Questions on Word Problems: Miscellaneous Word Problems answered by real tutors!

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remember when a constant like x in our case varies directly with another variable,s squared in our case it takes on this form and when a constant like x varies indirectly with t it takes on this form xt=k
: together we have xt/s^2=k or
:
so we have when s is doubled we have or meaning x increases by 4 when s is doubled.
:
when s and t are doubled we have or
meaning x doubles when s and t are doubled

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x varies directly as the square of s and inversely as t.
x = k[s^2/t]
-----------------------
How does x change when s is doubled?
x = k[(2s)^2/t] = 4k[s^2/t]
doubling s quadruples x
-----------------------
When both s and t are doubled?
x = k[(2s)^2/(2t)]
x = 4ks^2/(2t)
x = 2ks^2/t
doubling s and t doubles x
==============================
Cheers,
Stan H.

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so Jack can do 1/7 of the job per hour and Marlyn can do 1/4 of the job per hour
:
1/7(t)+1/4(t)=1(job)----->find common denominator---->28 --.so multiply by 28
:
4t+7t=28
:
11t=28
or 2 6/11 hours for both to mow the lawn

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Jack usually mows his lawn in 7 hours. Marilyn can mow the same yard in 4 hours. How much time would it take for them to mow the lawn together?
------------------------------------
Jack DATA:
Time = 7 hrs/job ; rate = 1/7 job/hr
----------------------------------------
Marilyn = 4 hrs/job ; rate = 1/4 job/hr
-----------------------------------------
Together DATA:
Time = x hrs/job ; rate = 1/x job/hr
----------------------------------------
EQUATION:
rate + rate = together rate
1/7 + 1/4 = 1/x
11/28 = 1/x
x = 28/11 = 2.545454... hrs
===============================
Cheers,
Stan H.

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BILL CAN BUILD A BOOKCASE IN 6 HOURS, BUT IT TAKES HIS FRIEND, DAVID, ONLY 4 HOURS TO BUILD THE SAME BOOKCASE. HOW LONG WILL IT TAKE THEM TO BUILD THE BOOKCASE IF THEY WORK TOGETHER?
-----------------------------
Bill DATA:
Time = 6hrs/job ; rate = 1/6 job/hr
--------------------------------------
Davis DATA::
Time = 4 hrs/job ; rate = 1/4 job/hr
---------------------------------------
Together DATA:
Time = x hrs/job ; rate = 1/x job/hr
----------------------------------------
EQUATION:
rate + rate = together rate
1/6 + 1/4 = 1/x
(10/24) = 1/x
x = 24/10 = 2.4 hrs
--------
0.4 hrs = 0.4*60 minutes = 24 minutes
----------
Answer: 2 hrs and 24 minutes (time they need to do the job together)
====================================
Cheers,
Stan H.

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the profit associated with making the craft is represented by P(x)=-x^2+130x-3000, where x represents the number of crafts sold and P represents the profilt. Determine the number of orders that could be acepted in order to make a profit.
------------------
Determine when P is > 0
--------------------------
Solve -x^2+130x -3000 = 0
-(x^2-130x+3000) = 0
(x-30)(x-100) = 0
----------------------
Profit interval: 30 < x < 100
=============================
Cheers,
Stan H.

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you need to double the amount of wristbands she had at the beginning of the day which makes 22, then you add the 7 she bought during the day and subtract out the 2 she gave her brother , and the 3 she gave her mother.....ending with 24 at the end of the day.

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X+Y=1882
X=Y-970 X BEING THE LOSER.
(Y-970)+Y=1882
2Y=1882+970
2Y=2852
Y=2852/2
Y=1,426 WAS THE NUMBER OF VOTE FOR THE WINNER.
PROOF:
1882-1426=456 FOR THE LOSER.
1426-456=970 IS THE DIFFERENCE.

