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Harold used a 3% iodine solution and a 20% iodine solution to make a 95- ounce solution in which was 19% iodine. How many ounces of the 3% iodine solution did he use?
Let x = the amount of 3% iodine solution. Let y = the amount of 20% iodine solution. Because the total amount of solution was 85 oz., then x + y = 85. The amount of each kind of solution added together and set equal to the amount of 19% solution could be expressed in the equation 0.03x + 0.20y = 0.19(85); Use both equations to solve for x. Multiply the second equation by 100 to remove the decimal point: 3x + 20y = 19(85). Simplify that equation: 3x + 20y = 1805. Multiply the ?rst equation by -20: -20x + -20y = -1700. Add the two equations to eliminate y: -17x + 0y = -85. Divide both sides of the equation by -17: -17x/-17 -85/-17 = x = 5. The amount of 3% iodine solution is 5 ounces.
Test of hypothesis about the population mean When the population standard deviation (S) is identified then the t statistic is defined as t = ¦(x¯ - µ)/ S x¯ ¦
128sinpower8=cos8-8cos6+28cos4-56cos2+35
Ten is decreased through four times the quantity of eight minus three. One is then added to in which result. What is the final answer? The area of a square whose side measures
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There are really three various methods for doing such integral. Method 1: This method uses a trig formula as, ∫sin(x) cos(x) dx = ½ ∫sin(2x) dx = -(1/4) cos(2x) + c
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