All Answers 358272 Answers

you can use shoelace

from there you would get

-1 -1

-3  4

1   7

6   5

3  -1

1  -1

so the answer is 47

Help Please!!!

I need it like RIGHT NOW

Please Do quickly

\(\text{It sounds like your asking for the number of ways you can partition the integer 4}\\ \text{using the digits 1, 2, 3, and 4}\\ (1,1,1,1)\\ (2,1,1) \text{ and it's other 2 permutations}\\ (2,2)\\ (1,3) \text{ and it's other permutation}\\ (4)\\ \text{for a total of $1+ 3+1+2+1 = 8$}\)

\(\text{a) $p = \dfrac{\dbinom{12}{3}\dbinom{16}{2}\dbinom{18}{2}}{\dbinom{46}{7}}$}\)

\(\text{b) $P[\text{at least 2 red balls withdrawn}] = 1 - P[\text{0 or 1 red balls withdrawn}]$}=\\ 1 - \dfrac{\dbinom{12}{0}\dbinom{34}{7}}{\dbinom{46}{7}} - \dfrac{\dbinom{12}{1}\dbinom{34}{6}}{\dbinom{46}{7}}\)

\(\text{c) $P[\text{all balls the same color}]$}=\\ \dfrac{\dbinom{12}{7}+\dbinom{16}{7}+\dbinom{18}{7}}{\dbinom{46}{7}}\)

\(\text{d) $P[\text{3 red or 3 blue withdrawn}]$}=\\ \dfrac{\dbinom{12}{3}\dbinom{34}{4} + \dbinom{16}{3}\dbinom{30}{4}}{\dbinom{46}{7}}\)

\(\text{The least common multiple of 6 and 7 is 42}\\ \text{It will be 42 days}\)

Let her constant speed on Monday = S
Let the distance from San Antonio to Dallas =1000 miles
[Note: The distance does not matter, since the portion of the distance where she traveled at 10% faster will be constant]

Let the portion of the distance where she drove 10% faster =D
The portion of the distance where she drove at 30% slower =1000 - D
1000/S =D/(1.1S) + (1000-D)/(0.7S), solve for D

D =825 miles
825 / 1000 =82.5% portion of the distance where she drove at the faster speed.

\(\text{Any set of 4 distinct integers can be put in strictly ascending order}\\ \text{We can choose $\dbinom{13}{4}=715$ sets of distinct integers in $[0,12]$}\)

\(\text{b) is the same idea as (a) but now the integers don't have to be distinct}\\ \text{we have a few flavors of how the selected integers might appear}\\ \text{i) 4 distinct integers, ii) 1 pair, ii) 2 pairs, iv) 3 of a kind, v) 4 of a kind}\\ \text{For all these flavors there is only a single way they can be distinctly sorted in ascending order}\\ \text{so $N=\dbinom{13}{4}+13 \dbinom{12}{2} + \dbinom{13}{2}+13\cdot 12 + 13 = 1820$}\\ \text{where each of those terms corresponds to the number of each flavor that occurs}\)

you can do (c)

(a):

Focusing on the chocolate chip cookies, we notice that dividing 8 cookies into 4 groups is a classic stars/stripes problem. Distributing 8 cookies among 4 friends is C(11,3) ways.

In each of the above realities, you can also similarly divide 7 cookies into 4 groups to find C(10,3).

C(11,3)*C(10,3) = 165*120 = 19800.

(b):

Ways to distribute among 3 other children...

If twins get 0 cookies: C(8,2) = 28

1 cookie: C(6,2) = 15

2 cookies: C(4,2) = 6

3 cookies: I mean, the twins already have every cookie... (it's C(2,2) = 1)

Sum = 28 + 15 + 6 + 1 = 50 ways total.

https://web2.0calc.com/questions/a-count-the-number-of-quadruples-a-b-c-d-of-nonnegative

oh that makes sense to use the hockey stick identity. Thanks for the help!!

Thanks so much!!

