All categories

3 years ago

In two to three sentences, describe how you would use the AutoFill shortcut.

1 answer:

4 0

When you fill in data based on adjacent cells by dragging the fill handle in Microsoft Excel, the Auto Fill Options<span> button is displayed just below and to the right of your filled selection. When you click the</span>
Auto Fill Options<span> button, a list of options appears that allows you to choose to fill the cells with text or data, and whether or not to include the format of the initial selection or to copy only the format.</span>

You might be interested in

Which QuickBooks Online report answers the common business question: "Who owes me money?"

ki77a [65]

Answer:

Accounts Receivable Aging report.

Explanation:

Because you're receiving money.

8 0

3 years ago

When the economy is on the short-run aggregate supply curve and to the left of the long-run aggregate supply curve, actual aggre

nirvana33 [79]

Answer:

Nominal wages will fall, and the short-run aggregate, supply curve, shifts to the right.

Explanation:

When the economy is on the short-run aggregate supply curve and to the left of the long-run aggregate supply curve, actual aggregate output will eventually equal potential output as nominal wages fall(s) and the short-run aggregate supply curve shifts to the right.

5 0

3 years ago

When a company pays a dividend, it isn't as simple as getting a paycheck from one's employer. There are several critical dates i

belka [17]

Answer: 1. Declaration Date

2. Payment Date

3. Holder-of-record date

4. Ex-dividend date

Explanation:

1. On the Declaration Date, the company's Director announces that they will pay a dividend as well as the amount of the dividend. This is recorded in the books by crediting it to Dividends payable.

2. On Payment day the dividends are disbursed amongst shareholders. Cash Account is credited and Dividends Payable is debited.

3. The Holder-of-record day is the day the company notes who the owners of it's stock are so that they may receive the dividend.

4. On the Ex-dividend date which is usually 2 days before the record date, any stock bought on or after this date will.not receive any Dividend payment.

6 0

3 years ago

Which part of the Business Cycle would this graph represent?

Archy [21]

bust is the correct answer

3 0

3 years ago

A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of pya, where a 5 pd2y4. if the load is kno

raketka [301]

Here is the correct question.

A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of P/A, where A= πd²/4. if the load is known with an uncertainty of ±10 percent, the diameter is known within ±5 percent (tolerances), and the stress that causes failure (strength) is known within ±15 percent, determine the minimum design factor that will guarantee that the part will not fail.

Answer:

the minimum design factor that will guarantee that the part will not fail. = 1.434

Explanation:

Looking at the uncertainty; loss of strength must be raised to $\dfrac{1}{0.85}$ due to the stress that causes the failure (strength) is known within ±15% uncertainty.

Looking at the uncertainty; the maximum allowable load must be reduced to $\dfrac{1}{1.1}$ because the load is known with an uncertainty of ±10.

Looking at the uncertainty; the diameter must be raised to $\dfrac{1}{0.95}$ because the diameter is known within an uncertainty of ±5.

The decrease in the maximum allowable stress can be estimated as:

$\sigma' = \dfrac{P'}{A'}$

where,

$\sigma =$ stress

P = load

A = cross-sectional area of the cylinder

∴

$\sigma' = \dfrac{P'}{\dfrac{\pi}{4}(d')^2}$

replacing P' with $\dfrac{1}{1.1}P$ and d' with $\dfrac{1}{0.95}d$ , we have:

$\sigma' = \dfrac{(\dfrac{1}{1.1})\times p }{\dfrac{\pi}{4}(\dfrac{1}{0.95 } d)^2 }$

$\sigma' =\dfrac{P}{\dfrac{\pi}{4}d^2} (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }$

$\sigma' =\sigma \times (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }$

$\sigma' =\sigma \times 0.82045$

$\dfrac{\sigma' }{\sigma } =0.82045$

Thus, the uncertainty in diameter and the load of the allowable stress needs to decrease to 0.82045

Now, the minimum design factor that will ascertain that the part will not fail can be computed as:

$n_d = \dfrac{loss \ of \ function \ parameter }{maximum \ allowable \ parameter}$

where;

the design factor = $n_d$

$n_d =\dfrac{\dfrac{1}{0.85} }{0.82045}$

the design factor $n_d$ = 1.434.

Thus, the minimum design factor that will guarantee that the part will not fail. = 1.434

7 0

3 years ago