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katrin [286]
3 years ago
14

On Wednesday afternoons, eight men play tennis on two courts. They know ahead of time which four will play on the north court an

d which four will play on the south court. The players arrive randomly at the tennis courts. What is the probability that the first four players that arrive are all assigned to (a)the north court, and (b)the same court
Mathematics
2 answers:
Bond [772]3 years ago
7 0

Answer:

50% for both

Step-by-step explanation:

but otherwise b they will go to the farthest court so the others can go to the closest coart

musickatia [10]3 years ago
6 0

Answer:

(A)

Step-by-step explanation:

You might be interested in
Rectangle ABCD has vertices at (-7,-2); (1, -2); (1, -8); and (-7, -8) respectively. If GHJK is a similar rectangle where G(2, 5
Gennadij [26K]

Answer:

Points J and K could be located at:

J(2,2)

K(6,2)

Step-by-step explanation:

Consider the vertices have a x and y coordinate:

A: x coordinate=-7  y coordinate=-2

B: x coordinate=1  y coordinate=-2

C: x coordinate=1  y coordinate=-8

D: x coordinate=-7  y coordinate=-8

G: x coordinate=2  y coordinate=5

H: x coordinate=6  y coordinate=5

Then it is possible to calculate the distance between the x and y coordinates:

x coordinate of Vertices AB:

x coordinate of B- x coordinate of A=1-(-7)=8

The distance between A and B is 8

y coordinate of Vertices AC:

y coordinate of A- y coordinate of C=-2-(-8)=6

The distance between A and C is 6

Then we know that the side AB of Rectangle ABCD measures 8 and the side AC, measures 6.

Repeat the analysis  with Rectangle GHJK. In this case, is only possible to calculate the distance with x coordinate.

x coordinate of Vertices GH:

x coordinate of H- x coordinate of G=6-(2)=4

The distance between G and H is 4

We can see that the distance in x of the Rectangle ABCD is 8, and the distance in x of the Rectangle GHJK is 4, it means that <em>the measure of ABCD is twice GHJK.</em>

Then, if the distance in y coordinate of Vertices AC is 6, we could say that the distance in<em> y coordinate of Vertices GJ is 3.</em>

<em />

Points J and K could be located at:

J(2,2)

K(6,2)

3 0
3 years ago
B -5 ≥ -45 answer and steps
Natalka [10]

Answer:

-50

Step-by-step explanation:

b-5 \geq -45

b-5 = -45 ( collect like terms)

b = -45 + -5

b = -45 -5  b = -50

8 0
2 years ago
Dave is making a recipe for soup that includes 12 cups of water and 3 cups of broth. how many times as much water as broth will
OlgaM077 [116]

Answer: There is 4 times as much water than broth in the soup recipe.

Step-by-step explanation:

12/3 = 4

3 0
3 years ago
Read 2 more answers
Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)
Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

x^{2} +y^{2}+4x-1=0

Step-by-step explanation:

<u>Explanation:</u>-

<u>Step 1:</u>-

The equation of the circle having center and radius is

(x-h)^2+(y-k)^2=r^2

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

      (\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2}  )

    (-2,0)

therefore center (h,k) = (-2,0)

<u>Step 2:-</u>

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}

r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}

r=\sqrt{5}

<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius r=\sqrt{5}

The standard form of circle equation

(x-h)^2+(y-k)^2=r^2

(x-(-2))^2+(y-0)^2=\sqrt{5} ^2

on simplification is

x^{2} +y^{2}+4 x-1=0

5 0
3 years ago
What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?
Akimi4 [234]

Answer:

25.6 units

Step-by-step explanation: From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).

First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:

d=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2}}

where

(x_1,y_1) are the coordinates of the first point

(x_2,y_2) are the coordinates of the second point

- For AB:

d=\sqrt{[1-(-5)]^{2}+(4-4)^2}

d=\sqrt{(1+5)^{2}+(0)^2}

d=\sqrt{(6)^{2}}

d=6

- For BC:

d=\sqrt{(3-1)^{2} +(-4-4)^{2}}

d=\sqrt{(2)^{2} +(-8)^{2}}

d=\sqrt{4+64}

d=\sqrt{68}

d=8.24

- For AC:

d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}

d=\sqrt{(3+5)^{2} +(-8)^{2}}

d=\sqrt{(8)^{2} +64}

d=\sqrt{64+64}

d=\sqrt{128}

d=11.31

Next, now that we have our lengths, we can add them to find the perimeter of our triangle:

p=AB+BC+AC

p=6+8.24+11.31

p=25.55

p=25.6

We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.

Read more on Brainly.com - brainly.com/question/12560433#readmore

8 0
3 years ago
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