0
nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!]
= n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1}
= n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]}
= n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]}
= (n+1)!/[r!(n+1-r)!]
= n+1Cr
1 answer
You would take the following steps for G = A / (1-R):
G = A / (1-R)
Multiply by (1-R):
G * (1-R) = A
Divide by G:
(1-R) = A/G
1-R = A/G
Subtract 1:
-R = (A/G) - 1
Divide By -1:
R = -((A/G) - 1)
Check Work:
Original Problem:
A = 12; R = 5
G = 12 / (1-5)
G = -3
Solving For R:
R = -((12/-3)-1)
R = 5
Therefore, R= -((A/G)-1)
1 answer
Consider a denominator of r;
It has proper fractions:
1/r, 2/r, ...., (r-1)/r
Their sum is: (1 + 2 + ... + (r-1))/r
The numerator of this sum is
1 + 2 + ... + (r-1)
Which is an Arithmetic Progression (AP) with r-1 terms, and sum:
sum = number_of_term(first + last)/2
= (r-1)(1 + r-1)/2
= (r-1)r/2
So the sum of the proper fractions with a denominator or r is:
sum{r} = ((r-1)r/2)/r = ((r-1)r/2r = (r-1)/2
Now consider the sum of the proper fractions with a denominator r+1:
sum{r+1} = (((r+1)-1)/2
= ((r-1)+1)/2
= (r-1)/2 + 1/2
= sum{r) + 1/2
So the sums of the proper fractions of the denominators forms an AP with a common difference of 1/2
The first denominator possible is r = 2 with sum (2-1)/2 = ½;
The last denominator required is r = 100 with sum (100-1)/2 = 99/2 = 49½;
And there are 100 - 2 + 1 = 99 terms to sum
So the required sum is:
sum = ½ + 1 + 1½ + ... + 49½
= 99(½ + 49½)/2
= 99 × 50/2
= 2475
2 answers
Resistors in parallel equation is
!/R = 1/r(1)_ + 1/r)2) + 1/r(3)
Since they are all 8 ohms.
Then
1/R = 1/8 + 1/8 + 1/8
1/R = 3/8
R = 8/3
R = 2 2/3 = 2.6666....
2 answers
There is not the factor: there are two factors: (m - 1) and (1 - r)
2 answers
r=0,
Tr-r = 0 = r(T-1), since T != 1, then T-1 is non zero so r must be zero.
1 answer
I think you mean r2 -7r -8 = 0. If so, then factor it: (r+1)(r-8) = 0
Then set r+1 = 0, r+1-1=-1, r = -1 Set r-8 =0, r-8+8 = 8 so r = 8
1 answer
It is a linear equation in one variable r.
The solution to the equation is r = 1.
1 answer
The terms in row 22 are 22Cr where r = 0, 1, 2, ..., 22.
22Cr = 22!/[r!*(22-r)!] where r! = r*(r-1)*...*3*2*1 and 0! = 1
1 answer
The formula for the sum of the series r(1/n2-1/n2) is r(1-1/n2).
1 answer
Stroker and Hoop - 2004 C-A-R-R- Trouble 1-1 was released on:
USA: 1 August 2004
1 answer
1 - r/7
Another way to write the same thing is 7/7-r/7 which is
(7-r)/7
2 answers
Each number separated by a dot is an eight bit number, so conversion requires converting four numbers. To convert a decimal number to binary, divide by two in succession until the number zero is reached.
121 / 2 = 60 r 1
60 / 2 = 30 r 0
30 / 2 = 15 r 0
15 / 2 = 7 r 1
7 / 2 = 3 r 1
3 / 2 = 1 r 1
1 / 2 = 0 r 1
Then, you read remainders backwards:
1111001
If there's less than 8 numbers, add zeroes to the left:
01111001
Then, do this for the other numbers.
