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babymother [125]
3 years ago
10

Twelve people each have a standard deck of cards. Each person shuffles the deck and draws two cards (without replacement). Anyon

e who draws at least one diamond wins a prize. To the nearest whole number, about how many of the twelve will win a prize? What is the probability that at least one person will win a prize? (Answer correct to four decimal places.) What is the probability that two people will win a prize?
Mathematics
1 answer:
Vlad [161]3 years ago
7 0

Answer:

5

0.9990

0.0401

Step-by-step explanation:

1) Probability(diamond) = 13/52 = 1/4

Probability(not diamond) = 1 - 1/4 = 3/4

Probability of atleast one diamond = 1 - probability of no diamonds

= 1 - (¾ × ¾) = 1 - 9/16 = 7/16

Expected no. = np = 12 × 7/16

= 5.25 = 5 (nearest whole no.)

2) let X be the no. of people winning the prize.

Then X ~ B(12, 7/16)

[ X follows a Binomial Distribution with 12 trials and probability of success (winning prize) = 7/16 ]

P(atleast 1 winning) = 1 - P(X=0)

= 1 - (9/16)^12

= 0.9989966087

= 0.9990 (4 dp)

3) P(X=2) = 12C2 × (7/16)² × (9/16)^10

= 0.0400613268

= 0.0401 (4 dp)

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I need help with this
9966 [12]

Answer:

$2.75

Step-by-step explanation:

You would divide the total (13.75) by the amount of crayons he got (5 packs) and you get your answer.

7 0
3 years ago
Subtract. −13−(−6) a−19 b−7 c-19
Elza [17]

Answer:

-7

Step-by-step explanation:

-13-(-6)

-13+6

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6 0
2 years ago
Read 2 more answers
The radius of a semi-circle measures 5<br> units. Find its area.
shutvik [7]

Answer:

≈39.27 units

Step-by-step explanation:

To find the area of a semi circle, you use the formula for finding the area of a circle, which is \pir^{2}, and then you divide that by 2 since it is half of a circle.

In this case, \pi(5)^{2} ≈ 78.54

\frac{78.54}{2} = 39.27

8 0
3 years ago
Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices
Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt

Where P, Q are scalar functions

We want to compute

\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths \large C_1,C_2,C_3,C_4 which represents the sides of the rectangle.

\large C_1 is the line segment from (0,0) to (5,0)

\large C_2 is the line segment from (5,0) to (5,1)

\large C_3 is the line segment from (5,1) to (0,1)

\large C_4 is the line segment from (0,1) to (0,0)

Then

\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize \large C_1 as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

\large \displaystyle\int_{C_1}xydx+x^2dy=0

 Now the second integral. We parametrize \large C_2 as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25

The third integral. We parametrize \large C_3 as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2

The fourth integral. We parametrize \large C_4 as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

\large \displaystyle\int_{C_4}xydx+x^2dy=0

So

\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2

Now, let us compute the value using Green's theorem.

According with this theorem

\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx

where A is the interior of the rectangle.

so A={(x,y) |  0≤ x≤ 5,  0≤ y≤ 1}

We have

\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x

so

\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2

3 0
3 years ago
Ik this is easy but i forgot how to do it help​
REY [17]

Answer:

7x

Step-by-step explanation:

4+3= 7

=7x

(since they both have the variable "x" we can combine them both)

have a good day! <3

8 0
3 years ago
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