A fibre has a total rms pulse broadening of 2 ns. The laser and phot.pdf
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A fibre has a total rms pulse broadening of 2 ns. The laser and photodetector connected to either end of the fibre have bandwidths of 400 MHz and 1.5 GHz respectively. What is the maximum recommended digital transmission rate for a non-return to zero data source connected to this optical fibre system? Rb= 296.5Mb/s Rb= 175Mb/s Rb= 200Mb/s Rb= 67.3Mb/s Rb= 750Mb/s Rb= 134.6Mb/s Solution Rb= 175Mb/s.
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A financial analyst collected useful information for 30 employees at.pdf
A financial analyst collected useful information for 30 employees at Gamma Technologies, Inc.
These data include each selected employee\'s citizenship status (i.e., U.S. citizen? yes/no), age in
years, number of years of relevant work experience prior to employment at Gamma, number of
years of employment at Gamma, promotion potential (1= low, 2 = average, 3 = high) , and
annual salary.
What is the type (scale) of data for each of the six variables included in this set?
Solution
Citizenship Status: Nominal Age in Years: Interval Number of Years of Work
Experience: Interval Number of Years of Employment: Interval Promotion Potential: Ordinal
(specifically, this is a Likert Scale) Annual Salary: Ratio There are four levels of variable
measurement: Nominal, Ordinal, Interval, and Ratio level of measurement. A nominal level
means the values are a label, name, or category and there is no ranking or specified order. An
ordinal level is a nominal level of measurement with a specified order, i.e., CEO, Manager,
Supervisor, Employee. An interval level of measurement is an ordinal level of measurement
when the difference in between values have meaning, and arithmetic operations such as addition
and subtraction can be performed; in addition, 0 does not mean the absence of the quantity. The
promotion potential is NOT an interval level since the difference doesn\'t really tell us anything
(2-1 =1 does not mean a candidate has 1 more \"promotion potential\" than another candidate). A
ratio level of measurement is an interval level of measurement with the properties that the ratios
of the values have meaning, and arithmetic operations such as multiplication and division can be
performed on the variable. Most physical quantities, such as mass, density, etc. are ratio levels of
measurement..
A Fill in the missing information for the balance sheet.B What.pdf
A: Fill in the missing information for the balance sheet.
B: What\'s the cash flow from assets?
C: What\'s the cash flow to creditors?
D: What\'s the current ratio?
E: What\'s the debt to equity ratio? Given is a partial balance sheet and income statement for
Louis, Inc. Balance Sheet 2014 2015 Curent Assets 225 Fixed Assets 150 Total Assets 557
Current Liabilities 234 Long-Term Liabilities 60 52 Stockholders\' Equity 22s Total liabilitiesSE
601 | Income Statement 2015 Revenue 1000 COGS 333 Depreciation 285 EBIT Interest Expense
49 EBT Taxes Expense @30% Net Income
Solution
Company income statement
Revenue
1000
Cost of goods sold
333
Depreciation
285
EBIT
382
Interest
49
Taxes (30%)
99.9
Net income
233.10
Balance sheet
2014
2015
Current assets
225
407
Fixed assets
376
150
Total assets
601
557
Current liabilities
318
234
Long term liabilities
60
52
Stock holders equity
223
271
Total liabilities/SE
601
557
B) Cash Flow from Assets
= Operating Cash Flow
- Capital Spending
- Additions to NWC
= 233.10 - (150-376) - (407 - 225)
= 277.10
C) Cash flow to creditors is equal to: beginning total liabilities minus ending total liabilities plus
interest paid
= 60 - 52 + 49
= 57
D: Current ratio = current assets / current liabilities
2015= 407 / 234
=1.739316
2014 = 225/318
= 0.707547
E) Debt to equity ratio = Debt / equity
2015= 52 /271
= 0.191882
2014 = 60/223
= 0.269058
Company income statement
Revenue
1000
Cost of goods sold
333
Depreciation
285
EBIT
382
Interest
49
Taxes (30%)
99.9
Net income
233.10.
A feather is dropped on the moon from a height of 1.40 meters. The a.pdf
A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on
the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.
Solution
Given: vi = 0 m/s d = -1.40 m a = -1.67 m/s2 Find: t = ?? d = vi*t + 0.5*a*t2 -
1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2 -1.40 m = 0+ (-0.835 m/s2)*(t)2 (-1.40 m)/(-0.835
m/s2) = t2 1.68 s2 = t2 t = 1.29 s.
A family thanksgiving dinner has 6 children and 6 adults present. Th.pdf
A family thanksgiving dinner has 6 children and 6 adults present. The host unintentionally feeds
the entire group a type of poisonous mushroom. This mushroom has a probability of .20 of
making an adult sick, and probability .30 of making a child sick. Assume all the family members
are independent regarding their reaction to the mushroom. What is the probability more adults
get sick than children get sick. Let Y1 represent the number of children that get sick, and Y2 the
number of adults that get sick. You should recognize the type of discrete random variables for
Y1 and Y2. Then use the fact that Y1 and Y2 are independent.
Solution
more adults sick than children: cases: adult-1, child-0 Probability:
0.2*0.8^5*0.7^6 adult=2, child=0 or 1: probability:0.2^2*0.8^4*0.7^5 adult:3 child:0,1,2
probability: 0.2^3*0.8^3*0.7^4 adult:4,chuld:0,1,2,3 P = 0.2^4*0.8^2*0.7^3
adult:5,child:0,1,2,3,4 P = 0.2^5*0.8*0.7^2 adult:6,child:0,1,2,3,4,5 P=0.2^6*0.7.
A family has 6 children. Find the probability that there are more g.pdf
A family has 6 children. Find the probability that there are more girls than boys. Assume that
the probabilities of girls and boys being born is equal.
Solution
P(all 6 girls ) = (1/2)^6 P(5 girls 1 boy) = 6C5x(1/2)^5 x1/2 P(4 girls 2 boy) =
6C4x(1/2)^4 x 1/2^2 P = total = 11/32 = 0.34375.
A family has 6 children. Find the probability that there are more gi.pdf
A family has 6 children. Find the probability that there are more girls that boys. Assume that the
probabilities of girls and boys being born are equal
Solution
P(all 6 girls ) = (1/2)^6 P(5 girls 1 boy) = 6C5x(1/2)^5 x1/2 P(4 girls 2 boy) =
6C4x(1/2)^4 x 1/2^2 P = total = 11/32 = 0.34375.
A fair coin is tossed two times in succession.The set of equally lik.pdf
A fair coin is tossed two times in succession.The set of equally likely outcomes is
{HH,HT,TH,TT}.find the probability of getting the same outcome on each toss.
Solution
P(getting same outcome) = P(getting either HH or TT) = number of samples having HH or TT /
total number of samples.
Number of samples having HH or TT = 2
Total number of samples = 4
therefore P(getting the same outcome) = 2/4 = 0.5.
A fair coin is tossed four times. Let Y denote the longest string of.pdf
A fair coin is tossed four times. Let Y denote the longest string of heads occurring. Find the
distribution, mean, variance and standard deviation of Y.
Solution
binomial distribution mean=p=0.5 variance=q/p^2=0.5/0.5^2=2 standard
deviation=sqrt(2)=1.414.
Recommended
A financial analyst collected useful information for 30 employees at.pdf
A financial analyst collected useful information for 30 employees at Gamma Technologies, Inc.
These data include each selected employee\'s citizenship status (i.e., U.S. citizen? yes/no), age in
years, number of years of relevant work experience prior to employment at Gamma, number of
years of employment at Gamma, promotion potential (1= low, 2 = average, 3 = high) , and
annual salary.
What is the type (scale) of data for each of the six variables included in this set?
Solution
Citizenship Status: Nominal Age in Years: Interval Number of Years of Work
Experience: Interval Number of Years of Employment: Interval Promotion Potential: Ordinal
(specifically, this is a Likert Scale) Annual Salary: Ratio There are four levels of variable
measurement: Nominal, Ordinal, Interval, and Ratio level of measurement. A nominal level
means the values are a label, name, or category and there is no ranking or specified order. An
ordinal level is a nominal level of measurement with a specified order, i.e., CEO, Manager,
Supervisor, Employee. An interval level of measurement is an ordinal level of measurement
when the difference in between values have meaning, and arithmetic operations such as addition
and subtraction can be performed; in addition, 0 does not mean the absence of the quantity. The
promotion potential is NOT an interval level since the difference doesn\'t really tell us anything
(2-1 =1 does not mean a candidate has 1 more \"promotion potential\" than another candidate). A
ratio level of measurement is an interval level of measurement with the properties that the ratios
of the values have meaning, and arithmetic operations such as multiplication and division can be
performed on the variable. Most physical quantities, such as mass, density, etc. are ratio levels of
measurement..
A Fill in the missing information for the balance sheet.B What.pdf
A: Fill in the missing information for the balance sheet.
B: What\'s the cash flow from assets?
C: What\'s the cash flow to creditors?
D: What\'s the current ratio?
E: What\'s the debt to equity ratio? Given is a partial balance sheet and income statement for
Louis, Inc. Balance Sheet 2014 2015 Curent Assets 225 Fixed Assets 150 Total Assets 557
Current Liabilities 234 Long-Term Liabilities 60 52 Stockholders\' Equity 22s Total liabilitiesSE
601 | Income Statement 2015 Revenue 1000 COGS 333 Depreciation 285 EBIT Interest Expense
49 EBT Taxes Expense @30% Net Income
Solution
Company income statement
Revenue
1000
Cost of goods sold
333
Depreciation
285
EBIT
382
Interest
49
Taxes (30%)
99.9
Net income
233.10
Balance sheet
2014
2015
Current assets
225
407
Fixed assets
376
150
Total assets
601
557
Current liabilities
318
234
Long term liabilities
60
52
Stock holders equity
223
271
Total liabilities/SE
601
557
B) Cash Flow from Assets
= Operating Cash Flow
- Capital Spending
- Additions to NWC
= 233.10 - (150-376) - (407 - 225)
= 277.10
C) Cash flow to creditors is equal to: beginning total liabilities minus ending total liabilities plus
interest paid
= 60 - 52 + 49
= 57
D: Current ratio = current assets / current liabilities
2015= 407 / 234
=1.739316
2014 = 225/318
= 0.707547
E) Debt to equity ratio = Debt / equity
2015= 52 /271
= 0.191882
2014 = 60/223
= 0.269058
Company income statement
Revenue
1000
Cost of goods sold
333
Depreciation
285
EBIT
382
Interest
49
Taxes (30%)
99.9
Net income
233.10.
A feather is dropped on the moon from a height of 1.40 meters. The a.pdf
A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on
the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.
Solution
Given: vi = 0 m/s d = -1.40 m a = -1.67 m/s2 Find: t = ?? d = vi*t + 0.5*a*t2 -
1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2 -1.40 m = 0+ (-0.835 m/s2)*(t)2 (-1.40 m)/(-0.835
m/s2) = t2 1.68 s2 = t2 t = 1.29 s.
A family thanksgiving dinner has 6 children and 6 adults present. Th.pdf
A family thanksgiving dinner has 6 children and 6 adults present. The host unintentionally feeds
the entire group a type of poisonous mushroom. This mushroom has a probability of .20 of
making an adult sick, and probability .30 of making a child sick. Assume all the family members
are independent regarding their reaction to the mushroom. What is the probability more adults
get sick than children get sick. Let Y1 represent the number of children that get sick, and Y2 the
number of adults that get sick. You should recognize the type of discrete random variables for
Y1 and Y2. Then use the fact that Y1 and Y2 are independent.
Solution
more adults sick than children: cases: adult-1, child-0 Probability:
0.2*0.8^5*0.7^6 adult=2, child=0 or 1: probability:0.2^2*0.8^4*0.7^5 adult:3 child:0,1,2
probability: 0.2^3*0.8^3*0.7^4 adult:4,chuld:0,1,2,3 P = 0.2^4*0.8^2*0.7^3
adult:5,child:0,1,2,3,4 P = 0.2^5*0.8*0.7^2 adult:6,child:0,1,2,3,4,5 P=0.2^6*0.7.
A family has 6 children. Find the probability that there are more g.pdf
A family has 6 children. Find the probability that there are more girls than boys. Assume that
the probabilities of girls and boys being born is equal.
Solution
P(all 6 girls ) = (1/2)^6 P(5 girls 1 boy) = 6C5x(1/2)^5 x1/2 P(4 girls 2 boy) =
6C4x(1/2)^4 x 1/2^2 P = total = 11/32 = 0.34375.
A family has 6 children. Find the probability that there are more gi.pdf
A family has 6 children. Find the probability that there are more girls that boys. Assume that the
probabilities of girls and boys being born are equal
Solution
P(all 6 girls ) = (1/2)^6 P(5 girls 1 boy) = 6C5x(1/2)^5 x1/2 P(4 girls 2 boy) =
6C4x(1/2)^4 x 1/2^2 P = total = 11/32 = 0.34375.
A fair coin is tossed two times in succession.The set of equally lik.pdf
A fair coin is tossed two times in succession.The set of equally likely outcomes is
{HH,HT,TH,TT}.find the probability of getting the same outcome on each toss.
Solution
P(getting same outcome) = P(getting either HH or TT) = number of samples having HH or TT /
total number of samples.
Number of samples having HH or TT = 2
Total number of samples = 4
therefore P(getting the same outcome) = 2/4 = 0.5.
A fair coin is tossed four times. Let Y denote the longest string of.pdf
A fair coin is tossed four times. Let Y denote the longest string of heads occurring. Find the
distribution, mean, variance and standard deviation of Y.
Solution
binomial distribution mean=p=0.5 variance=q/p^2=0.5/0.5^2=2 standard
deviation=sqrt(2)=1.414.
a fair coin is tossed five times. Which sequence is more likely, HT.pdf
a fair coin is tossed five times. Which sequence is more likely, HTTHHor HHHHH? Or are
they equally likely? Explain.
Solution
They are Equally Likely
Reason
This is because the Chances to get H or T = 1/2
Therefore
Chances to get HTTHH or HHHHH = (1/2)^5.
A fair coin is tossed 7 times. What is the probability that At most .pdf
A fair coin is tossed 7 times. What is the probability that At most 4 heads appear?
Solution
the total number of outcomes if a coin is tossed 7 times is 14, At most 4 heads
appear in the sense it may be 1,2,3 and 4. probability of getting one,two,three,four heads is
1/14,2/14,3/14,4/14 respectively.the probability of at most 4 heads is
1/14+2/14+3/14+4/14=10/14=5/7.
A fair coin is tossed 7 times. What is the probability that at least.pdf
A fair coin is tossed 7 times. What is the probability that at least two heads appear?
Solution
the total number of outcomes if a coin is tossed 7 times is 14, At most 4 heads
appear in the sense it may be 1,2,3 and 4. probability of getting one,two,three,four heads is
1/14,2/14,3/14,4/14 respectively.the probability of at most 4 heads is
1/14+2/14+3/14+4/14=10/14=5/7.
A fair coin is flipped 4 times. What is the probability of observing.pdf
A fair coin is flipped 4 times. What is the probability of observing (i) at least one head (ii)
no more than two heads, (iii) a run of two or more consecutive tails?
Solution
p(H) = 4/16 =1/4 p(not moe than 2 heads) =2/16 =1/8 p(two or more conse tails)
=3/16.
A fair coin flipped 9 times. Use Appendix Table for Binomial distrib.pdf
A fair coin flipped 9 times. Use Appendix Table for Binomial distribution to find the probability
that it lands on heads more than 5 times (5 is not included). Tip: first, figure out what is n, p, x
for this case, then use Appendix Table to find each probability, then add them together.
