1.00513 = $4020.52

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Triangle ABC has perimeter 22 cm.

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Answer: yes it is a right angle triangle

Step-by-step explanation:

a^2 + b^2 = c^2

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3 years ago

Read 2 more answers

If data for a time series analysis are collected on an annual basis only, which pattern can be ignored? a. horizontal b. seasona

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If data for a time series analysis are collected on an annual basis only, then the option(b) seasonal pattern can be ignored

Time series analysis is a specific way of analyzing a sequence of data points collected over an interval of time

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A horizontal pattern exists when the data fluctuate randomly around a constant mean over time.

A trend pattern exists when there is a long-term increase or decrease in the series.

The cyclical component of a time series refers to fluctuations around the trend, excluding the irregular component, revealing a succession of phases of expansion and contraction

Here, the data for a time series analysis are collected on an annual basis, so the seasonal pattern can be ignored

Learn more about Time series analysis here

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2 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

mojhsa [17]

Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, <em>MOE</em> = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of <em>z</em> for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, <em>MOE</em> = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of <em>z</em> for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

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3 years ago