Answer:
16. 137
17. 89
18. 168
19. 50
20. 8
21. 96
22. 39
23. 5
24. -2
25. 7
Step-by-step explanation:
We have to follow BODMAS which is the acronym for Bracket, Of, Division, Multiplication and Subtraction.
We have to perform our calculations in this order. i.e., Solve the calculations in a bracket first then of and so on.
a = 12; b = 9; c = 4
16. a² + b - c² = (12)² + 9 - 4² = 144 + 9 - 16 = 153 - 16 = 137
17. b² + 2a - c² = 81 + 2(12) - 16 = 81 + 24 - 16 = 105 - 16 = 89
18. 2c(a + b) = 2 · 4 (12 + 9) = 8(21) = 168
19. 4a + 2b - c² = 4(12) + 2(9) - 4² = (48 + 18 - 16 = 66 - 16 = 50
20. [a² ÷ (4b)] + c = [12² ÷ 4(9)] + 4 = [144 ÷ 36] + 4 = 4 + 4 = 8
21. c²(2b - a) = 4²(2(9) - 12) = 4²(18 - 12) = 16(6) = 96
22. [bc² + a] ÷ c = [9(4²) + 12] ÷ 4 = [9(16) + 12] ÷ 4 = 156 ÷ 4 = 39
23. [2c³ - ab] ÷ 4 = [2(4)³ - 12(9)] ÷ 4 = [2(64) - 108] ÷ 4
= [128 - 108] ÷ 4 = 20 ÷ 4 = 5
24. 2(a - b)² - 5c = 2(12 - 9)² - 5(4) = 2(3)² - 20 = 18 - 20 = -2
25. [b² - 2c²] ÷ [a + c - b]
= [9² - 2(4)²] ÷ [12 + 4 - 9] = [81 - 32] ÷ [16 - 9]
= 49 ÷ 7 = 7
One worker<span> produces an average of 84 units per </span>day<span> with a street </span>What is the probability<span> that in any </span>single day worker 1 will outproduce worker 2<span>? A) 0.1141.
</span>
Answer, factory worker productivity<span> is </span>normally distributed<span>. </span>One worker produces<span> an </span>average<span> of 75 </span>units per day<span> with a standar, day with a </span>standard deviation<span> of 20. </span>Another worker produces<span> at an </span>average rate<span> of 65 </span><span>per day.
</span>
Answer:
12.5 x 24=300
Step-by-step explanation:
