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Ne4ueva [31]
3 years ago
12

According to a study done by a university​ student, the probability a randomly selected individual will not cover his or her mou

th when sneezing is 0.267 . Suppose you sit on a bench in a mall and observe​ people's habits as they sneeze. ​(a) What is the probability that among 12 randomly observed individuals exactly 8 do not cover their mouth when​ sneezing? ​(b) What is the probability that among 12 randomly observed individuals fewer than 6 do not cover their mouth when​ sneezing? ​(c) Would you be surprised​ if, after observing 12 ​individuals, fewer than half covered their mouth when​ sneezing? Why?
Mathematics
1 answer:
Igoryamba3 years ago
5 0

Answer: kindly check explanation

Step-by-step explanation:

Probability of not covering mouth = p(success) = 0.267

Hence, p(covering mouth) = 1 - 0.267 = 0.733

a) What is the probability that among 12 randomly observed individuals exactly 8 do not cover their mouth when​ sneezing? ​

Number of samples (n) = 12

P(X = 8)

Using binomial distribution :

P(x) = nCx * p^x * (1-p)^(n-x)

P(x =8) = 12C8 * 0.267^8 * 0.733^(12-8)

P(x =8) = 12C8 * 0.267^8 * 0.733^4

= 0.00369

b) What is the probability that among 12 randomly observed individuals fewer than 6 do not cover their mouth when​ sneezing?

P(X < 6) = p(5) + p(4) + p(3) + p(2) + p(1) + p(0)

To save computation time, we can use an online binomial probability calculator :

P(X < 6) = 0.9275

C.) Yes, I will be surprised, because from the binomial probability obtained above, there is a high probability (0.9275) that fewer than half do not cover their mouth.

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An airplane is preparing to land at an airport. It is 44,8000 feet above the ground and is descending at the rate of 3,100 feet
Marina86 [1]

Answer:

Time: 8 minutes

Altitude: 20000ft

Method 1 is easiest

Method 3 is easiest

Step-by-step explanation:

Given

Airplane 1:

Height = 44800 ft

Descending\ Rate = 3100ft/min

Airplane 2:

Ascending\ Rate = 2500ft/min

Required

Determine when both planes would be at the same altitude?

Let the minute be represented by m

For Airplane 1, Its altitude at any height h is:

Airplane\ 1 = Height - Descending\ Rate * m

<em>It is minus (-) because the airplane is descending</em>

Airplane\ 1 = 44800 - 3100 * m

Airplane\ 1 = 44800 - 3100m

For Airplane 2, Its altitude at any height h is:

Airplane\ 2 = Ascending\ Rate * m

Airplane\ 2 = 2500 * m

Airplane\ 2 = 2500m

Method 1:

For both heights to be equal, we have that:

Airplane\ 1 = Airplane\ 2

This gives:

44800 - 3100m = 2500m

Collect Like Terms

44800 = 2500m + 3100m

44800 = 5600m

5600m = 44800

m = 44800/5600

m = 8\ min

<em>Hence, the time they will be at the same altitude is 8 minutes</em>

Substitute 8 for m in

Airplane\ 2 = 2500m

Airplane\ 2 = 2500 * 8

Airplane\ 2 = 20000\ ft

<em>Hence, they will be at the same altitude at 20000ft</em>

Method 2:

We have that:

Airplane\ 1 = 44800 - 3100m

Airplane\ 2 = 2500m

Since they are to be at the same altitude, then

The difference in their altitude must be 0

i.e.

Airplane\ 1 - Airplane\ 2 = 0

This gives

44800 - 3100m - 2500m = 0

44800 - 5600m = 0

Collect Like Terms

5600m = 44800

m = 44800/5600

m = 8\ min

Substitute 8 for m in

Airplane\ 1 = 44800 - 3100m

Airplane\ 1 = 44800 - 3100 * 8

Airplane\ 1 = 44800 - 24800

Airplane\ 1 = 20000\ ft

Method 3:

We have that:

Airplane\ 1 = 44800 - 3100m

Airplane\ 2 = 2500m

Since they are to be at the same altitude, then

The ratio of their altitudes must be 1

i.e.

\frac{Airplane\ 1}{Airplane\ 2} = 1

\frac{44800 - 3100m}{2500m} = 1

Cross Multiply

44800 - 3100m = 1 * 2500m

44800 - 3100m = 2500m

Collect Like Terms

44800 = 2500m + 3100m

44800 = 5600m

5600m = 44800

m = 44800/5600

m = 8\ min

Substitute 8 for m in

Airplane\ 1 = 44800 - 3100m

Airplane\ 1 = 44800 - 3100 * 8

Airplane\ 1 = 44800 - 24800

Airplane\ 1 = 20000\ ft

Hence;

<em>Their altitudes must be 20000ft</em>

<em>Though the three methods applied uses the same logic at some point, the first method applied is still the easiest and it is a straight forward method that could be applied in solving the question.</em>

<em>Method 3 is the most difficult.</em>

7 0
3 years ago
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