Mathematics College

What’s the correct answer for this?

Answers

Answer 1

Answer:

(0,2)

Step-by-step explanation:

2:4 means one part is 2/(2+4)=1/3 of AB and the other part is 2/3 of AB

Add 1/3 of the distance from -2 to 4. (1/3)(4+2)=2. -2+2=0 The x coordinate is 0

Subtract 1/3 of the distance from 6 to -6, (1/3)6+6)=4 6-4=2 The y coordinate is 2

The point is (0,2)

Related Questions

Point R is at (4, 1.5) and Point T is at (4, 2.9) on a coordinate grid. The distance between the two points is ____. (Input numbers and decimal point only, such as 8.2.)

Answers

Answer:

1.4

Step-by-step explanation:

Both points are on the vertical line x=4, so the distance between them is the difference of their y-coordinates:

2.9 -1.5 = 1.4

The points are 1.4 units apart.

A friend who works in a big city owns two cars, one small and one large. One-quarter of the time he drives the small car to work, and three-quarters of the time he takes the large car. If he takes the small car, he usually has little trouble parking and so is at work on time with probability 0.7. If he takes the large car, he is on time to work with probability 0.5. Given that he was at work on time on a particular morning, what is the probability that he drove the small car? (Give the answer to three decimal places.)

Answers

Answer:

The probability that he drove the small car is 0.318.

Step-by-step explanation:

We are given that a friend who works in a big city owns two cars, one small and one large. One-quarter of the time he drives the small car to work, and three-quarters of the time he takes the large car.

If he takes the small car, he usually has little trouble parking and so is at work on time with probability 0.7. If he takes the large car, he is on time to work with probability 0.5.

Let the Probability that he drives the small car = P(S) = [tex]\frac{1}{4}[/tex] = 0.25

Probability that he drives the large car = P(L) = [tex]\frac{3}{4}[/tex] = 0.75

Also, let WT = event that he is at work on time

So, Probability that he is at work on time given that he takes the small car = P(WT / S) = 0.7

Probability that he is at work on time given that he takes the large car = P(WT / L) = 0.5

Now, given that he was at work on time on a particular morning, the probability that he drove the small car is given by = P(S / WT)

We will use the concept of Bayes' Theorem for calculating above probability;

So, P(S / WT) = [tex]\frac{P(S) \times P(WT/S)}{P(S) \times P(WT/S)+P(L) \times P(WT/L)}[/tex]

= [tex]\frac{0.25 \times 0.7}{0.25 \times 0.7+0.75 \times 0.5}[/tex]

= [tex]\frac{0.175}{0.55}[/tex]

= 0.318

Hence, the required probability is 0.318.

o) x² – 18x - 19 + 20y - y2,

Answers

Answer:

19x-19+23y

Step-by-step explanation:

add the like terms

30 - -5n + 16 + 3n =

Answers

Answer:

8n + 46

Step-by-step explanation:

Combine like terms. First, simplify the signs. two negatives directly next to each other equals one positive:

- (-5n) = + 5n

30 + 5n + 16 + 3n = (30 + 16) + (5n + 3n)

(30 + 16) = 46

(5n + 3n) = 8n

8n + 46 is your answer.

A department store wraps gifts for its customers. What is the amount of wrapping paper it would take to wrap a gift box with a width of 8 inches, a height of 6 inches, and a length of 5 inches?

Answers

Answer:[tex]236\ in.^2[/tex]

Step-by-step explanation:

Given

Dimension of gift box is

Width [tex]b=8\ in.[/tex]

height [tex]h=6\ in.[/tex]

Length [tex]L=5\ in.[/tex]

So, box is in shape cubiod

and its surface area is given by

[tex]S.A.=2(lb+bh+hl)[/tex]

[tex]S.A.=2(5\times 8+8\times 6+6\times 5)[/tex]