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Of 80 people attending a dance 28 dance with their hands touching the floor. If four people at the dance are selected at random,

tensa zangetsu [6.8K]

Answer: The probability is 0.013

Step-by-step explanation:

Out of 80 people, 28 of them dance with the hands touching the ground.

Then, the probability of selecting at random someone that dances with the hands touching the floor is equal to the number of persons that dance in that way, divided by the total number of persons.

For the first selection, this is p1 = 28/80

now, the probability in the second selection will be 27/79, because we have selected 1 out of the 28 (and also 1 out of the 80)

for the third and the fourth we have:

p3 = 26/78

p4 = 25/77

Then the probability of the fourth events is equal to the product of the four probabilities:

P = (28/80)*(27/79)/(26/78)/(25/77) = 0.013

3 0

3 years ago

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Which of the following numbers is not divisible by 11 A)40579 B)109857 C)162383 D)110011

Vika [28.1K]

The answer is C

162,383/11= 14762.09091

8 0

3 years ago

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Please help me out. Will make as brainiest!

frutty [35]

Answer:

Quadratic function

Step-by-step explanation:

6 0

3 years ago

LORAN is a long range hyperbolic navigation system. Suppose two LORAN transmitters are located at the coordinates (−60,0) and (6

USPshnik [31]

LORAN follows an hyperbolic path.

The equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

The coordinates are given as:

$\mathbf{(x,y) = (-60,0)\ (60,0)}$

The center of the hyperbola is

$\mathbf{(h,k) = (0,0)}$

The distance from the center to the focal points is given as:

$\mathbf{c = 60}$

Square both sides

$\mathbf{c^2 = 3600}$

The distance from the receiver to the transmitters is given as:

$\mathbf{2a = 100}$

Divide both sides by 2

$\mathbf{a = 50}$

Square both sides

$\mathbf{a^2 = 2500}$

We have:

$\mathbf{b^2 = c^2 - a^2}$

This gives

$\mathbf{b^2 = 3600 - 2500}$

$\mathbf{b^2 = 1100}$

The equation of an hyperbola is:

$\mathbf{\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1}$

So, we have:

$\mathbf{\frac{(x - 0)^2}{2500} + \frac{(y - 0)^2}{1100} = 1}$

$\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

Hence, the equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$