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3 years ago

If abj=28, abc "congruent to" dbj, find jbc.

Mathematics

1 answer:

Rom4ik [11]3 years ago

8 0

DBJ + ABJ = 90 DBJ = ABC DBC + ABC = 90 JBC + DBC + ABJ = 90 And last but not least: ABJ = 28

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I need help with these 3 questions ASAP. please give me a step by step explanation.

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For number 2. The pilot flew 4 circuits and in total by the 5th day he flew 14 right?

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3 years ago

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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

A square outdoor patio with side length x feet is surrounded by a brick walkway that is 4 feet wide. What is the area of the wal

Masja [62]

Answer:

Area of walkway = 16(x+4) $ft^{2}$

Step-by-step explanation:

Length of the sides of the square outdoor = x feet

Area of the square outdoor = length x length

= x * x

= $x^{2}$ $ft^{2}$

The square outdoor is surrounded by 4 feet of walkway. So that,

length of the walkway = (x + 2(4))

= (x + 8)

Area of walkway with outdoor = (x + 8)*(x + 8)

= ( $x^{2}$ + 16x + 64) $ft^{2}$

Area of walkway = Area of walkway with outdoor - Area of the square outdoor

= ( $x^{2}$ + 16x + 64) - $x^{2}$

= 16x + 64

Area of walkway = 16(x+4) $ft^{2}$

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3 years ago

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Answer:

she would have to win 7 more games

Step-by-step explanation:

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Tiffany has $54. She is shopping for umbrellas that cost $12 each. She writes the following equation to model the situation. Pla

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Answer:

4 umbrellas

Step-by-step explanation:

Number of Umbrellas Total Money

1 $12

2 $24

3 $36

4 $48

She can't go over $48 because $48 + $12 is $60, which is more than she has ($54). So the max amount of umbrellas she can buy is 4.

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