SevenUP

U.S. Government bonds are sold at a discount and redeemed at maturity.

Par value * (1+APR)^-Y Y = years to maturity

Present value = 10000 * (1.0350)^-8

Present value = $7594.12

_7UP_

Here’s another way to do this. It’s a variation of Cramer’s rule. With this you don’t have to take the square root.

CPhill’s use of Heron's Formula is better because you are less likely to make a mistake.

Area of Triangle with vertices of A(-8,10) , B(6,17), C(2,-4)

Note that ABS is absolute value.

ABS(0.5[Xa(Yb –Yc) + Xb(Yc –Ya) + Xc(Ya –Yb) ])

ABS(0.5[-8(17-(-4)) + 6(-4-10) + 2(10-17))

ABS(0.5[8(21) + 6(-14) + (2(-7)])

ABS(0.5[-168 + (-84) + (-14)])

ABS(0.5[-266]

ABS( -133) = 133 acres

_7UP_

I like your avatar, justinas. It’s “just ‘n as” if I were in school. Look at my avatar, closely, in the profile image for my retort.

Vertically elevating a mass of 75Kg to 1.6m through a gravity field of 9.8m/s² requires a kinetic energy of

KE = m *g * h 

75kg *(9.8m/s²) * 1.6m =1176 Joules

The energy required to accelerate a mass of 75kg to a velocity of 0.8m/s is

KE= 0.5(m)*v²  

0.5(75kg) * (0.8m/s)² = 24 Joules

Total energy is the sum of these: 1176 + 24 = 1200 Joules

Extra energy will be required for overcoming wind resistance, and for curving and unfolding herself over the bar. The wind resistance should be easy to calculate. The rest not so much.

_7UP_

KE= 0.5m*v². Where m(a*s) is in kilo grams and v(elocity) is in meters per second.

Rearrange this formula to find velocity.

v= sqrt(2*KE/m)

v= sqrt(2*107/0.147)

v = 38.2 meters/second

That's 85.5 MPH (Typically considered a fast pitch for college and minor league. A skilled and practiced pitcher could throw this; most HS players coudn't come close).

_7UP_

Actually, I have three. One is a gift from my grandfather. It’s made from bamboo, and he used it when he was in college and my father did too.

Another is a gift from my cousin who goes to an Amish school in Pennsylvania, where electricity –even from a battery --is not used, so no calculators. He carved this himself, and it is more a work of art than practical.

The third one, my older brother found on eBay. It’s plastic and precise. That is the one I carry to school. It's a little longer than the others so it makes a better back scratcher. :)

_7UP_

Hi Ms. Melody,

My eyes are young and work well for reading fine print and old text books.:)

315's answers are correct. (IMHO)

The second answer can be reduced.

I was reading a similar question in an ancient text book that uses look up tables and slide rules.

The formula is just a variation on the circumference of a circle.

T= (2pi *L*sin(theta) )/Sqr(L*g*sin(theta)*tan(theta))

Rearranges to

T= 2pi *sqrt(L²*sin²(theta))/(g*L *sin(theta)*tan(theta))

Reduces to

T= 2pi *sqrt(L*cos(theta)/g)

Though mathematically the same, these ancient text books often simplified the equations as much as possible to make the use of a slide rule easer.

As far as I know, I’m the only student in my school who uses a slide rule. They never need batteries and it makes a great back scratcher. Hahaha

_7UP_

Thank You CPhill. I have learned a lot from you, Melody and Alan!

ncr(20,5)= 15504 ----Total number of ways to select 5 of twenty.

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

A.) Solution for at least 4 women

ncr(11,5)*ncr(9,0) = 462 Number of ways to select 5 of 11 females and select 0 of 9 males

ncr(11,4)*ncr(9,1) = 2970 Number of ways to select 4 of 11 females and select 1 of 9 males

Sum of combinations total = 3432. So there are 3432 ways to select this group of 5 where 4 or more are women.

(3432/15504 = 22.1%) Incidental probability of this occurring.

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B.) Solution for no more than 2 men.

ncr(11,5)*ncr(9,0)= 462 Number of ways to select 5 of 11 females and select 0 of 9 males.

ncr(11,4)*ncr(9,1) = 2970 Number of ways to select 4 of 11 females and select 1 of 9 males.

ncr(11,3)*ncr(9,2) = 5940 Number of ways to select 3 of 11 females and select 2 of 9 males.

Sum of combinations total = 9372. So there are 9372 ways to select this group of 5 where 2 or less are men.

(9372/15504 = 60.4%) Incidental probability of this occurring.

I doubled checked this, but Melody, CPhill, or Alan should verify the answers. 

PS This assumes males are men and females are women. My dad says it is not easy to tell any more. So far, I don’t seem to have a problem. LOL

~7UP~

Part A)  From the observer’s point of view it would seem like a free fall from rest, because the ceiling, floor and person in the elevator have the same velocity. So, standard gravity acceleration would apply here.

A(Z) =-9.8m/t²* Z₀ 3m

Z(f) = 0

0= -0.5gt² +Z₀

T=sqr(2Z₀/g)* 2*3.0m)/9.8ms² = 0.782s

 0.782 seconds to reach the floor.

Part B) 9.8ms²*0.782 = 7.67 m/s velocity at impact with floor

Part C) Observer at relative rest.* V of Z0 = +2.5m/s (initial)

A(Z) = -9.8m/s² * Zf =V(Z0) – gt = (2.5m/s) -(9.8m/s₂) (0.782s)

Zf = -5.16m/s. Velocity is (5.16m/s) for observer at relative rest.

Part D)  The floor will rise ((2.5m/s)(0.782s)) =1.96m during the fall of the bolt.

The bolt appears to fall (3.00m -1.96m) =1.04m

The bolt will appear to fall 1.04m for observer at relative rest.

Relative reference frames are a little confusing. I can hardly wait until I get into the quantum s**t. I doubled checked this, but Melody, CPhill, and Alan should confirm this before you accept it.

 --7UP--