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3 years ago

14

What’s the answer???

1 answer:

stellarik [79]3 years ago

7 0

Answer: (3,4)

Each input can only have one output for a function

The other pairs give a different output for the input so that would make it a relation not a function.

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The sum of two numbers is -6. The difference is 16. Find the two numbers.

Bess [88]

Answer:

$-5\:\mathrm{and}\: 11$

Step-by-step explanation:

We can write the following system of equations:

$\begin{cases}x+y=6\\x-y=16\\\end{cases}$

Adding both equations together, we isolate $x$ and get:

$2x=22,\: x=11$

Plugging $x=11$ in any of the equations, we can solve for $y$ :

$11+y=6,\: y=-5$

Verify that the solution pair $(11, -5)$ works $\checkmark$

Therefore, the two numbers are $\fbox{$-5\:\mathrm{and}\: 11$}$ .

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2 years ago

ANSWER THIS QUESTION FOR BRAINLEST AND 10 POINTS

EleoNora [17]

Answer:
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M(2,4)

N(4,2)

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2 years ago

Read 2 more answers

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JulsSmile [24]

The answer is 120 lol

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2 years ago

Need help please help me # 32

Simora [160]

Number 2 will have the highest percent in sales

4 0

2 years ago

A city hosted a music festival that included three concerts. According to the sales database, 28% of the audience attended the f

Vitek1552 [10]

Answer:

$P(X\cap Y\cap Z)=0.05$

Step-by-step explanation:

From the question we are told that:

Percentage of audience in first concert $P(X)=0.28$

Percentage of audience in second concert $P(Y)=0.42$

Percentage of audience in third concert $P(Z)=0.30$

Audience Percentage at at-least one concert $P(X \cup Y \cup Z)=0.80$

Percentage of audience at first & second concert $P(X \cap Y)=0.10$

Percentage of audience in first & third concert $P(X \cap Z)=0.08$

Percentage of audience in second & third concert $P(Y\cap Z)=0.07$

Generally the equation for probability of attending all concerts $P(X\cap Y\cap Z)$ is mathematically given by

$P(X \cup Y \cup Z)=P(X)+P(Y)+P(Z)-P(X \cap Y)-P(X \cap Z)-P(Y\cap Z)+P(X\cap Y\cap Z)$

$P(X\cap Y\cap Z)=P(X \cup Y \cup Z)-P(X)-P(Y)-P(Z)+P(X \cap Y)+P(X \cap Z)+P(Y\cap Z)$

$P(X\cap Y\cap Z)=0.80-0.28-0.42-0.30+0.10+0.80+0.70$

$P(X\cap Y\cap Z)=0.05$

Therefore the probability that a randomly selected audient attended all the concerts

$P(X\cap Y\cap Z)=0.05$

8 0

2 years ago