A certain candle is designed to last nine hours. However, depending on the wind, air bubbles in the wax, the quality of the wax, and the number of times the candle is re-lit, the actual burning time (in hours) is a uniform random variable with a = 5.5 and b = 9.5.
Suppose one of these candles is randomly selected.
(a) Find the probability that the candle burns at least six hours.
(b) Find the probability that the candle burns at most seven hours.
(c) Find the mean burning time. Find the probability that the burning time of a randomly selected candle will be within one standard deviation of the mean. (Round your answer to four decimal places.)
(d) Find a time t such that 25% of all candles burn longer than t hours.

Step-by-step explanation:

a) P(X > 6) = (9.5-6)/(9.5-5.5) = 3.5/4 = 0.875

b) P(X < 7) = (7-5.5)/(9.5-5.5) = 1.5/4 = 0.375

c) E(X) = (9.5+5.5)/2 = 7.5

Standard deviation = (9.5-5.5)/sqrt(12) =4/3.46 = 1.154

P= 1.156*2/(9.5-5.5) = 2.308/4 = 0.577

d) P(X > t) = 0.25

 (9.5-t) /(9.5-5.5) = 0.3

9.5-t = 1.2

t = 9.5-1.2 = 8.3

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40