Answer:
The answer is 0.8 m
Explanation:
According the attached diagram, if we take the momentum in the point 0:
Where
mD = mass of Diane = 64.2 kg
mJ = mass of Jack = 93.6 kg
Replacing:
A professional soccer player succeeds in scoring a goal on 84% of his penalty kicks. assume that the success of each kick is ind
ependent. (a) in a series of games, what is the probability that the first time he fails to score a goal is on his fifth penalty kick? (b) what is the probability that he scores on 5 or fewer of his next 10 penalty kicks? (c) suppose that our soccer player is out of action with an injury for several weeks. when he returns, he only scores on 5 of his next 10 penalty kicks. is this evidence that his success rate is now less than 84%? explain.
2 answers:
Geometric Distributions:Deals
with the number of trials
required until a single success.
Geometric random variable: the number of trials (y) it takes to get a success in a geometric setting.
When do you use it? To find first success or failure
Calculator Commands
When it takes more than n trials...1-geometcdf(p,n)
When it take n trials... geometpdf(p,x)
When it it takes a max of n trials...geometcdf(p,x)
a. geometcdf(.16,5) = .0797
b. binomcdf(10,.84,5) = .013
c. From par b. if the soccer player's succes rate were still 84%, the probability that he would score 5 or few goals in 10 penalty kicks is 0.0130. This is so low thatwe should be suspicious about whether he can still hit 84% of his shots. We have convincing that his penalty kick success rate has fallen below 84%
Geometric random variable: the number of trials (y) it takes to get a success in a geometric setting.
When do you use it? To find first success or failure
Calculator Commands
When it takes more than n trials...1-geometcdf(p,n)
When it take n trials... geometpdf(p,x)
When it it takes a max of n trials...geometcdf(p,x)
a. geometcdf(.16,5) = .0797
b. binomcdf(10,.84,5) = .013
c. From par b. if the soccer player's succes rate were still 84%, the probability that he would score 5 or few goals in 10 penalty kicks is 0.0130. This is so low thatwe should be suspicious about whether he can still hit 84% of his shots. We have convincing that his penalty kick success rate has fallen below 84%
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Answer:
(a) 19.71801m/s Velocity just before going up the ramp.
(b) 74.56338m.
Explanation:
We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.
Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.
Following formulas will be used.
(a) Calculating velocity right before going up the ramp.
Wooden block is going on a straightaway and has net for on it.
=
and this force produces acceleration of
With this acceleration, wooden block would reach at the foot of ramp in.
and final velocity will be
this velocity of wooden box just before going up the ramp.
(b) How far up the ramp will the wooden block go before stopping.
Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.
and deceleration along the ramp is
Where force of friction along the inclined plane.
is a component of g along normal of the inclined plane.
And with this deceleration time needed to get wooded block to stop is.
and in that time wooden block would travel
This is how up wooden box will go before coming to stop.