Solution and Explanation:
Step 1: Start Access. Open the downloaded Access file named exploring_a03_Grader_a1.accdb.
Step 2: Assume that there is a table Loans with the following attributes as shown in the screenshot.
Step 3: Create a query using Query Design. From the Clients table, display the client’s FirstName and LastName. From the Accounts table, select the Savings Balance and OpenDate. Sort the query by savings balance in descending order.
Add a calculated field named AccountTime that calculates the number of days each client’s accounts have been open. Assume today’s date is 12/31/2017. Recall dates must be enclosed in # to denote to Access it is a date. Format the results in General Number format. Save the query as Account Longevity, and close the query.
Step 4: Create a query using Query Design. From the Clients table, display the client first name and last name. From the Accounts table, select the savings balance.
Add appropriate grouping, so the client’s total retirement account savings balances are displayed. Add a sort so the highest total savings balances are displayed first.
Step 5: Switch to Datasheet view. Add a totals row displaying the count of the last name and the average of total savings balances. Save the query as Total Balances By Client and close the query.
Step 6: Create a copy of the Total Balances By Client query. Name the query Total Balances By State. Open the query in Design view and remove the client name from the query. Add grouping by the client’s state.
Sort by the client’s state in Ascending order and remove the sort on the savings balance. Add criteria so clients with retirement account savings balances of $10,000 or more are factored in to the query. Save and close the query.
Step 7: Create a new query using Query Design. From the Clients table, select the client first name, last name, and state. From the Accounts table, select the Savings Balance. Add criteria so only customers with balances under $15,000 are displayed.
Step 8: Enter the sample data (one record) as shown in the screenshot.
As present value is given as 25000, loan amount is taken as 25000.
As savings balance is given as 5000, savings balance is taken as 5000.
Step 9: SS
Step 10: Then an expression builder is opened as shown in the screenshot:
Step 11: Then enter the expression pmt(0.5/12, 2*12, - (Loan Amount] – [Savings balance]), 0, 0) as shown in the screenshot.
Answer:
The Break Even Point is the Sales Value that will cover the cost of production. Meaning the Sales Value that will bring profitability to Zero
Break Even sales for Company wide = $378,000
Break Even Value for Chicago is $111,429
And Break Even Value for Minneapolis is $120,000
The Addition of both Outlets/Offices Break Even Sales is less than the Company-wide because the Offices don't share in the Common Fixed Expense as these are specific to Group reporting.
Explanation:
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 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 because the load is known with an uncertainty of ±10.
Looking at the uncertainty; the diameter must be raised to because the diameter is known within an uncertainty of ±5.
The decrease in the maximum allowable stress can be estimated as:
where,
stress
P = load
A = cross-sectional area of the cylinder
∴
replacing P' with and d' with
, we have:
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:
where;
the design factor =
the design factor = 1.434.
Thus, the minimum design factor that will guarantee that the part will not fail. = 1.434