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2 years ago

Richard has a debt of $45. Express it in the form of absolute value.

1 answer:

6 0

The absolute value of debt of $45 is 45 because a balance of $0 on an account means there is no debt, and a negative of 45 value represents the debt.

It is defined as the amount one party needs to pay to another party as the first party borrowed an amount that will be credited by the second party. Debt occurs when one party cannot be able to purchase something under normal circumstances.

As we know,

The absolute value of a negative account balance is equal to the amount of debt. A balance of $0 on an account means there is no debt. Credit is shown by a positive account balance.

Richard has a debt of $45.

The negative of 45 value represents the debt.

-$45 = debt

Absolute value = |-45| = 45

Thus, the absolute value of debt of $45 is 45 because a balance of $0 on an account means there is no debt, and a negative of 45 value represents the debt.

Learn more about the debt here:

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Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%

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Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

$\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7$

$s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} } = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702$

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

$t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262$

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}$

$=176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)$

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

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