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2 years ago

How much less is -200m than 450m?

Mathematics

1 answer:

saveliy_v [14]2 years ago

4 0

Answer:

-200 metre is less 450m because - 200 metre is negative.

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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.

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2 years ago

What is the volume of this solid? Recall the formula V = Bh.

Viktor [21]

Answer:

280

Step-by-step explanation:

V=Bh

V=1/2(7x10)8 *it is a triangluar pyrmid so the base is a triangle*

V=1/2x70x8

V=35x8

V=280

Answer:280

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2 years ago

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Show that 7x + 1 is equivalence to 2(2x - 1) + 3(x + 1)

marissa [1.9K]

7x + 1 = 2(2x - 1) + 3(x + 1)

7x + 1 = 4x - 2 + 3x + 3

7x + 1 = 7x + 1

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1 year ago

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Drag the tiles to the boxes to form correct pairs. Find the probability for the given situations, and then match the situations

frozen [14]

Answer:

3 (die)
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Step-by-step explanation:

Josie rolls a six-sided die 18 times. What is the estimated number of times she rolls a two? 3 = (1/6)(18)

Slips of paper are numbered 1 through 10. If one slip is drawn and replaced 40 times, how many times should the slip with number 10 appear? 4 = (1/10)(40)

A spinner consists of 10 equal- sized spaces: 2 red, 3 black, and 5 white. If the spinner is spun 30 times, how many times should it land on a red space? 6 = (2/10)(30)

A card is picked from a standard deck of playing cards 65 times and replaced each time. About how many times would the card drawn be an ace? 5 = (4/52)(65)

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The probability of a given event is the number of ways it can occur divided by the number of possibilities. For example, a 2 is one of 6 numbers on a die, so we expect its probability of showing up to be 1/6. The expected number of times it will show up in 18 rolls of the die is (1/6)(18) = 3.

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3 years ago

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Answer:

landslide

Step-by-step explanation:

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2 years ago

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