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ipn [44]
3 years ago
6

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 . Suppose you sit on a bench in a mall and observe​ people's habits as they sneeze. ​(​a) What is the probability that among randomly observed individuals exactly do not cover their mouth when​ sneezing
Mathematics
1 answer:
solniwko [45]3 years ago
5 0

Complete Question

According to a study done by a university student, the probability a randomly selected individual will not cover his or her mouth 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 18 randomly observed individuals exactly 6 do not cover their mouth when sneezing?

Answer:

P(X=6)=(0.162)

Step-by-step explanation:

From the question we are told that:

Sample size n=18

Probability P=0.267

No. that do not cover their mouth when sneezing x=6

Generally the equation for The Binomial distribution is mathematically given by

Parameters

B(18,0.267)

Therefore

P(X=x)=(18..x)(0.267)^x(1-0.267)^{18-x}

Where

x=6

Therefore

P(X=6)=(18..6)(0.267)^6(1-0.267)^{18-6}

P(X=6)=(0.162)

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100 Points!!!
inessss [21]

Answer:

An object moving along the x-axis is said to exhibit simple harmonic motion if its position as a function of time varies as

x(t) = x0 + A cos(ωt + φ).

The object oscillates about the equilibrium position x0.  If we choose the origin of our coordinate system such that x0 = 0, then the displacement x from the equilibrium position as a function of time is given by

x(t) = A cos(ωt + φ).

A is the amplitude of the oscillation, i.e. the maximum displacement of the object from equilibrium, either in the positive or negative x-direction.  Simple harmonic motion is repetitive.  The period T is the time it takes the object to complete one oscillation and return to the starting position.  The angular frequency ω is given by ω = 2π/T.  The angular frequency is measured in radians per second.  The inverse of the period is the frequency f = 1/T.  The frequency f = 1/T = ω/2π of the motion gives the number of complete oscillations per unit time.  It is measured in units of Hertz, (1 Hz = 1/s).

The velocity of the object as a function of time is given by

v(t) = dx(t)/dt = -ω A sin(ωt + φ),

and the acceleration is given by

a(t) = dv(t)/dt = -ω2A cos(ωt + φ) = -ω2x.

4 0
3 years ago
a teacher promised a movie day to the class that did better, on average, on their test. The box plot shows the results of the te
natka813 [3]

The question is incomplete. The complete question is

A teacher promised a movie day to the class that did better, on average, on their test. The box plot shown in the below-mentioned figure shows the results of the test.

Which class should get the reward, and why?

- The 2nd-period class should get the reward. They have the highest score, a perfect 100.

- The 2nd-period class should get the reward. They have a higher median.

- The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

- The 4th-period class should get the reward. Their lowest score is an outlier and should be thrown out.

The box plot shown in the question provides the information that
In 2nd period the minimum of the data is 77, first quartile is 78, median of the data is 89, third quartile is 92 and the maximum of the data is 100.
Hence, the mean of the results of 2nd period is

\frac{77+78+89+92+100}{5}=87.2

In 2nd period the minimum of the data is 72, first quartile is 83, median of the data is 89, third quartile is 96 and the maximum of the data is 98.

Hence, the mean of the results of the 4th period is

\frac{72+83+89+96+98}{5}=87.6

Hence, obviously, the 4th period is having more mean.

Moreover, we can observe that if more of the marks are on the higher side the average will automatically be more. Hence, as the first and the third quartiles are more on the 4th-period, so the average marks for the 4th period must be better.

So, just by observing the box plot, we can give the statement that  "The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class. "

Therefore, the 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

Learn more about box plots here-

brainly.com/question/1523909

#SPJ10

6 0
2 years ago
A flying squirrel's nest is 6 meters high in a tree. From its nest, the flying squirrel glides 8 meters to reach an acorn that i
il63 [147K]

Answer:

2 meters

Step-by-step explanation:

7 0
3 years ago
gary has 30 beads. He puts all the beads on strings. each string has five beadshow many strings dose gary use. Here is another o
mariarad [96]

Answer:

6

Step-by-step explanation:

30 beads

5 per string

30/5 = 6

6 0
3 years ago
Which is an example of energy that is moving?
Advocard [28]

Answer:

C. football flying through a goal tee

Step-by-step explanation:

the football is moving therefore the energy is moving (kinetic energy)

8 0
3 years ago
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