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Illusion [34]
3 years ago
10

The automatic opening device of a military cargo parachute has been designed to open when the parachute is 200 m above the groun

d. Suppose opening altitude actually has a normal distribution with mean value 200 and standard deviation 34 m. Equipment damage will occur if the parachute opens at an altitude of less than 100 m. What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes? (Round your answer to four decimal places.)
Mathematics
1 answer:
nydimaria [60]3 years ago
6 0

Answer: 0.0082

Step-by-step explanation:

Let x be the random variable that represents the opening altitude.

We assume that opening altitude actually has a normal distribution .

As per given , we have

\mu=200 \ \ \sigma=34

Equipment damage will occur if the parachute opens at an altitude of less than 100 m.

z-score corresponds to x=100, z=\dfrac{x-\mu}{\sigma}

i.e.  z=\dfrac{100-200}{34}\approx-2.94

P-value = P(z

Let Y be a binomial variable that represents the opening altitude.

with parameters p=0.0016411 , n= 5

P(x)=^nC_xp^x(1-p)^{n-x}

Required probability :

P(x\geq1)=1-P(x

Hence, the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes = 0.0082

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4 years ago
Please Help with math word problem.
motikmotik

Answer:

Adult Tickets: 173

Student Tickets: 43

Explanation:

  • To solve this problem, you'll need to set up a system of equations.
  • Assume a = # of adult tickets sold
  • Assume s = # of student tickets sold

2: a + s = 216; 1: 10.25a+ 8s= 2117.25

2: <<As the total number of tickets sold from both sides is equal to 216>>

1: <<Each adults ticket (a) costs $10.25 and each student ticket (s) costs $8, and the total amount of money earned (2117.25) from sales is the combination of these two))>>

  • Note that there are two ways to solve systems of equations (by elimnation and substitution), in this case I'll use elimnation as substitution requires one of the variables in one of the two equations to be isolated.

  • In this case, I'll elimnate a.

a + s = 216

10.25a + 8s = 2117.25

  • In order to elimnate a, it has to be equal to - 10.25 so that it cancels out + 10.25 (so you have to multiply everything on the first equation by 10.25 ((what you do to one part, you'll do to all the other parts)).

-10.25a -10.25s = -2214

10.25a + 8s = 2117.25

  • a cancels, and now you solve accordingly.

-2.25s/-2.25 = -96.75/-2.25

s = 43

  • You could solve for a using this same method, but it's easier to use the first formula <<a+s=216>> to find a.

a + 43 = 216

- 43 -43

a = 173

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