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3 years ago

Chee's family took a road trip to Niagara Falls. Chee fell asleep 19% of the way

Mathematics

1 answer:

12345 [234]3 years ago

7 0

190 miles
You just needed to find 19% of 1000

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Given the following exponential function, identify whether the change represents growth or decay, and determine the percentage r

postnew [5]

Answer: -34738.4259228

Step-by-step explanation:

It’s a decrease because your dividing 1600e -0.1252

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3 years ago

Which equation has the soluion x=4

AfilCa [17]

Answer:

3x=12

Step-by-step explanation:

3 times X with x being 4 making it 12 so X=4 and 3x=12

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3 years ago

Chicago has a temperature of -8°F. Seattle has a temperature colder than Chicago. Select all value that could represent the temp

mihalych1998 [28]

Answer:

-10°F, -13°F, and -21°F

Step-by-step explanation:

we are looking for a temperature that is less than -8

-10°F, -13°F, and -21°F are all less than -8

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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.

4 0

3 years ago

Please help again I need it

Ilya [14]

The two angles are verticals angles so they are congruent, so you can solve for x by writing and solving an equation

6x-82=3x-23

Combine like terms

3x=105

And simplify to get x=35

4 0

3 years ago

Read 2 more answers