May 16, 2013

what is the best way to hook up rockford fosgate p8002 amp?

Specs:400 rms x2 @ 2ohms
800 rms x1 @ 4ohms (bridged)
right now i have kicker cvr15s @2 ohms hooked up to each channel…would there be a benefit to bridge the amp and run at 4omhs?

Go to rockford fosgate.com and click on the index type cards on the page,you will find wiring diagrams and other use full information there.hope this helps.you best make sure your subs and your charging system can handle a 2 ohm load.most sub woofers are rated to run on a 4 ohm load but not all.its better to be safe than sorry.and you are going to need an alternator upgrade if you plan on running your system at a 2 ohm load.they call it the BIG 3 UPGRADE look into this.good luck.

How to find P given variance and sample size of two independent samples. Here is the question:?

Two different types of glass bottles are suitable for use by a soft drink beverage bottler. Internal pressure strength of the bottle is an important quality characteristic. It is known that ?1=?2=3.0 psi. From a random sample of n1=n2=16 bottles, the mean pressure strengths are observed to be
X1_bar = 175.8 psi and X2_bar = 181.3 psi. The company will not use bottle design 2 unless its pressure strength exceeds that of bottle design 1 by at least 5 psi. Based on the sample data , should they use bottle design 2? If we use ?= 0.05, What is the P-Value of the test?

Z = (181.3-180.8)/ sqrt(3/16+3/16) = 0.82
p = 1- 0.2939 = 0.7061 (from normal table)
don’t use the new bottles.

Aces. A standard deck of 52 cards is shuffled and dealt. Let X1 be the number of cards appearing before the f?

Aces. A standard deck of 52 cards is shuffled and dealt. Let X1 be the number of cards appearing before the first ace, X2 the number of cards between the first and second ace (not counting either ace), X3 the number between the second and third ace, X4 the number between the third and forth ace, and X5 the number after the last ace. it can be shown that each of these random variables Xi had the same distribution, i=1,2,…,5, and you can assume this to be true.

a)Write down a formula for P(Xi=k), 0<k<48.

Part a)

Well, the first one, X1, is probably the easiest to tackle, so let’s give that a try.

P(X1 = 0) = 4/52 = 1/13 (since it’s P(first card is Ace))
P(X1 = 1) = 48/52 (no ace first) * 4 / 51 (Ace second)
P(X1 = 2) = 48/52 * 47/51 * 4/50 (no ace, no ace, Ace)
P(X1 = 3) = 48/52 * 47/51 * 46/50 * 4/49 (no, no, no, Ace)

and so on up to P(X1 = 48) = 48/52 * 47/51 * 46/50 * ….. * 1/5 * 4/4

We can write those as
4/52 * 48/51 * 47/50 * … * however many factors
but simply rearranging the numerators, since multiplication is commutative
and associative.

So we have
P(X1 = k) = 4/52 * 48! / (48 – k)! * (51-k)! / 51! ? Formula (answer to part a)

For the above that gives:
k = 0: 4/52 * 48! / (48!) * 51! / 51! = 4/52 = 1/13
k = 1: 4/52 * 48! / (47!) * 50! / 51! = 4/52 * 48 / 0! * 50! / 51! = 4/52 * 48/51
k = 2: 4/52 * 48! / (46!) * 49! / 51! = 4/52 * 48 * 47 / (51 * 50) = 4/52 * 48/51 * 47/50
and so on

Part b)
The 4 aces divide the deck of 52 cards into 5 groups:
g1 ..A1 … G2 … A2 … G3 …. A3 ….g4 …. A4 …. G5
(where A* = aces and g* = other cards (0 or more) )

On average, the length of g[i] = 48/5 = 9.6
so no matter where the Aces fall, and all places are equally likely,
for each g[i] which is less than 9.6, there will be a compensating
length in one of the others to make up for it.

Hence E(Xi) = 9.6

Part c)
Sorry, I cannot help with this part.

I suspect they are not independent, since,
for any Xi above average, there has to be another one (at least)
which is below average, since their sum is always 48.

To take an extreme example,
if X1 = 48 (all the Aces are the last 4 cards in the deck),
then X2 = X3 = X4 = X5 = 0 necessarily.

Similarly, if X1 = 0 (Ace is first card dealt),
then necessarily one of the others will be above the average
to “make up for the shortfall”.

What Would Be The Best Motherboard For My Processor?

I am getting an Intel Core i7-2600k Processor, and was looking for a fairly good motherboard, with at least 2 PCI x16 slots and 1 PCI Express x1 slot. I am also planning on overclocking my cpu (not sure if im going to internally or on a workbench, just some info that might be usefull) Thanks.

>I was just looking at some motherboards on the ASUS site and I think this might be a really good one:

http://usa.asus.com/Motherboards/Intel_Socket_1155/P8Z68M_PRO/

Also, Asus has this other one that I like a lot after looking at its specs:

http://usa.asus.com/Motherboards/Intel_Socket_1155/P8Z68V_PRO/

Both of them are mid-range priced and for the fan user or gamer these would be a good purchase.

You can find them both on Newegg for sale.

Help me form my constraints with this linear programming model?

Its an advertising problem that has cost, audience reached and exposure quality of internet, radio, and bulletin ads. There are budget constraints but I know how to do that, but there is one line that reads, the number of internet ads cannot exceed the number of bulletin ads by more than 4. Internet ads are my X1 and bulletin ads are my X2.

Ok,
so the difference in the number of internet ads and bulletin ads can’t be more than 4.

Thus, X1 – X2 <= 4.
Solving this for X1, you get X1 <= X2 + 4.

My computer keeps freezing when I try to switch applications or minimize windows while using Word and Endnote?

I’m running Word 2007 and Endnote X1. All the word docs without references work fine, but as soon as I add a CWYW reference, the poo hits the fan. Any compatibility patches or settings anyone knows about?

How would you set up this Probability & Statistics Problem?

If x1+x2+x3 = 20 and each xi is greater than or equal to 1, how many ways can you arrange the values?

Firstly preplace 1 each in x1, x2 & x3
we now need a total of 17
apply the “stars and bars” formula to get
# of ways = (17+3-1)C(3-1) = 19C2 = 171 <———
http://jhyun95.hubpages.com/hub/Stars-and-Bars-Combinatorics

What is the best gaming laptop you can buy at Singapore at the moment?

I’ll be going to Singapore this coming November 4 for vacation and I’m planning on buying a gaming laptop. I’d just like to ask what is the best gaming laptop you can buy at the moment under 4000 SGD. So far I’m thinking of the Asus w90Vn. I’m also considering the Asus w90vp-x1 or the MSI gt725 if they are already available in the market. Thanks.