Marcel is nearing graduation. He just met with his counselor and she showed him this information.
2 answers:
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Answer:
(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.
(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.
(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.
Step-by-step explanation:
The complete question is:
A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is
(a) exactly 5
(b) 2 or fewer
(c) more than 1.
Let <em>X</em> = number of unspoiled eggs in the bad carton of eggs.
Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.
The probability of selecting an unspoiled egg is:
A randomly selected egg is unspoiled or not is independent of the others.
It is provided that a chef picks 5 eggs at random.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.
The success is defined as the selection of an unspoiled egg.
The probability mass function of <em>X</em> is given by:
(a)
Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:
Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.
(b)
Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.
(c)
Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:
P (X > 1) = 1 - P (X ≤ 1)
= 1 - P (X = 0) - P (X = 1)
Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.
Answer:
1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.
2) See attached picture.
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
Step-by-step explanation:
1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:
and we set the first derivative equal to zero so we get:
95-18t=0
and solve for t
-18t=-95
t=5.28s
so now we calculate the position of the car after 5.28 seconds, so we get:
so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.
2) For answer 2 I take the second derivative of the function so I get:
j"(t)=-18
and then we graph them. (See attached picture)
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
Answer:
77.76 times
Step-by-step explanation:
The average distance of Neptune from the sun
= 4.503 × 10 ⁹ k m .
and Mercury = 5.791 × 10 ⁷ k m .
Hence neptune is ( 4.503 × 10 ⁹) ÷ (5.791 × 10 ⁷ ) times farther from the sun than mercury
i.e.( ) × 10⁹⁻⁷ times
= 0.7776 × 10 ² times
= 77.76 times.