1. Motherboard (you plug all other components to it)
2. Processor (CPU)
3. Memory (RAM)
4. Graphics card
5. Sound card (sometimes integrated with motherboard)
6. Hard Disk Drive (HDD)
<span>7. Power supply unit</span>
Answer:
b. instantiated.
Explanation:
In computer programming, to be instatiated or when variables are instatiated; this simply refers to the acts of making example out of a group variables or a form of template. For example in a situation of class of objects.
Hence, in this case, the correct answer to the question above is the option B, in which Variables set equal to patterns are said to be INSTATIATED
Answer:
The procedure in SQL is created as follows
Explanation:
--creating the procedure
CREATE OR REPLACE PROCEDURE prc_inv_amounts (W_IN IN NUMBER)
AS
--defining variables
W_CK NUMBER := 0;
W_SUBT NUMBER := 0;
W_TAX NUMBER := 0;
BEGIN
--Authentication process of the invoice
SELECT COUNT(*) INTO W_CK FROM INVOICE WHERE INV_NUMBER := W_IN;
--Transaction confirmation
IF W_CK = 1 THEN
SELECT SUM(LINE_TOTAL) INTO W_SUBT FROM LINE
WHERE
--checking the invoice for the desired invoice number
LINE.INV_NUMBER = W_IN;
W_TAX :=W_SUBT * 0.08;
--updating the invoice
UPDATE INVOICE
--setting the new values
SET INV_SUBTOTAL = W_SUBT,
INV_TAX = W_TAX,
INV_TOTAL =W_SUBT + W_TAX
WHERE INV_NUMBER = W_IN;
--ending the if statement
END IF;
--ending the procedure
END;
Death by PowerPoint simply refers to a phenomenon that's caused as a result of the poor use of a presentation software.
It should be noted that the phrase was coined by Angela Garber. It simply means when an individual uses the presentation software poorly.
The main contributors to death by PowerPoint include confusing graphics, slides that had too much text, etc. It's important to present one's work logically with good information.
Learn more about PowerPoint on:
brainly.com/question/25795152
Complete Question:
Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 4096-byte cache using 32-bit memory addresses, 8-byte cache blocks and a 8-way associative design. The cache has :
Cache size = 1024 bytes, sets t = 26.8, tag bits, s = 3.2, set index bit =2
Answer:
Check below for explanations
Explanation:
Cache size = 4096 bytes = 2¹² bytes
Memory address bit = 32
Block size = 8 bytes = 2³ bytes
Cache line = (cache size)/(Block size)
Cache line = 
Cache line = 2⁹
Block offset = 3 (From 2³)
Tag = (Memory address bit - block offset - Cache line bit)
Tag = (32 - 3 - 9)
Tag = 20
Total number of sets = 2⁹ = 512
