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3 years ago

Two lines, A and B, are represented by equations given below:

Mathematics

1 answer:

frozen [14]3 years ago

4 0

9514 1404 393

Answer:

(b) (−3, −5), because the point satisfies both equations

Step-by-step explanation:

Any solution to a system of two equations <em>must satisfy both equations</em>. The only reasonable explanation of "why" is the one associated with the answer shown above.

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What is the solution to this equation?

Zolol [24]

Answer:

Step-by-step explanation:

7 0

3 years ago

Let X be from a geometric distribution with probability of success p. Given that P(X > y) = (1 ???? p)y for any positive inte

nevsk [136]

Full Question

Let X be from a geometric distribution with probability of success p.

Given that P ( X > a + b | X > a ) = q ^ { b } = P ( X > b ) for any positive integer x.

Show that for positive integers a and b, P(X > a + b | X > a) = P(X > b).

Answer:

P(X > a + b | X > a) = P(X > b)

Proved --- See Explanation

Step-by-step explanation:

Given

P ( X > a + b | X > a ) = q ^ { b } = P ( X > b )

From the above.

We can derive the following

P(X > a), P(X > b) and P(X > a + b)

P(X > a) = q^a

P(X > b) = q^b

P(X > a + b) = q^(a + b)

Using the definition of conditional probability

P(X > a + b | X > a) can be represented by P(X > a + b and X > a)/ P(X>a)

X>a+b and X>a is equivalent to X>a+b since a+b is larger than a

So,

P(X > a + b and X > a)/ P(X>a) can be rewritten as

P(X>a + b)/P(X > a)

Bringing both sides together, we're left with

P(X > a + b | X > a) = P(X>a + b)/P(X > a)

By substituton

P(X > a + b | X > a) = q^(a+b)/q^a

P(X > a + b | X > a) = q^(a + b - a)

P(X > a + b | X > a) = q^(a - a + b)

P(X > a + b | X > a) = q^b

Since P(X > b) = q^b

So,

P(X > a + b | X > a) = P(X > b)

4 0

3 years ago

PLEASE HELP

77julia77 [94]

Answer:

Step-by-step explanation:

Vertical is usually the answer to functions because of how it is made.

Brainlist Pls!

5 0

3 years ago

What two estimates is the quotient 345\8 between

grandymaker [24]

345/8 is the another one yes

4 0

3 years ago

Read 2 more answers

How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.

Svetllana [295]

Naturally, any integer $n$ larger than 127 will return $127\equiv127\mod n$ , and of course $127\equiv0\mod127$ , so we restrict the possible solutions to $1\le n$ .

Now,

$127\equiv7\mod n$

is the same as saying there exists some integer $k$ such that

$127=nk+7$

We have

$\implies 120=nk$

which means that any $n$ that satisfies the modular equivalence must be a divisor of 120, of which there are 16: $\{1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\}$ .

In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

$127=21\cdot6+1\iff127\equiv1\equiv7\mod6$

(If we're allowing $n=1$ , then I see no reason we shouldn't also allow 2, 3, 4, 5, 6.)

5 0

3 years ago