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scoray [572]
3 years ago
14

The automatic opening device of a military cargo parachute has been designed to open when the parachute is 135 m above the groun

d. Suppose opening altitude actually has a normal distribution with mean value 135 and standard deviation 35 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? (Give your answer to four decimal places.)
Mathematics
1 answer:
laiz [17]3 years ago
8 0

Answer:

the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes is 0.4215

Step-by-step explanation:

Let consider Q to be the opening altitude.

The mean μ = 135 m

The standard deviation = 35 m

The probability that the equipment damage will occur if the parachute opens at an altitude of less than 100 m can be computed as follows:

P(Q

P(Q

P(Q

P(Q

If we represent R to be the number of parachutes which have equipment damage to the payload out of 5 parachutes dropped.

The probability of success = 0.1587

the number of independent parachute n = 5

the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes can be computed as:

P(R ≥ 1) = 1 - P(R < 1)

P(R ≥ 1) = 1 - P(R = 0)

The probability mass function of the binomial expression is:

P(R ≥ 1) = 1 - (^5_0)(0.1587)^0(1-0.1587)^{5-0}

P(R ≥ 1) =1 - (\dfrac{5!}{(5-0)!})(0.1587)^0(1-0.1587)^{5-0}

P(R ≥ 1) = 1 - 0.5785

P(R ≥ 1) = 0.4215

Hence, the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes is 0.4215

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oee [108]
8/2.79 = 1/2.867, or $2.86 per each oz
15/5.54 = 2.707/1, or 2 oz for every 1 dollar

the 15 oz jar of salsa has the better unit rate

hope this helps
7 0
3 years ago
Jason is mikes older brother. the sum of their ages is 25. the differnece in their ages is 7 write the system of equation that r
Jet001 [13]

Answer:

Equation 1: m+j=25

Equation 2: j=m+5


I hope this helps! Good luck! <3


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5 0
3 years ago
The cost of downloading songs from iTunes can be represented by the expression 1.29s, where s is the number
FromTheMoon [43]
The answer is A trust me
6 0
3 years ago
Read 2 more answers
Find the probability of at least 6 failures in 7 trials of a binomial experiment in which the probability of success in any one
ziro4ka [17]

The probability of at least 6 failures in 7 trials of a binomial experiment in which the probability of success in any one trial is 9% will be <u>0.000343311 %</u>

<h3>What is the probability?</h3>

Probability is synonymous with possibility. It is concerned with the occurrence of a random event.

Probability can only have a value between 0 and 1. Its simple notion is that something is very likely to occur. It is the proportion of favorable events to the total number of events.

No of failure,n = 7

No of trials,x≥6

A binomial probability is represented as;

\rm P(x) = nC_x p^z(1-p)^{n-r}

Substitute the given data;

P(x \geq 6) = 7C_6 \times 9^6 \times 91-9)^{7-6} +7C_7 (9)^7+(1-9)^{(7-7)}

\rm P(x \geq 6)  = 0.000000343311 \\\\ P(x \geq6)=0.000343311 \%

Hence,the probability of at least 6 failures in 7 trials of a binomial experiment in which the probability of success in any one trial is 9% will be <u>0.000343311 %</u>

To learn more about probability, refer to the link: brainly.com/question/795909.

#SPJ1

4 0
2 years ago
The breaking strengths of cables produced by a certain manufacturer have a mean, u, of 1750 pounds, and a standard deviation of
AlladinOne [14]

Answer:

We accept the null hypothesis that the  breaking strength mean is less and equal to 1750 pounds and has not increased.

Step-by-step explanation:

The null and alternative hypotheses are stated as

H0:  u ≥ 1750   i.e the mean is less and equal to 1750

against the claim

Ha: u > 1750  ( one tailed test)  the mean is greater than 1750

Sample mean = x`= 1754

Population mean = u = 1750

Population deviation= σ = 65 pounds

Sample size= n = 100

Applying the Z test

z= x`- u / σ/ √n

z= 1754- 1750 / 65/ √100

z= 4/6.5

z= 0.6154

The significance level alpha = 0.1

The z - value at 0.1 for one tailed test is ± 1.28

The critical value is z > z∝.

so

0.6154 is < 1.28

We accept the null hypothesis that the  breaking strength mean is less and equal to 1750 pounds and has not increased.

8 0
3 years ago
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