Solution
n = 9
p = probablity of getting head = 1/2
x = 5
the probability that it lands on heads more than 5 times
now we need P(x>5)
P(x=6) + P(x=7) + P(x=8) + P(x=9)
9C6 (1/2)^6(1/2)^3 + 9c 7 (1/2)^7 (1/2)^2 + 9c8 (1/2)^8(1/2) + 9c9 (1/2)^9
= 0.25390625.
A fair coin (i.e. P(Heads)=P(Tails)=½) is tossed repeatedly. Find Pn.pdf
A fair coin (i.e. P(Heads)=P(Tails)=½) is tossed repeatedly. Find Pn, the probability that the
number of heads appearing in #n tosses is even (Note: 0 is an even number).
Solution
Pn=nc0 0.5^n+nc2 0.5^n +------------------
=1/2=.5.
A fair 6 sided die is rolled 6 independent times. Let Ai be the even.pdf
A fair 6 sided die is rolled 6 independent times. Let Ai be the event that side i is observed on the
i\'th roll, called a match on the i\'th trial, i = 1, 2, 3, 4, 5, 6. If we let B denote the event that at
least one match occurs, find P(B).
Solution
p(B) = 1-(5/6)^6 =0.665.
A fair coin (HT) is tossed three times. Let E = exactly one tail (T.pdf
A fair coin (H/T) is tossed three times. Let E = exactly one tail (T) appears. Let F = a tail (T)
appears on the first toss.
Compute P(F|E).
Answer choices:
A. 1/4
B. 1/3
C. 3/8
D. None of the above
Please explain because I am SO confused! thanks!!!
Solution
P(F/E)=P(F intersecton E)/P(E) =(0.5*0.5*0.5)/(3*0.5*0.5*0.5) =1/3.
A die is rolled and then a coin is tossed. Draw a tree diagram to in.pdf
A die is rolled and then a coin is tossed. Draw a tree diagram to indicate the possible results.
How many results are possible?
Solution
There are total 6 possibilities in die, that is from 1 to 6 There are two possibilities
with toss, heads/tail Total result possible = 6*2 = 12 Heads / Tails 1 1 2 2 3 3 4 4 5 5 6 6.
A dolphin moves in water towards an obstacle at a speed of 15 ms. A.pdf
A dolphin moves in water towards an obstacle at a speed of 15 m/s. At 2m from the obstacle, it
emits a signal whose speed in water is 1500 m/s.
How much time does it take to receive the echo? If the response time of the dolphin is 0.1 s, can
it avoid the obstacle?
Solution
Let t be the time at which the dolphin receives the return signal, and let d be the distance that the
dolphin travels in that time. Then,
d = t*15 m/s
The original distance D from the dolphin to the object is 2 m. When the dolphin recieves the
signal, he is at position D - d from the object. So the total distance traveled by the ultrasonic
signal is:
D + D - d = 2D - t*15 m/s = t * 1500 m/s
4 m = t*1485 m/s
t = 2.69 ms
At 2.69 ms, the dolphin has traveled d = 2.69 ms * 15 m/s = 0.04 m. So the object is 1.96 m
away from the dolphin. At 2.69 ms + 100 ms = 102.69 ms, the dolphin has traveled d = 102.69
ms * 15 m/s = 1.54 m. So, having responded with 0.46 m to spare, the dolphin easily misses the
object..
A differential equation a.can always be solved explicitly b..pdf
A differential equation
a.can always be solved explicitly
b.can always be solved implicitly
c.can have no solution
d.can have imaginary solutions
e.can have infinite order
choose one option
Solution
A differential equation can be of infinite order.other option are partially correct...only some
equations can be solved explicitly and implicitly. It can have no solution but analyticaly it can be
solved..
A die is rolled 10 times. Find the probability of rolling exactly 1 .pdf
A die is rolled 10 times. Find the probability of rolling exactly 1 five.
What is the probability?
Solution
Given all these, probability of rolling a five on roll will be 1/6 and anything else on
one roll will be 5/6 Then, If you roll it 10 times, (1/6) * (5/6)^9 * 10 or ==> 1.24605326 × 10-8
is an answer.
A disc jockey has 10 songs to play. Four are slow songs, and six are.pdf
A disc jockey has 10 songs to play. Four are slow songs, and six are fast songs. Each song is to
be played only once. In how many ways can the disc jockey play the 10 songs if
The songs can be played in any order.
The first song must be a slow song and the last song must be a slow song.
The first two songs must be fast songs.
Solution
11! 9! 9!.
A die is rolled 20 times and the number ofSolutionA binomial i.pdf
A die is rolled 20 times and the number of
Solution
A binomial is nothing but repeated bernouli. Where Bernoulli is a binomial and we can perform
the experiment only once,i.e., simple \"success/failure\" experiment. Since the probability of
success is fixed. Here probability of getting 5\'s is 1/6 and you\'re repeating the experiment 20
times..
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a fair coin is tossed five times. Which sequence is more likely, HT.pdf
a fair coin is tossed five times. Which sequence is more likely, HTTHHor HHHHH? Or are
they equally likely? Explain.
Solution
They are Equally Likely
Reason
This is because the Chances to get H or T = 1/2
Therefore
Chances to get HTTHH or HHHHH = (1/2)^5.
A fair coin is tossed 7 times. What is the probability that At most .pdf
A fair coin is tossed 7 times. What is the probability that At most 4 heads appear?
Solution
the total number of outcomes if a coin is tossed 7 times is 14, At most 4 heads
appear in the sense it may be 1,2,3 and 4. probability of getting one,two,three,four heads is
1/14,2/14,3/14,4/14 respectively.the probability of at most 4 heads is
1/14+2/14+3/14+4/14=10/14=5/7.
A fair coin is tossed 7 times. What is the probability that at least.pdf
A fair coin is tossed 7 times. What is the probability that at least two heads appear?
Solution
the total number of outcomes if a coin is tossed 7 times is 14, At most 4 heads
appear in the sense it may be 1,2,3 and 4. probability of getting one,two,three,four heads is
1/14,2/14,3/14,4/14 respectively.the probability of at most 4 heads is
1/14+2/14+3/14+4/14=10/14=5/7.
A fair coin is flipped 4 times. What is the probability of observing.pdf
A fair coin is flipped 4 times. What is the probability of observing (i) at least one head (ii)
no more than two heads, (iii) a run of two or more consecutive tails?
Solution
p(H) = 4/16 =1/4 p(not moe than 2 heads) =2/16 =1/8 p(two or more conse tails)
=3/16.
A fair coin flipped 9 times. Use Appendix Table for Binomial distrib.pdf
A fair coin flipped 9 times. Use Appendix Table for Binomial distribution to find the probability
that it lands on heads more than 5 times (5 is not included). Tip: first, figure out what is n, p, x
for this case, then use Appendix Table to find each probability, then add them together.
Solution
n = 9
p = probablity of getting head = 1/2
x = 5
the probability that it lands on heads more than 5 times
now we need P(x>5)
P(x=6) + P(x=7) + P(x=8) + P(x=9)
9C6 (1/2)^6(1/2)^3 + 9c 7 (1/2)^7 (1/2)^2 + 9c8 (1/2)^8(1/2) + 9c9 (1/2)^9
= 0.25390625.
A fair coin (i.e. P(Heads)=P(Tails)=½) is tossed repeatedly. Find Pn.pdf
A fair coin (i.e. P(Heads)=P(Tails)=½) is tossed repeatedly. Find Pn, the probability that the
number of heads appearing in #n tosses is even (Note: 0 is an even number).
Solution
Pn=nc0 0.5^n+nc2 0.5^n +------------------
=1/2=.5.
