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STatiana [176]
3 years ago
10

A standard deck of 52 playing cards has 4 suits with 13 different cards in each suit. How many different five-card hands are pos

sible?
A) 260 hands
B) 2,598,960 hands
C) 24,380 hands
D) 311,875,200 hands
Mathematics
2 answers:
jok3333 [9.3K]3 years ago
7 0
I think the answer is B
AlexFokin [52]3 years ago
7 0

Answer:

B) 2,598,960 hands

Step-by-step explanation:

This is a combination problem. So we can use the next formula:

C(n,k)=nCk=\frac{n!}{k!(n-k)!}

Where:

n=Number\hspace{3}of\hspace{3}elements=52\\k=Number\hspace{3}of\hspace{3}combinations= 5

So:

52C_5=\frac{52!}{5!(47!)}=2598960\hspace{3}hands

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3 years ago
A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far
ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3} (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2}) (1b)

Where:

x, y - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

r_{1}, r_{2} - Length of each vector, in kilometers.

\theta_{1}, \theta_{2} - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that r_{1} = 42\,km, r_{2} = 28\,km, \theta_{1} = 32^{\circ} and \theta_{2} = 154^{\circ}, then the resulting coordinates of the final position of the surveyor is:

(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})

(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]

(x,y) = (10.452, 34.531)\,[km]

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}

\theta \approx 16.840^{\circ}

And the distance from the camp is calculated by the Pythagorean Theorem:

r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}

r = 36.078\,km

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

5 0
3 years ago
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