The following pseudocode algorithm uses nested IF statements. ``` IF Response = 1 THEN X ← X + Y ELSE IF Response = 2 THEN X ← X - Y ELSE IF Response = 3 THEN X ← X * Y ELSE IF Response = 4 THEN X ← X / Y ELSE OUTPUT "No response" ENDIF ENDIF ENDIF ENDIF ``` (a) Name the type of statement demonstrated by the use of IF ... THEN ... ELSE ... ENDIF \_\_\_\_ \_\_\_\_ [1] (b) Re-write the pseudocode algorithm using a CASE statement. \_\_\_\_ \_\_\_\_ \_\_\_\_ \_\_\_\_ \_\_\_\_ \_\_\_\_ \_\_\_\_ \_\_\_\_ \_\_\_\_ \_\_\_\_ [4]

2019 - O/N - 0478/23 | Q4

0/7

0 correct · 0 wrong

The following pseudocode algorithm uses nested IF stateme...