Answer:
65.3658 inches
Step-by-step explanation:
Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value 
such that 
, but, 
, i.e., 
is the 76th percentile of the standard normal distribution. So, 
, 
. Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.
12 ft * 10 ft = 120 sq ft
½ * 6 ft * 7 ft = 21 sq ft
120 sq ft - 21 sq ft = 99 sq ft (it is area of the composite shape)
Answer:
ano ba gagawin
parang Ang Dali pero ano gagawin dyan
<span>Solving the equation for Y = 1.
5 * (3 * 1 - 2) = 15 * 1 - 10
5 * (3 - 2) = 15 -10
5 * 1 = 5
5 = 5
Solving the equation for Y = 2.
5 * (3 * 2 - 2) = 15 * 2 - 10
5 * (6 - 2) = 30 -10
5 * 4 = 20
20 = 20
Solving the equation for Y = 4.
5 * (3 * 4 - 2) = 15 * 4 - 10
5 * (12 - 2) = 60 -10
5 * 10 = 50
50 = 50
Solving the equation for Y = 8.
5 * (3 * 8 - 2) = 15 * 8 - 10
5 * (24 - 2) = 120 -10
5 * 22 = 110
110 = 110
Solving the equation for Y = 9.
5 * (3 * 9 - 2) = 15 * 9 - 10
5 * (27 - 2) = 135 -10
5 * 25 = 125
125 = 125
This proves that the equation holds good for at least 5 values of 'y', which are 1, 2, 4, 8 and 9.
However, it can be proved that the equation holds good for any value of y.
Expression 5(3y-2) can be simplified to 15y -10 which is the same expression on the right had side of the equation provided.
So, equation 5(3y-2)=15y-10 is actually 15y-10=15y-10 and since this is true for all values of y, it has been proved that it is true for at least 5 values of y.</span>
