In the xy-coordinate plane shown below, point P has coordinates (8,-6). Which

Find an equation of the line that has the given slope and contains the given point. m=7/3 (1,6)

iris [78.8K]

The point-slope form:

$y-y_1=m(x-x_1)$

We have the point (1, 6) and the slope m = 7/3. Substitute:

$y-6=\dfrac{7}{3}(x-1)$ use distributive property

$y-6=\dfrac{7}{3}x-\dfrac{7}{3}$ add 6 to both sides

$y=\dfrac{7}{3}x+\dfrac{11}{3}$ multiply both sides by 3

$3y=7x+11$ subtract 7x from both sides

$-7x+3y=11$ change the signs

$7x-3y=-11$

Answer:

point-slope form: $y-6=\dfrac{7}{3}(x-1)$

slope-intercept form: $y=\dfrac{7}{3}x+\dfrac{11}{3}$

standard form: $7x-3y=-11$

5 0

3 years ago

You are traveling to Japan and need Japanese Yen (JPY). How much JPY could you get for $100 USD if the exchange rate is USD/JPY

Vanyuwa [196]

Answer:

107.15

Step-by-step explanation:

100/.9333=107.15

3 0

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.

$A = 0.2$

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

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

So

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

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

$A \cap B = 0.1$

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

2 years ago

-13=m-7 how can i solve for m

mart [117]

Answer: m=-6

Step-by-step explanation:

Add 7 to the other side

5 0

2 years ago

What is the coefficients and Constants for 5x-10

Artist 52 [7]

Answer:

coefficient is 5 and

constant is -10

7 0

3 years ago

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