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Note: I'm going to let a=first digit, b=second digit, c=third digit etc, until I get to j=tenth digit


So the telephone number looks like this:

(a b c) d e f - g h i j


Now let's do some translations:

"The Product of the sixth and seventh numbers is the third number. "

translates to

f*g=c

--------------------

"The fourth, eigth, ninth, and tenth numbers are multiples of 3"

translates to

d, h, i, j are multiples of 3 (giving us the possible values 0, 3, 6, 9)


Note: 0 is a multiple since 3*0=0

--------------------
"The sum of the fourth and sixth numbers is the same as the sum of the fifth and eigth numbers. "

translates to

d+f = e+h


--------------------
"The second, third, sixth, and seventh numbers are powers of 2."

translates to

b, c, f, g are powers of 2 which gives the possible values: 1, 2, 4, 8


Note: 1 is a power of 2 since

--------------------

"The first, fifth, seventh, and tenth numbers are prime."

translates to

a, e, g, j are prime

List of (single digit) prime numbers: 2, 3, 5, 7


You will need to refer back to this list so it might help to write the list on a separate sheet of paper and cross out any digits that we've used

===============================================================



Note: remember, each digit is different which means that NO repeats are allowed.


Since digit j is both a multiple of 3 AND is prime, this means that the only possible digit for j is 3 (since 3 is both a multiple of 3 AND prime)


So j=3

The same can be done with g. Since g is both a power of 2 AND is prime, the only candidate is 2 (since 2 is both a power of 2 AND is prime)


So g=2


Looking at f*g=c, we know that all three are powers of 2. So this means that f*2=some power of 2


So let's try some values of f

f=1 (CANNOT do since 1*2=2; a repeat would occur where g=c)
f=2 (CANNOT do since 2 is already taken)
f=4 (CAN do since 4*2=8; all three are powers of 2)
f=8 (CANNOT do since 8*2=16; 16 is comprised of 2 digits)

So f=4 and c=8


Since we know that g=2, f=4, and c=8 (and there are ONLY 4 digits that are powers of 2), this means that b=1 (by elimination)

Just to recap, we've taken the digits 3, 2, 4, 8, and 1 so far (5 digits taken already).


--------------------------------------

Now let's move onto the equation

d+f = e+h

We know that f=4, so let's plug that in:

d+4 = e+h


Here's where we have to try some combinations

The possible values for d are 0, 6, and 9 (remember d is a multiple of 3 and the digit 3 is already taken)

The possible values for e are 5 and 7 (e is prime and the digits 2 and 3 are already taken)

The possible values for h are 0, 6, and 9 (remember h is a multiple of 3 and the digit 3 is already taken)


So we could have the following:

d=0 OR d=6 OR d=9

e=5 OR e=7

h=0 OR h=6 OR h=9


So let's add some combinations:


Combination 1:
If d=0, e=5, and h=6 (0 already taken), then

0+4=5+6
4=11 ... which is false (cross this combination out)


Combination 2:
If d=0, e=5, and h=9, then

0+4=5+9
4=14 ... which is false (cross this combination out)


Combination 3:
If d=0, e=7, and h=6 (0 already taken), then

0+4=7+6
4=13 ... which is false (cross this combination out)


Combination 4:
If d=0, e=7, and h=9, then

0+4=7+9
4=16 ... which is false (cross this combination out)


Combination 4:
If d=6, e=5, and h=0 (6 already taken), then

6+4=5+0
10=5 ... which is false (cross this combination out)


Combination 5:
If d=6, e=5, and h=9 (6 already taken), then

6+4=5+6
10=11 ... which is false (cross this combination out)


Combination 6:
If d=6, e=7, and h=0 (6 already taken), then

6+4=7+0
10=7 ... which is false (cross this combination out)


Combination 7:
If d=6, e=7, and h=9 (6 already taken), then

6+4=7+9
10=16 ... which is false (cross this combination out)


Combination 8:
If d=9, e=7, and h=0 (9 already taken), then

9+4=7+0
13=7 ... which is false (cross this combination out)


Combination 9:
If d=9, e=7, and h=6 (9 already taken), then

9+4=7+6
13=13 ... which works



So after trying 8 combinations of numbers, we get
d=9, e=7 and h=6

-----------------------------------------------------------

So to recap, we've taken the digits 3, 2, 4, 1, and 8 (from the first part) and the digits 9, 7, and 6 (from the second part)

Altogether, we know that

j=3, g=2, f=4, b=1, and c=8 (from part 1)

d=9, e=7 and h=6 (from part 2)


------------------------------------------------------

Since d=9, h=6, and j=3, using elimination we get i=0 (note: I'm looking at the "multiples of 3" group)



Finally, since e=7, g=2, and j=3, this means that a=5 (note: I'm looking at the "prime numbers" group)


================================================================


Answer:

So we have the following numbers:

a=5
b=1
c=8

d=9
e=7
f=4

g=2
h=6
i=0
j=3


which means that the telephone number is

(518) 974-2603

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Let x=amount of rice he starts with and y=amount he has leftover

After First Station

Since "At the first he gives up 1/3 of his rice", this tells us that after the first station, he has pounds of rice leftover

Note: I've subtracted 1/3 of the amount "x" from "x"



Combine like terms to get . So he has pounds of rice leftover after the first station (this makes sense because the guard took 1/3 so the man must have 2/3 leftover).