55 / 2 = 22 r 1
22 / 2 = 11 r 0
11 / 2 = 5 r 1
5 / 2 = 2 r 1
2 / 2 = 1 r 0
1 / 2 = 0 r 1
00101101
7 / 2 = 3 r 1
3 / 2 = 1 r 1
1 / 2 = 0 r 1
00000111
15 / 2 = 7 r 1
7 / 2 = 3 r 1
3 / 2 = 1 r 1
1 / 2 = 1 r 1
00001111
Once you have all the binary digits, simply stick them back together using dots:
01111001.00101101.00000111.00001111
1 answer
Surface area of sphere is 4 x pi x r x r
Volume of sphere is (4/3) x pi x r x r x r
Hence ((4/3) x pi x r x r x r) / (4 x pi x r x r) = 1/2
((4/3) x r) / (4) = 1/2
(1/3) x r = 1/2
r = 1 and 1/2
1 answer
This browser is totally bloody useless for mathematical display but...
The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]
Let n -> infinity while np = L, a constant, so that p = L/n
then
P(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)
= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)
= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)
= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)
Now lim n -> infinity of (1 - L/n)^n = e^(-L)
and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1
lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)
So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.
1 answer
factor out the common monomial, 5.
5(r^3-1)
Factor as a difference of cubes.
5(r-1)(r^2+r+1)
4 answers
The r+1 th term is nCr(-x)r where r = 0, 1, 2, ... , n.
and where nCr = n!/[r!*(n-r)!]
1 answer
If you start with an investment of I and the interest rate is r% per annum (compounded), then you want a solution to
2I = I(1 + r/100)24
or I = (1 + r/100)24
That is ln(2) = 24*ln(1 + r/100)
so that ln(1 + r/100) = ln(2)/24 = 0.02888
or (1 + r/100) = exp(0.02888) = 1.0293
and so r/100 = 0.0293 so that r = 2.93%
If you start with an investment of I and the interest rate is r% per annum (compounded), then you want a solution to
2I = I(1 + r/100)24
or I = (1 + r/100)24
That is ln(2) = 24*ln(1 + r/100)
so that ln(1 + r/100) = ln(2)/24 = 0.02888
or (1 + r/100) = exp(0.02888) = 1.0293
and so r/100 = 0.0293 so that r = 2.93%
If you start with an investment of I and the interest rate is r% per annum (compounded), then you want a solution to
2I = I(1 + r/100)24
or I = (1 + r/100)24
That is ln(2) = 24*ln(1 + r/100)
so that ln(1 + r/100) = ln(2)/24 = 0.02888
or (1 + r/100) = exp(0.02888) = 1.0293
and so r/100 = 0.0293 so that r = 2.93%
If you start with an investment of I and the interest rate is r% per annum (compounded), then you want a solution to
2I = I(1 + r/100)24
or I = (1 + r/100)24
That is ln(2) = 24*ln(1 + r/100)
so that ln(1 + r/100) = ln(2)/24 = 0.02888
or (1 + r/100) = exp(0.02888) = 1.0293
and so r/100 = 0.0293 so that r = 2.93%
2 answers
Yes.
y = e^x
dy/dx = e^x
Note: e^x = 1 + x + x²/2! + x³/3! + ... + x^r/r! + ...
→ d/dx(e^x) = 1 + x + x²/2! + x³/3! + ... + x^(r-1)/(r-1)! + ...
(The font is extremely bad: r! is r-exclamation mark = r factorial = r × (r-1) × (r-2) × ... × 2 × 1)
1 answer
If we have y=a(b)^t as the equation then take b from this equation
case !: If b <1
then b=1-r
r=1-b
this r is the decay factor
case 2:If b >1
then b=1+r
r=b-1
this is the growth factor
1 answer
The terms in row 29 are:
29Cr = 29!/[r!*(29-r)!] for r = 0, 1, 2, ... 29
where r! denotes 1*2*3*...*r
and 0! = 1
1 answer
Given T = R + RS
Lateral inversion makes it to be R + RS = T
Taking R as common factor, we get R(1+S) = T
Now dividing by (1+S) both sides, R = T / (1+S)
Hence the solution R = T/(1+S)
1 answer
r is worth - 0.542 mathematically if 1 2 r plus 6.5.
1 answer
1/R = 1/60 + 1/60+ 1/60+ 1/60 + 1/60
1/R = 5 / 60
R = 60/5
R = 12 ohms.
