Member: Rom

Quote:
If a friend, at 10 am, drives at a rate of 30 mph, then stops at s red light for 4 minutes then proceeds for 6 more miles at 20 mph, what is the total travel time for him?

drives how far at a rate of 30 mph? As written this problem can't be answered.

RomJan 30, 2014

+6251

Quote:
4+x-2(x-6)=8 what is x?

4+x-2(x-6) = 8

4+x-2x+12 = 8

-x + 16 = 8

-x = -8

x = 8

RomJan 30, 2014

+6251

where exactly is the stereo located? Suppose your room is ABCD and the speaker is up in the corner at B.

RomJan 30, 2014

+6251

Quote:
A room has 32 columns, each 13 feet high and 2 feet in circumference. How much ribbon do I need to wrap each column 8 times?

how wide is the ribbon?

RomJan 30, 2014

+6251

what do you mean?

the interval between a<b is usually denoted by either

[a, b) if a is included but b isn't

[a,b] if both a and b are included

(a, b) if neither a nor b are included

(a,b] if b is included but a isn't

RomJan 30, 2014

+6251

http://web2.0calc.com/questions/n-and-k

RomJan 30, 2014

+6251

Quote:
Two cards are drawn at random from a pack of 8 cards numbered from 1 to 8. Find the conditional probability that:
(a) the sum of the two numbers is odd given that their product is even.
(b) the product of the two numbers is odd given that their sum is even.

First we note that
ExE -> E
ExO -> E
OxO -> O

E+E -> E
E+O -> O
O+O -> E

There are 4 even cards. We can choose any of 4, and then there are 3 left so we can choose any of 3. That gives 12 total EE pairs. But we don't care about the order so there are only 6 different pairs.

The same applies to the OO pairs. There are 6 different OO pairs.

The total number of different pairs is (8*7)/2 because we can choose any of 8, then any of 7, but again we don't care about the order, so there are 28 total different pairs.

Thus there must be 28 - 2*6 = 16 EO pairs

so in sum

6 different EE pairs
6 different OO pairs
16 different EO pairs

a) the product is even so we know we have either an EO or EE pair. Of those only the EO pairs give an odd sum.

so the conditional probability Pr[odd sum |even product]= (#EO pairs)/(#EO pairs + #EE pairs) = 16/(6+16) = 16/24 = 2/3

b) the sum is even here so we are looking at the EE and the OO pairs and we want to know how many of those the product is odd. The product being odd means it's only the OO pairs.

so the conditional probability Pr[odd product |even sum] = (#OO pairs)/(#EE pairs + #OO pairs) = 6/(6+6) = 1/2

RomJan 30, 2014

+6251

Quote:
find the vaule of x when b=5 and c=2
x = 2(b-c power of 2)

c^2 = 2^2 = 4

b-c^2 = 5-4 = 1

x = 2(b-c^2) = 2(1) = 2

RomJan 30, 2014

+6251

Quote:
When you simplify ,3 to the 56 power what number is in the ones place

well let's take a look

3 9 27 81 243 729 2187 ...

so the last digit repeats as
1 2 3 4 5 6 7 8 (power)
3 9 7 1 3 9 7 1 ...

so
3^(4n+0) -> 1

3^(4n+1) -> 3

3^(4n+2) -> 9

3^(4n+3) -> 7

56 = 4*14 + 0

so 3^56 will end in 1

RomJan 30, 2014

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