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3 years ago

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To get to his office from home, Greg walks 7 blocks north and then 3 blocks east. After work he meets some friends at a cafe; to

get there he walks 2 blocks south and 6 blocks west. All blocks are 660 feet long. What is the straight-line distance?

1 answer:

Natali [406]3 years ago

7 0

Answer:

The straight-line distance in feet is 4.174,20

Step-by-step explanation:

To get this straight-line distance we can imagine both locations in a chart.

Work will be point (3, 7) (7 blocks north and then 3 blocks east)

Cafe must be calculate from the first point. Greg walks 2 blocks south (5)and 6 blocks west (-3)

Cafe= (-3, 5)

Formula of Distance between two points

d (P₁ y P₂)=√ ((x₂-x₁ )²+ (y₂-y₁ )²)

d (P₁ y P₂)=√ ((-3-3 )²+ (5-7 )²)

d (P₁ y P₂)=√ ((-6 )²+ (-2 )²)

d (P₁ y P₂)=√ (36+ 4)=√40= 6,3245 blocks

Each block is 660 feet long

The straight-line distance in feet is

d=6,3245 blocks *660 feet/block=4.174,20 feet

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A group consists of seven men and five women. Three people are selected to attend a conference.

Dima020 [189]

Answer:

a) 220

b) 10

c) 4.55% probability that the selected group will consist of all women.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the people are selected is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

a. In how many ways can three people be selected from this group of twelve?

$C_{12,3} = \frac{12!}{3!(12-3)!} = 220$

b. In how many ways can three women be selected from the five women?

Three women, from a set of 5. So

$C_{5,3} = \frac{5!}{3!(5-3)!} = 10$

c. Find the probability that the selected group will consist of all women.

Desired outcomes:

3 women from a set of 4. So

$D = C_{5,3} = \frac{5!}{3!(5-3)!} = 10$

Total outcomes:

3 people, from a set of 12. So

$T = C_{12,3} = \frac{12!}{3!(12-3)!} = 220$

Probability:

$p = \frac{D}{T} = \frac{10}{220} = 0.0455$

4.55% probability that the selected group will consist of all women.

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3 years ago

Ratios that are equivalent to 8 : 20

Sidana [21]

Answer: 16 and 40. 24 60

Step-by-step explanation:

6 0

3 years ago

If A(4 -6) B(3 -2) and C (5 2) are the vertices of a triangle ABC fine the length of the median AD from A to BC. Also verify tha

Gnoma [55]

Answer:

a) The median AD from A to BC has a length of 6.

b) Areas of triangles ABD and ACD are the same.

Step-by-step explanation:

a) A median is a line that begin in a vertix and end at a midpoint of a side opposite to vertix. As first step the location of the point is determined:

$D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2} \right)$

$D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2} \right)$

$D(x,y) = (4,0)$

The length of the median AD is calculated by the Pythagorean Theorem:

$AD = \sqrt{(x_{D}-x_{A})^{2}+ (y_{D}-y_{A})^{2}}$

$AD = \sqrt{(4-4)^{2}+[0-(-6)]^{2}}$

$AD = 6$

The median AD from A to BC has a length of 6.

b) In order to compare both areas, all lengths must be found with the help of Pythagorean Theorem:

$AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}$

$AB = \sqrt{(3-4)^{2}+[-2-(-6)]^{2}}$

$AB \approx 4.123$

$AC = \sqrt{(x_{C}-x_{A})^{2}+ (y_{C}-y_{A})^{2}}$

$AC = \sqrt{(5-4)^{2}+[2-(-6)]^{2}}$

$AC \approx 4.123$

$BC = \sqrt{(x_{C}-x_{B})^{2}+ (y_{C}-y_{B})^{2}}$

$BC = \sqrt{(5-3)^{2}+[2-(-2)]^{2}}$

$BC \approx 4.472$

$BD = CD = \frac{1}{2}\cdot BC$ (by the definition of median)

$BD = CD = \frac{1}{2} \cdot (4.472)$

$BD = CD = 2.236$

$AD = 6$

The area of any triangle can be calculated in terms of their side length. Now, equations to determine the areas of triangles ABD and ACD are described below:

$A_{ABD} = \sqrt{s_{ABD}\cdot (s_{ABD}-AB)\cdot (s_{ABD}-BD)\cdot (s_{ABD}-AD)}$ , where $s_{ABD} = \frac{AB+BD+AD}{2}$

$A_{ACD} = \sqrt{s_{ACD}\cdot (s_{ACD}-AC)\cdot (s_{ACD}-CD)\cdot (s_{ACD}-AD)}$ , where $s_{ACD} = \frac{AC+CD+AD}{2}$

Finally,

$s_{ABD} = \frac{4.123+2.236+6}{2}$

$s_{ABD} = 6.180$

$A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ABD} \approx 3.004$

$s_{ACD} = \frac{4.123+2.236+6}{2}$

$s_{ACD} = 6.180$

$A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ACD} \approx 3.004$

Therefore, areas of triangles ABD and ACD are the same.

4 0

4 years ago

15.60 to 11.70 percent of change

Marizza181 [45]

15.6 - 11.7 = 3.9

3.9 ÷ 15.6 = .25

25% of change

8 0

4 years ago

Solve the equation. −10x + 1 + 7x = 37

pantera1 [17]

−10x + 1 + 7x = 37

combine like terms

-3x + 1 = 37

subtract 1 from both sides

-3x = 36

divide both sides by (-3)

x = - 12

Answer

x = - 12

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3 years ago