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3 years ago

Solve by using Elimination.

Mathematics

1 answer:

Alexxx [7]3 years ago

6 0

<h3>Answer:</h3>

It is (3, -1)

<em>Hope this helped :)</em>

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Please help me with this question

cricket20 [7]

They give you: <RSP = 90 degrees

If you look in the diagram, <RSP is vertical to <UPW and vertical angles have the same measure. <RSP = <UPW

So the correct answer is A.

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3 years ago

What is the equation to the quadratic function that has zeros at -12 and 7?

icang [17]

I believe the answer to this question is the 3rd one

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3 years ago

Let’s write the possible subsets of the following sets and list the proper subsets separately

Arisa [49]

Step-by-step explanation:

(a) subsets are {apple}, { }

(b) subsets are {1}, {2}, {1,2}, { }

proper subsets are {1}, {2}

{a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}, { }

proper subsets are {a}, {b}, {c}, {a,b}, {a,c}, {b,c}

(d) subsets are {1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4},{ }

proper subsets are {1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4}

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3 years ago

One household is to be selected at random from a town. The probability that the household has a cat is 0.20.2 . The proba

dimulka [17.4K]

Answer:

There is a 50% probability that the household has a dog, given that the household has a cat.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a household has a cat.

B is the probability that a household has a dog.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a household has a cat but not a dog and $A \cap B$ is the probability that a household has both a cat and a dog.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that the household has a cat or a dog is 0.5

$a + b + (A \cap B) = 0.5$

The probability that the household has a dog is 0.4

$B = 0.4$

$B = b + (A \cap B)$

$b = 0.4 - (A \cap B)$

The probability that the household has a cat is 0.2.

$A = 0.2$

$A = a + (A \cap B)$

$a = 0.2 - (A \cap B)$

$a + b + (A \cap B) = 0.5$

$0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5$

$A \cap B = 0.1$

What is the probability that the household has a dog, given that the household has a cat?

20% of the households have a cat, and 10% have both a cat and a dog. So

$P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5$

There is a 50% probability that the household has a dog, given that the household has a cat.

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3 years ago

Do you estimate that 200 people attend a school concert. The actual attendance was 240 people find the percent error

harkovskaia [24]

The percent error is 20%.

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3 years ago