Solve for d: 2d+1/6=6(1/3d+1)

2 answers:

7 0

2d+1/6=2d+6 (I distributed the 6 to what was in the parentheses)
2d-2d=6-1/6 (I subtracted 2d from both sides to get the variables on one side and 1/6 from both sides to get the variables by themselves)
0=35/6 (combined like terms)
0=35/6 is an untrue statement so x doesn't exist

Colt1911 [192]3 years ago

3 0

2d + 1/6 = 6(1/3d + 1)
2d + 1/6 = 2d + 6
2d - 2d = 6 - 1/6
0d = 35/6

$Ans: \O$

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assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the

drek231 [11]

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 $x_{0}$ such that $P(X < x_{0}) = 0.76$ , but, $0.76 = P(X < x_{0}) = P((X-63.6)/2.5 < (x_{0}-63.6)/2.5) = P(Z < (x_{0}-63.6)/2.5)$ , i.e., $(x_{0}-63.6)/2.5$ is the 76th percentile of the standard normal distribution. So, $(x_{0}-63.6)/2.5 = 0.7063$ , $x_{0} =63.6+(2.5)(0.7063) = 65.3658$ . Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.