Measurement error that is continuous and uniformly distributed from millivolts is added to the true voltage of a circuit. then the measurement is rounded to the nearest millivolt so that it becomes discrete. suppose that the true voltage is 219 millivolts. what is the mean and variance of the measured voltage

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Measurement error that is continuous and uniformly distributed from –3 to +3 millivolts is added to a circuit’s true voltage. Then the measurement is rounded to the nearest millivolt so that it becomes discrete. Suppose that the true voltage is 219 millivolts. What is the mean and variance of the measured voltage

Mean = 219

Variance = 4

Step-by-step explanation:

Given

Let X be a random variable measurement error.

X has a discrete uniform distribution as follows

a = 219 - 3 = 216

b = 219 + 3 = 222

Mean or Expected value is calculated as follows;

E(x) = ½(216+222)

E(x) = ½ * 438

E(x) = 219

Variance is calculated as follows;

Var(x) = ((b-a+1)²-1)/12

Var(x) = ((222-216+1)²-1)/12

Var(x) = (7²-1)/12

Var(x) = 48/12

Var(x) = 4

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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