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