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3 years ago

8 / (-8) = ? Could you guys help i am bad at math

Mathematics

2 answers:

timurjin [86]3 years ago

4 0

Answer:

-1

Step-by-step explanation:

Anything divided by itself is going to be 1 and since we are dividing it into a negative number its going to be a negative output so the answer would be -1

djverab [1.8K]3 years ago

4 0

Answer:

-1

Step-by-step explanation:

you can write 8/(-8) as -(8/8) as they would share the negative (both numerator and denominator).

Then solve, 8/8 = 1 so -(8/8) is -1.

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Give an example of each of the following, or argue that such a request is impossible.

Masteriza [31]

Answer and Step-by-step explanation:

Solution:

(a) Let (bn) be a bounded sequence of (an) then, Bolzano- Weierstrass theorem,

(bn) contains a subsequence that is converges.

(b) The sequence :

Xn = 1 + (-1) n / 2 + 1/n

Clearly, (xn) does not contain 1 or 0.

If we take sequence (x2n), then it converges to 1 and when we take sequence (x2n+1),

Then it converges to 0.

1,1/2

1, 1/2, 1/3

1, 1/2, 1/3, 1/4

i.e

(xn) = (1,1/2, 1, 1/2, 1/3, …)

Then, the subsequence (xn) that is identically ½ in second column and 1/3 is third column and so on.

It has sub sequence that converges to every point in infinite set {1, 1/2, 1/3, 1/4, 1/5…}.

(d) The sequence

{1, 1/2, 1/3, 1/4, 1/5…}

if we take sequence of given set, then, this set converges to zero. This is not contained in it.

If (xn) converges to every point in set {1, 1/2, 1/3 …} then there exists subsequence which converges to 0. And 0 does not belong to given set.

3 0

3 years ago

A computer chess game and a human chess champion are evenly matched. They play twelve games. Find probabilities for the followin

sleet_krkn [62]

Answer:

The probability that they each win six games is 0.225.

Step-by-step explanation:

Given : A computer chess game and a human chess champion are evenly matched. They play twelve games.

To find : The probability that they each win six games?

Solution :

Applying binomial distribution,

Here n=12 and p=0.5

$P(X=k)=\frac{n!}{k!(n-k)!}\times p^k\times (1-p)^{n-k}$

The probability that they each win six games is k=6.

$P(X=6)=\frac{12!}{6!(12-6)!}\times 0.5^6\times (1-0.5)^{12-6}$

$P(X=6)=\frac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6\times 5\times 4\times 3\times 2\times 6!}\times 0.015625\times 0.015625$

$P(X=6)=11\times 2\times 3\times 2\times 7\times 0.015625\times 0.015625$

$P(X=6)=0.225$

Therefore, The probability that they each win six games is 0.225.

3 0

3 years ago

What is the initial value of the equation shown?

Ksju [112]

Answer:

Y = 2

Step-by-step explanation:

if you were looking for y's value

3 0

3 years ago

If f(x) = <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D" id="TexFormula1" title="\frac{1}{9}" alt="\frac{1}{9}" align=

boyakko [2]

Answer:

Solution given:

f(x) = $\frac{1}{9}$ x-2

let f(x)=y

y = $\frac{1}{9}$ x-2

interchanging role of x and y

x= $\frac{1}{9}$ y-2

x+2= $\frac{1}{9}$ y

y=9(x+2)

y=9x+18

F-¹(x)=9x+18

and

F-¹(-1)=9*-1+18=-9+18=9

6 0

3 years ago

If two dice are rolled, what is the probability that their sum will be 10, given that both dice are odd?

8090 [49]

Answer:

I think it is either 1/6 or 1/8

Hope it helps :)

5 0

3 years ago