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Rina8888 [55]
3 years ago
9

A random survey of 100 students asked each student to select the most preferred form of recreational activity from five choices.

Following are the results of the survey:Recreational Choice Baseball JoggingGender Basketball Softball Swimming Running Tennis TotalsMale 21 5 9 12 13 60Female 9 3 1 15 12 40Totals 30 8 10 27 25 1001. Test whether the choice is independent of the gender of the respondent. Approximate the p-value of the test.2. Would we reject the null hypothesis at α = 0.05?
Mathematics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

No , we would not reject the null hypothesis at α = 0.05

Step-by-step explanation:

Given that a  random survey of 100 students asked each student to select the most preferred form of recreational activity from five choices. Following are the results of the survey.

H_0: Independent\\H_a: Not independent.

(Two tailed chi square test at 5% level for testing independence between gender and sports activities)

We assume H0 to be true and prepare expected values and then calculate chi square as (O-E)^2/E where O = observed value, E = expected value

Basketball Softball Swimming running Tennis  

Males 21  (18.00)  [0.50] 5  (4.80)  [0.01] 9  (6.00)  [1.50] 12  (16.20)  [1.09] 13  (15.00)  [0.27] 60

Females 9  (12.00)  [0.75] 3  (3.20)  [0.01] 1  (4.00)  [2.25] 15  (10.80)  [1.63] 12  (10.00)  [0.40] 40

     

     

chisquare 8.4097      

p value 0.077      

df 4      

Since p > 5% we accept null hypothesis.

No , we would not reject the null hypothesis at α = 0.05

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An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
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Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

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