2 answers
It is
48/2= 24 r 0 ^
24/2= 12 r 0 |
12/2 = 6 r 0 |
6/2 = 3 r 0 |
3/2 = 1 r 1 |
1/2 = 0 r 1 |
So your read it from 1 to 48
:D Enjoy
1 answer
R/4 = 1/3
r/4*4 = 1/3 *4
r=4/3
1 answer
R-2 Has all the same laws as R-1 plus 1 more. R-2 allows for multiple family homes, such as apartment complexes.
1 answer
PIVFA=(1-(1/(1+R)^n)/R PIVFA(10%10)=(1-(1/(1+0.1)^10)/0.1=6.1446
1 answer
Use Pythagoras
3^2 = r^2 + (2 + r)^2
9 = r^2 + 4 + 4r + r^2
2r^2 + 4r = 5
Complete the Square
r^2 + 2r = 5/2
(r + 1)^2 - 1^2 = 5/2
(r + 1)^2 = 5/2 +1 = 7/2
r + 1 = +/- sqrt)7/2)
r = -1 +/-sqrt)7/2)
r = -1 +/- sqrt(3.5
r = -1 +/- 1.8708....
r = -2.8708.... (Unresolved!!!! Philosophically you cannot have a negative length).
& r = 0.8708.... The answer!!!!
1 answer
Emi = l * r * ((1 + r)^n / (1 + r)^n - 1) * 1/12
where l = loan amt
r = rate of interest
n = no of terms
1 answer
PVIFA = (1 - 1 / (1 + r)n) / r where n is the number of payment periods; r is the nominal interest rate for one period
1 answer
NPV=NFV/(1+r)^n The role of the "(1+r)^n" is to discount the future money to what it is worth in todays dollars. The 1 accounts to the sum itself and the plus r takes into account the interest rate. NPV=NFV/(1+r)^n The role of the "(1+r)^n" is to discount the future money to what it is worth in todays dollars. The 1 accounts to the sum itself and the plus r takes into account the interest rate.
1 answer
p = 50q/100 = 1/2 q
r = 40q/100 = 2/5 q
p = (1/2)/(2/5) = (1/2)(5/2) = 5/4 r or 1 1/4 r Thus, p is 125% of r.
1 answer
No. It more than doubles it.
It the interest rate is r%, the total interest increases, per unit invested, from
[(1+r/100)^2t - 1] - [(1+r/100)^t - 1] = (1+r/100)^2t - (1+r/100)^t
1 answer
Let the three numbers in GP: a/r, a, ar---------(A)
Where '^' is power of . . .
Sum of these numbers are:
a/r +a +ar = 38
a(1\r+1+r) = 38 ---- (1)
Product of these numbers are:
a^3 = 1728
= (12)^3
a = 12
Putting the value of a in (1) you will get:
12(1\r+1+r) = 38
And factorising, we get
r = 2/3 or r = 3/2
Sub. the r and a value in (A), we get
8,12,18 or 18,12,8. When a = 12
And smallest no. is 8.
1 answer
values of general gas constant
R = 0.082 atm. ℓ.mol-1.K-1
R = 8.314 Pa.m3.mol-1.K-1
R = 1.99Cal.mol-1.K-1
1 answer
The total value after 2 years is 15000 + 2496 = 17496.
So 17496 = 15000*(1 + r/100)2That is, (1 + r/100)2= 17496/15000 = 1.1664
(1 + r/100) = sqrt(1.1664) = 1.08
so r = 8% pa.
The total value after 2 years is 15000 + 2496 = 17496.
So 17496 = 15000*(1 + r/100)2That is, (1 + r/100)2= 17496/15000 = 1.1664
(1 + r/100) = sqrt(1.1664) = 1.08
so r = 8% pa.
The total value after 2 years is 15000 + 2496 = 17496.
So 17496 = 15000*(1 + r/100)2That is, (1 + r/100)2= 17496/15000 = 1.1664
(1 + r/100) = sqrt(1.1664) = 1.08
so r = 8% pa.
The total value after 2 years is 15000 + 2496 = 17496.
So 17496 = 15000*(1 + r/100)2That is, (1 + r/100)2= 17496/15000 = 1.1664
(1 + r/100) = sqrt(1.1664) = 1.08
so r = 8% pa.
2 answers