\(\text{a) is equivalent to putting 50 identical balls into 4 distinct boxes}\\ \text{This is the well known stars and bars problem}\\ N=\dbinom{50+4-1}{4-1} = \dbinom{53}{3} = 23426\)

\(\text{b) is essentially the same as (a) but now there is a 5th box}\\ N=\dbinom{50+5-1}{5-1} = 316251\)

\(\text{Using the hockey stick identity}\\ \sum \limits_{n=13}^{54}\dbinom{n}{13} = \dbinom{54+1}{13+1} = \dbinom{55}{14}\)

Let x be the number of 5 kg weights and y be the number of 8 kg weights. Because there are 14 disks, x+y = 14.

Now, looking to weight; if there are x 5 kg weights, then they weigh 5x kg total. Similarly, the y 8 kg weights weigh 8y kg total.

So, if their total weight is 100 kg, 5x + 8y = 100.

From here, you can multiply both sides of the equation 'x+y = 14' 5 to get '5x+5y=70'.

You subtract '5x+5y=70' from '5x+8y=100' to get '3y=30', so dividing by 3 gets us y=10. There are 10 '8 kg' disks.

Going back to x+y = 14, we can substitute y=10 to find that x+10 = 14; subtracting 10 from both sides gets us x = 4. There are 4 '5 kg' disks.

Will this help?

Jacqueline has = J dollars

Rosselyn has    = J + 12 dollars.

Susan has         = J + 15 dollars.

J + J + 12 + J + 15 =$57

3J + 27 = $57

3J =$57 - $27

J =$10 what Jacqueline has.

$10 + $12 =$22 what Rosselyn has.

$10 + $15 =$25 what Susan has.

S * 17/60 =D,    
S * 32/60 =D+12, solve for D, S
S = 48 mph - the speed of the 2 cars
D =13.6 miles driven by one car
13.6 + 12 =25.6 miles driven by the 2nd car.

a -  [1 +2 +3+ 4 +5] =15 / 6^2 =5 / 12

b = 2 nCr 1 * 5^1 + 2 nCr 2  * 5^0 =11 / 6^2 =11 / 36

Assume the quadrilateral is a square, then AB = BC = CD = DA =a for some non-negative a

AM = BM = BE =EC = CF = FD = 1/2*a (given)

Triangle MAD is congruent to triangle EBA also congruent to FCB (Side-angle-side)

Therefore, angle EAB = angle MDA, angle AMD = angle BEA (corresponding angles are equal)

angle EAB + angle BEA = 90 degrees \(\Rightarrow\)angle MDA + angle BEA = 90 degrees \(\Rightarrow\) angle AND and angle MNA are right angle

That is triangle MNA is similar to triangle AND and triangle MAD

MA / AD =1/2  \(\Rightarrow\) ND = 2*AN and AN = 2* MN \(\Rightarrow\) DN / MN =4

Similarly, you can prove angle BGE is a right angle, which yield Trinagle BGE is similar to BCF (Angle-Angle)

BF=\(\frac{\sqrt{5}}{2}a\)  by pythagorean theorem, cos of angle CBF = BC / BF = \(\frac{2}{\sqrt{5}}\), BG = BE *cos CBF =1/2 * a*2/sqrt(5)=\(\frac{\sqrt5}{5}a\) 

GF= BF - GF =\(\frac{3\sqrt5}{10}a\), GE=1/2*BG=\(\frac{\sqrt5}{10}a\),FG/GE = 3 , Triangle BGE is congruent to triangle ANM, which yield MN =EG

Therefore FG/MN= FG/GE = 3

Connect BN, AG = AB*cos BAG = AB * cos CBF = \(\frac{2}{\sqrt{5}}a\)

Triangle BAG is similar MAN. AM/AB=1/2 \(\Rightarrow\) AN= 1/2 AG \(\Rightarrow\) GN= 1/2*AG = \(\frac{\sqrt5}{5}a\)

BN= \(\sqrt{{​​(\frac{\sqrt5}{5}a)}^{2}+{​​(\frac{\sqrt5}{5}a)}^{2}}=​​\frac{\sqrt{10}}{5}a\) by  pythagorean theorem, and MN =EG=\(\frac{\sqrt5}{10}a\)

BN/MN= \(2\sqrt{2}\)

Edit: highlighted answer

Thank you Guest and Omi67!

ohh, ok, so would we get k+m=7?

As long as  a queen  wasn't  the first draw, the probability  for  drawing a queen on the second draw is the same no matter what the  first  card was.....

This equals    4 / 51