How do you find weight =mass×gravitational acceleration
2 answers:
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<h3>Question:5</h3>
The given data can be written in following way in coordinates axes.
(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).
a) Average velocity for first 4 seconds
Average velocity = Total Displacement/ Time taken
= (4-0)/(4 - 0)
=4/4
= 1/1
<h3> =1 m/s</h3>b) Average velocity for 4 to 8 seconds
Average velocity = Total Displacement/ Time taken
= (4 - 4)/(8-4)
= 0/4
<h3> = 0</h3>c) Average velocity for last 6 seconds
Last 6 seconds = from 8 to 14 seconds
Average velocity = Total Displacement/Time taken
= (2 - 4)/(14 - 8)
= -2/6
<h3> = -1/3 m/s</h3><h3>Question 6:</h3>The given data can be written in following way in coordinates axes.
(0,0), (10,20), (20,20), (30,20), (40,0)
a) State the kind of motion from Os to 10s and from 30s to 40s
It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.
b) What is the velocity of the body after 10s and 40s ?
It is clear from the graph and table as well that,
<h3>Velocity after 10s is 20m/s</h3>and
<h3>Velocity after 40s is 0 m/s</h3>c) Calculate the distance covered by the body between 10s to 30s.
Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD
Area of rectangle ABCD= (30 - 10) × (20 - O)
=20×20
<h3> = 400 m</h3>
Answer:
a. μ 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Explanation:
a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:
μ±(t*s)/sqrt(n)
where:
μ = is the 95% confidence interval estimate
x_ = mean of the sample = 3
s = standard deviation of the sample = 5.8
n = size of the sample = 41
t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021
substituting all the variable, we have:
μ 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.