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C-HRHFC-2405 Pass Exam & Hot C-HRHFC-2405 Questions - SAP Certified Associate - Implementation Consultant - SAP SuccessFactors Full Cloud/Core Hybrid Materials - Sugakumaster

C-HRHFC-2405

Exam Code: C-HRHFC-2405

Exam Name: SAP Certified Associate - Implementation Consultant - SAP SuccessFactors Full Cloud/Core HybridCertification

Version: V16.75

Q & A: 400 Questions and Answers

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NEW QUESTION: 1
Given the code fragment:

Assume that dbURL, userName, and password are valid.
Which code fragment can be inserted at line n1 to enable the code to print Connection Established?
A. con = DriverManager.getConnection (dbURL);con.setClientInfo ("user", userName);con.setClientInfo
("password", password);
B. con = DriverManager.getConnection (userName, password, dbURL);
C. Properties prop = new Properties();prop.put ("userid", userName);prop.put ("password",
password);prop.put("url", dbURL);con = DriverManager.getConnection (prop);
D. Properties prop = new Properties();prop.put ("user", userName);prop.put ("password", password);con =
DriverManager.getConnection (dbURL, prop);
Answer: D

NEW QUESTION: 2
Risoner Company plans to purchase a machine with the following conditions:
Purchase price = US $300,000
The down payment = 10% of purchase price with remainder financed at an annual interest rate of 16%. The financing period is 8 years with equal annual payments made every year.
The present value of an annuity of US $1 per year for 8 years at 16% is 4.3436.
The present value of US $1 due at the end of 8 years at 16% is 3050
The annual payment rounded is:
A. US $43,200
B. US $82.350
C. US $39,150
D. US $62,160
Answer: D
Explanation:
The periodic payment is found by dividing the amount to be accumulated US $300,000 price - $30.000 down payment = US $270,000) by the interest factor for the present value of an ordinary annuity for 8 years at 16%. Consequently, the payment is US $62.160$270,000 - 4.3436).

NEW QUESTION: 3
A customer wants to reproduce the following SQL statement in SAP HANA:
SELECT P.PROD_NAME, P.PROD_ID, PT.PROD_DESC FROM PRODUCTS P, PROD_TEXTS PT WHERE P.PROD_ID=PT.PROD_ID;
Which of the following SQLScript constructs is equivalent to this SQL statement and yields the best performance?
A. ot_products= CE_COLUMN_TABLE(PRODUCTS);
ot_prodtexts= CE_COLUMN_TABLE(PROD_TEXTS);
ot_all=CE_JOIN(:ot_products, :ot_prodtexts, [PROD_ID]);
output=CE_PROJECTION(:ot_all, [PROD_NAME, PROD_ID, PROD_DESC]);
B. SELECT P.PROD_NAME, P.PROD_ID, PT.PROD_DESC FROM PRODUCTS P JOIN
PROD_TEXTS PT ON P.PROD_ID=PT.PROD_ID;
C. ot_products= CE_COLUMN_TABLE(PRODUCTS);
ot_prodtexts= CE_COLUMN_TABLE(PROD_TEXTS);
ot_all=CE_UNION_ALL(:ot_products, :ot_prodtexts);
output=CE_PROJECTION(:ot_all, [PROD_NAME, PROD_ID, PROD_DESC]);
D. SELECT P.PROD_NAME, P.PROD_ID, NULL as PROD_DESC FROM PRODUCTS P
UNION SELECT NULL as PROD_NAME, PT.PROD_ID, PT.PROD_TEXTS FROM
PROD_TEXTS PT;
Answer: A

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