<span>The answer would be:
B. The frequency is LESS THAN OR EQUAL TO the cumulative frequency.
</span>
Answer: 490 grams of the first alloy should be used.
30 grams of the second alloy should be used.
Step-by-step explanation:
Let x represent the weight of the first alloy in grams that should be used.
Let y represent the weight of the second alloy in grams that should be used.
A chemist has two alloys, one of which is 15% gold and 20% lead. This means that the amount of gold and lead in the first alloy is
0.15x and 0.2x
The second alloy contains 30% gold and 50% lead. This means that the amount of gold and lead in the second alloy is
0.3y and 0.5y
If the alloy to be made contains 82.5 g of gold, it means that
0.15x + 0.3y = 82.5 - - - - - - - - - - - -1
The second alloy would also contain 113 g of lead. This means that
0.2x + 0.5y = 113 - - - - - - - - - - - - -2
Multiplying equation 1 by 0.2 and equation 2 by 0.15, it becomes
0.03x + 0.06y = 16.5
0.03x + 0.075y = 16.95
Subtracting, it becomes
- 0.015y = - 0.45
y = - 0.45/- 0.015
y = 30
Substituting y = 30 into equation 1, it becomes
0.15x + 0.3 × 30 = 82.5
0.15x + 9 = 82.5
0.15x = 82.5 - 9 = 73.5
x = 73.5/0.15
x = 490
Mc012-2.jpg ........................ 10/90%= 11.1111111111111
Slope= (y2-y1)/(x2-x1)= (-3-(-1))/(0-4)= -2/-4 =2/4 =1/2
Answer: slope =1/2.
2x + 100 = 5x + 55
3x = 45
x = 15
2(15) + 100 = 130
180-130 = 50
so angle 2 is 50 degrees
