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3 years ago

What's the answer please help

Mathematics

2 answers:

Alina [70]3 years ago

4 0

55/550x100=55/55x10=1x10=10

Talja [164]3 years ago

3 0

$\frac{55}{550} * \frac{100}{1} =x\\
\frac{55*100}{550*1}=x\\
\frac{5500}{550}=x\\
x=10$

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Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,

vladimir1956 [14]

Answer:

(a) $C(6,0) = 1$ , $C(6,1) = 6$ , $C(6,2) = 15$ , $C(6,3) = 20$ , $C(6,4) = 15$ , $C(6,5) = 6$ , $C(6,6) = 1$ .

(b) $C(7,0) = 1$ , $C(7,1) = 7$ , $C(7,2) = 21$ , $C(7,3) = 35$ , $C(7,4) = 35$ , $C(7,5) = 21$ , $C(7,6) = 7$ , $C(7,7)=1$ .

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$ .

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$ , for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

$C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

$C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

$C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$ .

(b) The explanations to the solutions is just the calculations.

$C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

$C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

$C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$

$C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35$

$C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21$

$C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1$

For all the calculations just recall that 4! =24 and 5!=120.

6 0

3 years ago

The solution system to 3y-2x=-9 and y=-2x+5

Alecsey [184]

Answer:

$\boxed{(3,-1)}$

Step-by-step explanation:

Hey there!

Well to find the solution the the given system,

3y - 2x = -9

y = -2x + 5

So to find x lets plug in -2x + 5 for y in 3y - 2x = -9.

3(-2x + 5) - 2x = -9

Distribute

-6x + 15 - 2x = -9

-8x + 15 = -9

-15 to both sides

-8x = -24

Divide -8 to both sides

x = 3

Now that we have x which is 3, we can plug in 3 for x in y = -2x + 5.

y = -2(3) + 5

y = -6 + 5

y = -1

So the solution is (3,-1).

Hope this helps :)

4 0

3 years ago

In a class of 47 students, the number of males is 7 less than twice the number of females. How many females and how many males a

Rom4ik [11]

X=number of females
47-x=number of males
# of males is 2*(number of females)-7:
47-x= 2x-7
add x to both sides:
47=3x-7
add 7 to both sides:
54=3x
divide both sides by 3:
x= 18 females
47-18= 29 males

6 0

3 years ago

What does x equals to if 3( x + 7 ) =57

zepelin [54]

3(x + 7) = 57 |÷3

x + 7 = 57/3

x + 7 = 19 |-7

x = 19 - 7

x = 12

7 0

2 years ago

If you shift the quadratic parent function, f(x)=x^2, right 10 units, what is the equation of the new function?

ololo11 [35]

F(x)=(x-10)^2

Right is negative
Left is positive

Remember that - those become handy later on. Try to become really good at transformations!

Good luck!

8 0

3 years ago

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