Note: remember, the "x" is just a number (a number in which we do NOT know). However, to help you get an idea of what is going on, let (ie let's say that he started at 12 lbs of rice). After the first station, he would have lbs of rice.


--------------------------------------------

After Second Station

Keep in mind that the man still has lbs of rice.



Since at "the second he is allowed to keep 1/3 of what is left", this means that they will take 2/3 of what he has left.



So this means that you need to subtract like this

Amount he has left - (2/3)*(Amount he has left )





Multiply


Combine the fractions.



So after the second station, he has lbs of rice left.
--------------------------------------------

After Third Station

Finally, because "at the third he gives up 1/4 of what remains", this tells us that we subtract 1/4 of what he has left like this:

Amount he has left - (1/4)*(Amount he has left )


Which algebraically looks like:





Multiply


Combine the fractions.



So after the third station, he has lbs of rice left.



----------------------------------------------------------


Since "he has 6 pounds less than he started with." this means that he ended up with lbs. Also, we know that he ended up with lbs of rice (after passing through the third station)


So this means that the two are equal, so...




Multiply EVERY term by the LCD 6 to clear the fraction.


Add to both sides.


Subtract from both sides.


Combine like terms on the left side.


Divide both sides by to isolate .


Divide.


=============================================================

Answer:


So the man started with 7.2 lbs of rice and the amounts after each station are:


1st station: so he had 4.8 lbs after the 1st station


2nd station: so he had 1.6 lbs after the 2nd station


3rd station: so he had 1.2 lbs after the 3rd station



Since the difference between 7.2 and 1.2 is this helps confirm our answer.

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follow how much the man keeps __ let x=starting amount

"At the first, he gives up 1/3 of his rice" __ keeps (2/3)x

"at the second he is allowed to keep 1/3 of what is left" __ keeps 1/3 of (2/3)x __ now has (2/9)x

"at the third, he gives up 1/4 of what remains" __ keeps 3/4 of (2/9)x __ now has (1/6)x

"At the end, he has 6 pounds less than what he started with" __ (1/6)x=x-6

multiplying by 6 __ x=6x-36

adding 36-x __ 36=5x

dividing by 5 __ 7.2=x

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follow how much the man keeps __ let x=starting amount

"At the first, he gives up 1/3 of his rice" __ keeps (2/3)x

"at the second he is allowed to keep 1/3 of what is left" __ keeps 1/3 of (2/3)x __ now has (2/9)x

"at the third, he gives up 1/4 of what remains" __ keeps 3/4 of (2/9)x __ now has (1/6)x

"At the end, he has 6 pounds less than what he started with" __ (1/6)x=x-6

multiplying by 6 __ x=6x-36

adding 36-x __ 36=5x

dividing by 5 __ 7.2=x

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200/1=x/39
x=39*200
x=$7,800 ans.

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There is no "other dollar"
1. The men originally paid $30----it went into the manager's cash register
2. The manager realized that he charged too much, so he gave the bellboy five dollars to return to the men ($25 in cash register $5 in bellboys's hand)
3. The bellboy gave each man one dollar back and kept 2 dollars for himself ($25 in register, each man has $1 =$28 plus the $2 the bellboy kept=$30)
Another way to look at it is as follows: The statement that the men paid $27 for the room is incorrect---they paid $25 for the room and got fleeced out of $2 by the bellboy ($27-$2=$25 and that jives with whats in the cash register)
Gotta keep track of where the $$$ are, otherwise the fingers may leave the hand.


Hope this helps---ptaylor

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This was already solved, I copied it fro Stan H, or stanbon.
------------------------------
Use a proportion:
[9 infected/15 sampled] = [x infected/all the PCs]
(9/15) = (x/202)
x = 202(9/15)
x = 202(3/5)
x = 121.2
Round down to get x = 121 infected PC's out of 202

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THE ELECTRIC CURRENT I, IN AMPERES, IN A CIRCUIT VARIES DIRECTLY AS THE VOLTAGE V. WHEN 12 VOLTS ARE APPLIED, THE CURRENT IS 4 AMPERES. WHAT IS THE CURRENT WHEN 18 VOLTS ARE APPLIED?
-----------
I = kV
Find "k":
4 = k*12
k = 4/12 = 1/3
-----------
Equation: I = (1/3)V
If 18 volts are applied you get:
I = (1/3)*18
I = 6 amps.
==============
Cheers,
Stan H.

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A truck carrying 3600 cubic feet of cargo consisting of washing machines and refrigerators was hijacked. The washing machine are worth $300 each and are shipped in 36 cubic-foot cartons. The refrigerators are worth $900 each and are shipped in 45 cubic-foot cartons. If the total value of the cargo is $51,000, then how many of each were there on the truck?
-----------------
value equation: 300w + 900r = 51000
volume equation: 36w + 45r = 3600
=======================================
Rearrange:
w + 3r = 170
4w + 5r = 400
-------------------
Multiply thru 1st by 4 to get:
4w + 12r = 4*170
4w + 5r = 400
--------------------
Subtract 2nd from 1st to solve from r:
7r = 4*170-400
r = 40 (# of refrigerators)
w = 170-3r = 170-3*40 = 50 (# of washers)
===================================
Cheers,
Stan H.

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what are all of the trigonomic values given that cot=-3 and sin= square root 10/10
---------------------
Keep in mind that x^2 + y^2 = r^2
------------
cot=-3 = -3/1
Since cot = x/y, x = -3 and y = 1
Therefore r = sqrt(-3)^2 + 1^2) = sqrt(10)
---------------------
Then:
sin = y/r = 1/sqrt(10) = sqrt(10)/10
cos = x/r = -3/sqrt(10)
tan = y/x = -1/3
csc = r/y = sqrt(10)
sec = r/x = -sqrt(10)/3
cot = -3
=========================
Cheers,
Stan H.

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Re-writing:
Via Reciprocal Identites: ;
Subst. & becomes:


Via Quotient Identities:
Subst. & becomes:


Thank you,
Jojo

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Note: I'm ONLY manipulating the left side. I'm showing the right side for comparison


Start with the given equation


Use the identities and to get the left side in terms of sine and cosine


Multiply the first fraction by the reciprocal of the second fraction.


Combine and multiply the fractions.


Use the identity


Use the identity


Reduce


Use the identity

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It follows Ideal Gas Law: --->
where--->
Subst.:

--->

.
We know that ---->
Then,
,of Cyclohexane (Answer)
.
Thank you
Jojo

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"One number is 6 less than another." translates to


"The product of the number is 72" translates to


Start with the second equation


Plug in


Distribute.


Subtract 72 from both sides.


Notice we have a quadratic equation in the form of where , , and


Let's use the quadratic formula to solve for x


Start with the quadratic formula


Plug in , , and


Negate to get .


Square to get .


Multiply to get


Rewrite as


Add to to get


Multiply and to get .


Take the square root of to get .


or Break up the expression.


or Combine like terms.


or Simplify.


So the answers are or


This means that the numbers are:


12 and 6

OR...

-6 and -12

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4g + 2d = 316
And? Do you have a question?

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cos^2-sin^2=2cos^2-1 prove that this is true
----------
Note: sin^2(x) = 1 - cos^2(x)
------------------------------------
Substitute that into the left side to get:
cos^2 -(1-cos^2) = 2cos^2-1
cos^2 -1 + cos^2 = 2cos^2-1
2cos^2 - 1 = 2cos^2 - 1
=============================
Cheers,
Stan H.

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(1-r^3)(1-r)^-3
= [(1-r)(1+r+r^2)]/[(1-r)^3]
= (1+r+r^2]/[(1-r)^2]
=======================
Cheers,
Stan H.

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Simplify
t^4-1 over t^3+t^2+t+1
-------------

=
Cancel the (t^2+1)
=
=
Cancel (t+1)
= t-1

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With our working formula ----> ,
see highlighted "n" equals the number of moles. But the problem states that "each" has 1.35 moles. Having said that, isn't your only solving for "1" Volume? (take note, STP, -- no change in Pressure & Temp.)
Considering too, each gases stated have different Molecular weight, how could they be then have the same number of moles?
.
For the given, we solve for ?



, Answer
Thank you,
Jojo

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Group terms in the denominator:



Factoring the numerator:



Applying factor for "difference of cubes" in the denominator:

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Working Eqn ----->
where--->
*Note: At STP, is atmospheric Pressure & is Freezing Point.
Solving for ?

where,
Via Periodic Table, Atomic Weight of = "19."
So,
Subst. ------>

.
Going back Working Eqn,


, Answer
.
Thank you,
Jojo

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---------> atm?

, Answer
.
Thank you,
Jojo

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.
It follows Ideal Gas Law --------> , Working Eqn
where --->
Solving for :

where,
In , there are "2" atoms of Oxygen and "1" atom of Carbon.
Look on Periodic Table, we see has Atomic Weight --->. Therefore, its Molecular Weight is expressed as
.
For Carbon, look on Periodic Table and the Atomic Weight is ---->.
It's Molecular Weight is expressed as
.
Therefore,
Subst.:

.
Going Back our Working Eqn,


----------> mmHg?

, Answer
.
Thank you,
Jojo

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.
This follows the Ideal Gas Law -----> , Working Eqn
where--->
Highlighted above is not given, so we have to calculate:
Look thru Table of Elements or Periodic Table and you find Molecular Weight for Argon:
So,
, Subst. in our Working Eqn:

------>
, Answer
.
Thank you,
Jojo

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Same as Q# 169508
See Answer# 125296. Done
Thank you

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By Charles' Law--> it's stated the relation of and :

Given------>
*Note: Celsius need to be converted to Kelvin when dealing with Ideal Gas.
Susbt.


, Answer
.
Thank you,
Jojo

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recalling the standard formula p*v=k
where p=pressure,v=volume and k=constant
we have p1*v1=p2*v2 since temparature is contant we have
p1=0.180atm
v1=4.00L
p2=?
v2=9.30L
making p2 to be the subject of our formula we have

p2=(0.180*4.00)/9.30
p2=0.077
hence the new pressure=0.077atm

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COFFEE AT $3.75 PER POUND IS BLENDED WITH COFEE AT $4.25 A POUND. HOW MUCH OF EACH MUST BE USED TO MAKE 50 POUNDS OF A MIXTURE AT $3.95 A POUND?
-----------------------
Value Equation: 3.75x + 4.25(50-x) = 3.95*50
375x + 425*50 - 425x = 395*50
-50x = -30*50
x = 30 lbs. (# of lbs of 3.75 coffee in the mixture)
50-x = 20 lbs. (# of lbs of 4.25 coffee in the mixture)
===================================================
Cheers,
Stan H.

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A MAN HAS 5 MORE ONE-DOLLAR BILLS THAN FIVE DOLLAR-BILLS, AND ONE MORE FIVE-DOLLAR BILL THAN TEN-DOLLAR BILLS. ALTOGETHER HE HAS $139 HOW MANY ONE-DOLLAR BILLS DOES HE HAVE?
--------------------------
Let # of 5 dollar bills be "x".
Then # of 1 dollar bills is "x+5".
And # of 10 dollar bills "x-1".
-----------------------------------
Value Equation:
5x + 1(x+5) + 10(x-1) = 139
5x + x + 5 + 10x - 10 = 139
16x = 144
x = 9 (# of 5 dollars bills)
# of 1 dollar bills = x+5 = 14
# of 10 dollar bills = x-1 = 8
=================================
Cheers,
Stan H.

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N(t) = -3t^3 + 23t^2 + 8t.
a) Rewrite the formula by factoring the right-hand side completely.
N(t) = t(-3t^2 + 23t + 8)
---------------------------------

b) Use the factored version of the formula to find N(3).
N(3) = 3(-3^3^2 + 23*3 + 8)
N(3) = 3(-27 + 69 + 8)
N(3) = 3(50) = 150
=======================
Cheers,
Stan H.

