4. Suppose a copy machine is set to reduce a picture that's 8.5 inches by 11 inches to picture dimensions one-half the original
2 answers:
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Answer:
(a) 5.714s
(b) 0.625m
Step-by-step explanation:
Acceleration due to gravity=a=±9.8m/s/s (+ when falling, - when going upwards)
(a)
STAGE A: between start and first bounce
Initial speed=u=0m/s, Distance=s=40m, and Final velocity=v
v²=u²+2as
v²=0²+2(10)(40)
v²=794
v=28m/s (As the ball hits the ground)
v=u+at
28=0+9.8t
t=28/9.8=2.85714285714s
STAGE B: from the first bounce until it starts to fall
u=half the speed it hit the ground with=(28/2)=14m/s, and v=0m/s(when it starts to fall)
v=u+at
0=14-9.8t
t=14/9.8=1.42857142857s
STAGE C: between when it starts to fall after bounce 1, and bounce 2
We can assume that the time it takes to go from the ground to max height after bounce 1 is equal to the time it takes to fall from that same height to the ground. Therefore, t =1.42857142857s
The time from when the ball was released to when it hit the ground for the second time = t
t = t + t
+ t
t=2.85714285714+1.42857142857+1.42857142857
t=5.71428571428s≈5.714s
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(b)
Before bounce 1, u=28m/s (see stage a)
After bounce 1, u= 28/2= 14m/s
After bounce 2, u= 14/2 = 7m/s
After bounce 3, u= 7/2 =3.5m/s
u=3.5m/s and v=0m/s(when it starts to fall)
v²=u²+2as
0²=3.5²+2(-9.8)s
19.6s=12.25
s=<u>0.625m (max height after third bounce)</u>
Answer:
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
For this case we select a sample of n =100
From the central limit theorem we know that the distribution for the sample mean is given by:
So then the sample mean would be:
And the standard deviation would be: