One card is randomly selected from a deck of cards. Find the odds against drawingnbsp a king.

1 answer:

3 0

<span>There are 4 kings out of 52 cards.
So P(king) = 4/52 = 1/13
Odds against getting a king are 1 - P(king) = 1 - 1/13 = 12/13

hope this helps!</span>

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Pls help me someone

Lynna [10]

Y=-1/3x+4

x2-x1/ y2-y1
(0,4)(2,-2)
2-0/-2-4
1/-3

y-intercept is 4

y=-1/3x+4

3 0

3 years ago

The governor is making a visit to your school and you need to assign seating for 14 students. unfortunately there are only six s

Serjik [45]

First seat: 14 candidates to seat
Second seat: 13 candidates to seat
Third seat : 12 candidates
4th seat: 11 candidates
5th seat: 10 candidates
6th seat: 9 candidates

Number of different variations: 14*13*12*11*10*9 = 2,162,160 different ways,

Observe that is 14P6 = 14! / (14-6)! = 14! / 8! = 14*13*12*11*10*9*8! / 8! =

= 14*13*12*11*10*9

Answer: 2,162,160

4 0

3 years ago

You have 100 yards of fencing available to create an enclosure for farm animals. As the

Sunny_sXe [5.5K]

1. area of a pentagon when 1 side is x is $\frac{x^2}{4}\sqrt{5(5+2\sqrt{5})}$
so if perimiter is 100 then one side is 100/5 or 20
area will be $\frac{20^2}{4}\sqrt{5(5+2\sqrt{5})}$ ≈<span>688.2 square yards

2. area of a hexagon when 1 side is length x is $\frac{3x^2\sqrt{3}}{2}$
so if permiter is 100 then length of one side is 100/6 or about 16.666666666666
the area will be $\frac{3(16.6666666)^2\sqrt{3}}{2}$ ≈721.7 square yards

3. area of a regular octagon with side lenghts x is $2(1+\sqrt{2})x^2$
so if 8 sides then side length of each is 100/8 or 12.5
area will be $2(1+\sqrt{2})12.5^2$ ≈754.4 square yards

octagon wil have most space
you can tell because the more sides it has, the more it gets to a circle and the more area it encloses with a given perimiter
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3 0

3 years ago

F(x) = 3x2²2 = 13

snow_lady [41]

Answer: 5/13

Step-by-step explanation:

$f(2)=\frac{3(2)^{2}-7}{4(2)^{2}-3}=\boxed{\frac{5}{13}}$