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Decreasing cube. Each of the three dimensions of a cube with sides of length s centimeters is decreased by a whole number of centimeters. The new volume in cubic centimeters is given by V(s) = s^3 -13s^2 +54s - 72.
a) Find V(10).
Sub 10 for s
V(10) = 10^3 - 13*10^2 + 54*10 - 72
V(10) = 1000 - 1300 + 540 - 72
V(10) = 1540 - 1372 = 168 cc
---------------------
b) If the new width is s - 6 centimeters, then what are the new length and height?
The product of L x H = (s^3 -13s^2 +54s - 72)/(s-6) = s^2 - 7s + 12
Since that's integers, that's (s-3)*(s-4). There's no way to know which is L and which is H.
c) Find the volume when s = 10 by multiplying the length, width, and height.
-------------------
?? Wasn't that part a? Maybe it means:
(10-6)*(10-3)*(10-4)
= 4*7*6
= 168 cc
Must be, it matches.

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Demand for pools. Tropical Pools sells an aboveground model for p dollars each.
The monthly revenue for this model is given by the formula R(p) = -0.08p^2 -300p. Revenue is the product of the price p and the demand (quantity sold).
a) Factor out the price on the right-hand side of the formula.
R(p) = p(-0.08p-300)
----------------------------------
b) Write a formula D(p) for the monthly demand.
D(p) = -0.08p-300
-----------------------
c) Find D(3000).
D(3000) = -0.08*3000-300 = -540
====================================
Cheers,
Stan H.

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Area of a sail. The area in square meters for a triangular sail is given by A(x) = x^2 + 5x + 6.
a) Find A(5).
A(5) = 5^2 + 5*5 + 6 = 56 m^2
--------------------------------------
b) If the height of the sail is x + 3 meters, then what is
the length of the base of the sail?
Area = (1/2)base*height
A = (1/2)b*(x+3)
2A = b(x+3)
base = 2A/(x+3)
==================
Cheers,
Stan H.

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THe coins Phil has are worth $4.15. If they are just quarters and dimes, how many of each coin does he have?

There are 8 possible solutions:

19, 22, 25, 28, 31, 34, 37, or 40 coins:

  1 quarter makes $0.25 plus 39 dimes, $3.90 makes $4.15 with 40 coins.
 3 quarters makes $0.75 plus 34 dimes, $3.40 makes $4.15 with 37 coins.
 5 quarters makes $1.25 plus 29 dimes, $2.90 makes $4.15 with 34 coins.
 7 quarters makes $1.75 plus 24 dimes, $2.40 makes $4.15 with 31 coins.
 9 quarters makes $2.25 plus 19 dimes, $1.90 makes $4.15 with 28 coins.
11 quarters makes $2.75 plus 14 dimes, $1.40 makes $4.15 with 25 coins.
13 quarters makes $3.25 plus 9 dimes,   $.90 makes $4.15 with 22 coins.
15 quarters makes $3.75 plus 4 dimes,   $.40 makes $4.15 with 19 coins.

Edwin

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4.15 in quarters & dimes
Q----D--=4.15
-------------------
15---4--=4.15
13---9--=4.15
11---14-=4.15
9----19-=4.15
7----24-=4.15
5----29-=4.15
3----34-=4.15
1----39-=4.15
PICK THE ANSWER YOU LIKE.

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THe coins Phil has are worth $4.15. If they are just quarters and dimes, how many of each coin does he have?
---------------
He might have 1 quarter and 39 dimes. Or 3 quarters and 34 dimes. Or 15 quarters and 4 dimes. There's no way to know.

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BELIEVE YOU LEFT OUT THE HOUR FIGURE????
The speed of the current in a river is 2mph. Jay travels 20 miles upstream and then 20 miles downstream in a total time of 5 1/3 hours. Find the speed of the boat. Let s be the speed of the boat.
D=ST
T=D/S
5.33=20/(S-2)+20/(S+2)
5.33=(20[S+2]+20[S-2])/(S+2)(S-2)
5.33=(20S+40+20S-40)/(S^2-4)
5.33=40S/(S^2-4)
5.33S^2-21.333=40S
5.33S^2-40S-21.333=0
uSING THE QUADRATIC EQUATIONWE GET:
S=(40+-SQRT[-40^2-4*5.33*-21.333])/2*5.333
S=(40+-SQRT[1,600+455])/10.6667
S=(40+-SQRT2,055)/10.6667
S=(40+-SQRT45.333)/10.667
S=(40+45.333)/10.667
S=85.333/10.667
S=8 MPH. ANS FOR THE SPEED OF THE BOAT IN CALM WATERS.