Answer:
The larger the sample size, the more accurate the estimation of the true population value.
Step-by-step explanation:
- As large will be the sample size more data will be shown and more are the c c changes of it being an estimate of a true population. The sample size can be determined on the basis of use of experience, target variance, confidence level, and target for power.
By taking the average number of eggs for both barns, we will see that the correct option is A.
<h3>
How to get the average?</h3>
For a set of N elements {x₁, x₂, ..., xₙ} the mean is given by:

For barn 1, the set is {2, 5, 5, 6}
So the mean is:
M1 = (2 + 5 + 5 + 6)/4 = 4.5
For the barn 2, the set is {3, 6, 8, 9}
So the mean is:
M2 = (3 + 6 + 8 + 9)/4 = 6.5
Now, taking the quantity 100%(M2 - M1)/M1 we get:
100%*(6.5 - 4.5)/4.5 = 44%
So the average in barn 2 is around 50% larger than the average in barn 1. Then the correct option is A.
If you want to learn more about averages, you can read:
brainly.com/question/20118982
A. The answer is 6$ because 3 x 2 is 6
B. The answer is 12
Answer:
Incomplete question
Complete question: Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker point with a cross-court passing shot. The 57.5-gram ball hit her racket with a northward velocity of 26.7 m/s. Upon impact with her 331-gram racket, the ball rebounded in the exact opposite direction (and along the same general trajectory) with a speed of 29.5 m/s.
a. Determine the pre-collision momentum of the ball.
b. Determine the post-collision momentum of the ball.
c. Determine the momentum change of the ball.
Answer:
A. 1.5353kgm/s
B. 1.6963kgm/s
C. 0.161kgm/s
Step-by-step explanation:
A. The pre-collision momentum of the ball = mass of ball × velocity of ball
Mass of ball = 57.5g = 0.0575kg
Velocity of ball = 26.7m/s
Pre-collision momentum of ball = 0.0575×26.7
= 1.5353kgm/s
B. Post collision momentum of the ball = mass of ball × velocity of ball after impact
Velocity of ball after impact = 29.5m/s
Post collision momentum of ball after impact = 0.0575×29.5
= 1.6963kgm/s
C. Momentum change of ball = momentum after impact - momentum before imlact
= 1.6963kgm/s - 1.5353kgm/s
= 0.161kgm/s