A fair 6 sided die is rolled 6 independent times. Let Ai be the even.pdf
A fair 6 sided die is rolled 6 independent times. Let Ai be the event that side i is observed on the
i\'th roll, called a match on the i\'th trial, i = 1, 2, 3, 4, 5, 6. If we let B denote the event that at
least one match occurs, find P(B).
Solution
p(B) = 1-(5/6)^6 =0.665.
A fair coin (HT) is tossed three times. Let E = exactly one tail (T.pdf
A fair coin (H/T) is tossed three times. Let E = exactly one tail (T) appears. Let F = a tail (T)
appears on the first toss.
Compute P(F|E).
Answer choices:
A. 1/4
B. 1/3
C. 3/8
D. None of the above
Please explain because I am SO confused! thanks!!!
Solution
P(F/E)=P(F intersecton E)/P(E) =(0.5*0.5*0.5)/(3*0.5*0.5*0.5) =1/3.
A die is rolled and then a coin is tossed. Draw a tree diagram to in.pdf
A die is rolled and then a coin is tossed. Draw a tree diagram to indicate the possible results.
How many results are possible?
Solution
There are total 6 possibilities in die, that is from 1 to 6 There are two possibilities
with toss, heads/tail Total result possible = 6*2 = 12 Heads / Tails 1 1 2 2 3 3 4 4 5 5 6 6.
A dolphin moves in water towards an obstacle at a speed of 15 ms. A.pdf
A dolphin moves in water towards an obstacle at a speed of 15 m/s. At 2m from the obstacle, it
emits a signal whose speed in water is 1500 m/s.
How much time does it take to receive the echo? If the response time of the dolphin is 0.1 s, can
it avoid the obstacle?
Solution
Let t be the time at which the dolphin receives the return signal, and let d be the distance that the
dolphin travels in that time. Then,
d = t*15 m/s
The original distance D from the dolphin to the object is 2 m. When the dolphin recieves the
signal, he is at position D - d from the object. So the total distance traveled by the ultrasonic
signal is:
D + D - d = 2D - t*15 m/s = t * 1500 m/s
4 m = t*1485 m/s
t = 2.69 ms
At 2.69 ms, the dolphin has traveled d = 2.69 ms * 15 m/s = 0.04 m. So the object is 1.96 m
away from the dolphin. At 2.69 ms + 100 ms = 102.69 ms, the dolphin has traveled d = 102.69
ms * 15 m/s = 1.54 m. So, having responded with 0.46 m to spare, the dolphin easily misses the
object..
A differential equation a.can always be solved explicitly b..pdf
A differential equation
a.can always be solved explicitly
b.can always be solved implicitly
c.can have no solution
d.can have imaginary solutions
e.can have infinite order
choose one option
Solution
A differential equation can be of infinite order.other option are partially correct...only some
equations can be solved explicitly and implicitly. It can have no solution but analyticaly it can be
solved..
A die is rolled 10 times. Find the probability of rolling exactly 1 .pdf
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What is the probability?
Solution
Given all these, probability of rolling a five on roll will be 1/6 and anything else on
one roll will be 5/6 Then, If you roll it 10 times, (1/6) * (5/6)^9 * 10 or ==> 1.24605326 × 10-8
is an answer.
A disc jockey has 10 songs to play. Four are slow songs, and six are.pdf
A disc jockey has 10 songs to play. Four are slow songs, and six are fast songs. Each song is to
be played only once. In how many ways can the disc jockey play the 10 songs if
The songs can be played in any order.
The first song must be a slow song and the last song must be a slow song.
The first two songs must be fast songs.
Solution
11! 9! 9!.
A die is rolled 20 times and the number ofSolutionA binomial i.pdf
A die is rolled 20 times and the number of
Solution
A binomial is nothing but repeated bernouli. Where Bernoulli is a binomial and we can perform
the experiment only once,i.e., simple \"success/failure\" experiment. Since the probability of
success is fixed. Here probability of getting 5\'s is 1/6 and you\'re repeating the experiment 20
times..
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