Homework help...

[Maximus]

The length of each piece in a model railroad built in the HO scale is 1/87 of the actual length. Another popular model is the N scale, for which the scale is 1/160.

#1 Each car on a full length passenger train is 80 ft long. What is the length in inches of a model car in the HO scale? In the N scale?


#2 A diesel locomotive is 60 ft long. What is the length in inches of a model of the locomotive in the N scale.


Please show your work.

[Maximus]

[MattyFos.]

Quit grave digging........ Sir.

[Maximus]

Quit grave digging........ Sir.

I can't believe you hate my kid...no old school rep fo you.

[Scholar]

The length of each piece in a model railroad built in the HO scale is 1/87 of the actual length. Another popular model is the N scale, for which the scale is 1/160.

#1 Each car on a full length passenger train is 80 ft long. What is the length in inches of a model car in the HO scale? In the N scale?


#2 A diesel locomotive is 60 ft long. What is the length in inches of a model of the locomotive in the N scale.


Please show your work.

1) 80(12) = 960/87 = 11" ; 960/160 = 6"

2) 60(12) = 720/160 = 4.5"

The 11" is a rounding to the nearest one tenth of an inch.

[MattyFos.]

I can't believe you hate my kid...no old school rep fo you.

I hate your kid?



































Prove it.

[SuperSurge]

The length of each piece in a model railroad built in the HO scale is 1/87 of the actual length. Another popular model is the N scale, for which the scale is 1/160.

#1 Each car on a full length passenger train is 80 ft long. What is the length in inches of a model car in the HO scale? In the N scale?


#2 A diesel locomotive is 60 ft long. What is the length in inches of a model of the locomotive in the N scale.


Please show your work.

#1- O scale is the most popular model railroad scale in current use. Its ratio is 1:87 -- or one actual foor (12 inches) equals 87 HO scale feet. Using a conventional inch ruler, an HO scale foot would be about 1/8 of an inch long.

Example: a "forty foot" long boxcar model would be about 5-1/2 inches long in HO scale. (so 80 ft. would be 11 inches)


N scale is the second most popular model railroad scale in current use. Its ratio is 1:160 -- or one actual foot (12 inches) equals 160 N scale feet. Using a conventional inch ruler, an N scale foot would be about 1/16 of an inch long.

Example: a "forty foot" long boxcar model would be about three inches long in N scale. (so 80 ft. would be 6 inches)

Example: a "two story" house model would be about two inches tall in N scale.



#2-

1/160 is our scale, so I divide 1 by 160 = .00625 (easy enough) so .00625 = 1 inch

SOOO - .00625 x 720 (60 feet) = 4.5 inches!

[bob2the2nd]

1) 80(12) = 960/87 = 11" ; 960/160 = 6"

2) 60(12) = 720/160 = 4.5"

The 11" is a rounding to the nearest one tenth of an inch.

double checked.

[Maximus]

Thank you!

[JiggleIt]

Thank you!

Prove it.

[MattyFos.]

1) 80(12) = 960/87 = 11" ; 960/160 = 6"

2) 60(12) = 720/160 = 4.5"

The 11" is a rounding to the nearest one tenth of an inch.

#1- O scale is the most popular model railroad scale in current use. Its ratio is 1:87 -- or one actual foor (12 inches) equals 87 HO scale feet. Using a conventional inch ruler, an HO scale foot would be about 1/8 of an inch long.

Example: a "forty foot" long boxcar model would be about 5-1/2 inches long in HO scale. (so 80 ft. would be 11 inches)


N scale is the second most popular model railroad scale in current use. Its ratio is 1:160 -- or one actual foot (12 inches) equals 160 N scale feet. Using a conventional inch ruler, an N scale foot would be about 1/16 of an inch long.

Example: a "forty foot" long boxcar model would be about three inches long in N scale. (so 80 ft. would be 6 inches)

Example: a "two story" house model would be about two inches tall in N scale.



#2-

1/160 is our scale, so I divide 1 by 160 = .00625 (easy enough) so .00625 = 1 inch

SOOO - .00625 x 720 (60 feet) = 4.5 inches!

double checked.

Proven

[chrisrich91]

This is obviously developing into a classic "long troll". Max will continue to post math problems of increasing difficulty until he has us trying to solve Bernoulli differential equations for his fifth grader.

[Chardon]

Maxy, did you provide a workcited page? Something that shows where you got your information? Like which poster provided you the answer? If not, then this goes against everything you have taught us.

[Dave K]

The Crown Colony Condo Association is required by law to set aside funds to replace its roof. The current cost to replace the roof is $200000 and it will need to be replaced in 15 years. The cost of a roof is expected to increase at the rate of 4% per year. If the Condo invests in Treasuries yielding 10% paid semiannually, what semiannual payment is required to have enough funds to replace the roof in 15 years?

I cant get this one to work for the life of me...



formula to get cost of roof:

a=c(1+a)^t
A = amount accumulated.
c = current cost
a = rate %
t = years
formula to get payment required:

a=p* ((1+i)^n) -1
----------------------
i
n = number of payments
i = rate per payment period

[Jon]

The Crown Colony Condo Association is required by law to set aside funds to replace its roof. The current cost to replace the roof is $200000 and it will need to be replaced in 15 years. The cost of a roof is expected to increase at the rate of 4% per year. If the Condo invests in Treasuries yielding 10% paid semiannually, what semiannual payment is required to have enough funds to replace the roof in 15 years?

First you need to calculate the future value of the roof. And then you can calculate the present value of that. What you've got there is the future value formula. Keep in mind that the interest rate used in these formulas is always divided by the number of payments in the year.