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Mkey [24]
3 years ago
13

The astronauts from apollo 17 completed 3 space walks while on the moon for a total duration of 22 hour 4 minutes how many minut

es did the astronauts walk in space
Mathematics
1 answer:
Vadim26 [7]3 years ago
7 0

Answer:

The duration of each of the three spacewalk is between 364 minutes and 480 minutes with a total duration of 1324 minutes

Step-by-step explanation:

A spacewalk also termed an EVA or extravehicular activity, is defined as the general activity of an astronaut outside of a vehicle in space

Spacewalks can last up to 5 to 8 hours

Here we have a word problem, and are required to find the number of minutes equivalent to 22 hour and 4 minutes

Number of minutes per hour = 60 minutes

Number of minutes in 22 hours = 60 × 22 = 1320 minutes

Total number of minutes in 22 hours 4 minutes = 1320 + 4 = 1324 minutes.

Whereby the duration of a space walk is up to 5 to 8 hours gives;

22 hours 4 minutes - 8 hour - 8 hour -x = 0

Hence x = 6 hour 4 minutes

Therefore, the duration of each spacewalk is between 6 hour 4 minutes and 8 hours or in minutes we have

The duration of each spacewalk is between  (6 × 60 + 4) 364 minutes and 480 minutes.

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n a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of inches and a stand
kotykmax [81]

Answer:

(a) The probability that a study participant has a height that is less than 67 inches is 0.4013.

(b) The probability that a study participant has a height that is between 67 and 71 inches is 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is 0.0401.

(d) The event in part (c) is an unusual event.

Step-by-step explanation:

<u>The complete question is:</u> In a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches. A study participant is randomly selected. Complete parts​ (a) through​ (d) below. ​(a) Find the probability that a study participant has a height that is less than 67 inches. The probability that the study participant selected at random is less than inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(b) Find the probability that a study participant has a height that is between 67 and 71 inches. The probability that the study participant selected at random is between and inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(c) Find the probability that a study participant has a height that is more than 71 inches. The probability that the study participant selected at random is more than inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.

We are given that the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches.

Let X = <u><em>the heights of men in the​ 20-29 age group</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean height = 67.5 inches

            \sigma = standard deviation = 2 inches

So, X ~ Normal(\mu=67.5, \sigma^{2}=2^{2})

(a) The probability that a study participant has a height that is less than 67 inches is given by = P(X < 67 inches)

 

      P(X < 67 inches) = P( \frac{X-\mu}{\sigma} < \frac{67-67.5}{2} ) = P(Z < -0.25) = 1 - P(Z \leq 0.25)

                                                                 = 1 - 0.5987 = <u>0.4013</u>

The above probability is calculated by looking at the value of x = 0.25 in the z table which has an area of 0.5987.

(b) The probability that a study participant has a height that is between 67 and 71 inches is given by = P(67 inches < X < 71 inches)

    P(67 inches < X < 71 inches) = P(X < 71 inches) - P(X \leq 67 inches)

    P(X < 71 inches) = P( \frac{X-\mu}{\sigma} < \frac{71-67.5}{2} ) = P(Z < 1.75) = 0.9599

    P(X \leq 67 inches) = P( \frac{X-\mu}{\sigma} \leq \frac{67-67.5}{2} ) = P(Z \leq -0.25) = 1 - P(Z < 0.25)

                                                                = 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 1.75 and x = 0.25 in the z table which has an area of 0.9599 and 0.5987 respectively.

Therefore, P(67 inches < X < 71 inches) = 0.9599 - 0.4013 = <u>0.5586</u>.

(c) The probability that a study participant has a height that is more than 71 inches is given by = P(X > 71 inches)

 

      P(X > 71 inches) = P( \frac{X-\mu}{\sigma} > \frac{71-67.5}{2} ) = P(Z > 1.75) = 1 - P(Z \leq 1.75)

                                                                 = 1 - 0.9599 = <u>0.0401</u>

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.9599.

(d) The event in part (c) is an unusual event because the probability that a study participant has a height that is more than 71 inches is less than 0.05.

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