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3 years ago

In which quadrant is the ordered pair ( 7, -8 ) located? *

Mathematics

2 answers:

Semmy [17]3 years ago

6 0

The ordered pair is located in Quadrant IV

Paul [167]3 years ago

4 0

Answer:

i think IV

Step-by-step explanation:

because of the picture

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2×-y when ×= when 5 and y=12

Damm [24]

25-12=13 because if x=5 and y=12 just replace the letters with the numbers. Hope it helps mate

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3 years ago

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4(x-3)-(2x+5)=-3x-27

Georgia [21]

4(x-3)-(2x+5)=-3x-27
4*x - 4*3 - 2x - 5 = -3x - 27
4x - 12 - 2x - 5 = -3x - 27
2x - 17 = -3x - 27
2x + 3x = -27 + 17
5x = -10 / : 5
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3 years ago

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Df = g + 32 solve for d

LekaFEV [45]

Answer and Step-by-step explanation:

To solve for d, divide f from both sides of the equation.

df ÷ f = (g + 32) ÷ f

d = $\frac{g+32}{f}$ is the answer.

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2 years ago

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What is the perimeter of this figure? fractions

Alona [7]

Answer:

Step-by-step explanation:

You can add the numbers to get the perimeter

10 1/3 + 15 2/3 = 26

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26 + 42 1/2= 68 1/2

That is answer choice A

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2 years ago

A football team has a probability of .75 of winning when playing any of the other four teams in its conference. If the games are

Alexeev081 [22]

Answer:

0.3164 = 31.64% probability the team wins all its conference games

Step-by-step explanation:

For each conference game, there are only two possible outcomes. Either the team wins it, or they lose. The probability of winning a game is independent of any other game. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A football team has a probability of .75 of winning when playing any of the other four teams in its conference.

The probability means that $p = 0.75$ , and four games means that $n = 4$

If the games are independent, what is the probability the team wins all its conference games?

This is P(X = 4). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{4,4}.(0.75)^{4}.(0.25)^{0} = 0.3164$

0.3164 = 31.64% probability the team wins all its conference games

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3 years ago