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Yakvenalex [24]
4 years ago
6

One card is randomly selected from a deck of cards. Find the odds against drawing a three​

Mathematics
1 answer:
Burka [1]4 years ago
5 0

The answer is 12:1

note: this is often read as "12 to 1"

==============================================

Explanation

We have exactly four card that are labeled "3" and 52 cards total. So there are 52-4 = 48 cards that are not labeled "3".

From here you write the ratio 48:4 to indicate the odds against drawing a 3. There are 48 ways to lose, 4 ways to win. You list the number of ways to lose first when it comes to "odds against" type of problems.

Next step is to reduce the ratio 48:4 to get 12:1, note how I divided both parts by the GCF 4

48/4 = 12

4/4 = 1

So thats how 48:4 reduces to 12:1

If you focus on one suit, say the spades, then we have 12 cards that arent a 3 of spades while exactly one card is a 3 of spades. This is one way to see why 12:1 is the answer. If you were saying this out loud to a friend, then you might say something like "the odds against drawing a three are 12 to 1".

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NEED IT ASAP What is the value of x in the diagram below? A.18/ ROOT 3 B.18 ROOT 2/ROOT 3 C.18 ROOT 3/ ROOT 2 D. 18 ROOT 2
Ostrovityanka [42]

Answer:

x = B. 18 ROOT 2/ROOT 3

Step-by-step explanation:

There are two Triangles in the diagram above.

Step 1

We would solve for Triangle A First

Using Trigonometric function Cosine

cos θ = adjacent /hypotenuse

θ = 30°

Adjacent = 9 units

Hypotenuse = unknown

cos 30= 9/ Hypothenuse

Cross Multiply

cos 30 × Hypotenuse = 9

Hypotenuse = 9 / cos 30

cos 30 in surd form = √3/2

Hypotenuse = 9/√3/2

Hypotenuse = 9 × 2/√3

= 18/√3 units

Step 2

We would solve for the upper triangle = Triangle B

We are looking for x

θ = 45°

For Triangle B, the Hypotenuse we solved for in Triangle A is equivalent to the adjacent in Triangle B

Therefore ,

Hypotenuse for Triangle A = Adjacent side for Triangle B

θ = 45°

Adjacent = 18/√3 units

Hypotenuse = x = unknown

We would solve for this using Trigonometric function cosine

cos 45 = 18/√3 units / Hypothenuse

Cross Multiply

cos 45 × Hypotenuse = 18/√3 units

Hypotenuse = 18/√3 units / cos 45

cos 45 in surd form = 1/√2

Subtituting, we have

Hypotenuse (x) = 18/√3 units / 1/√2

Hypotenuse (x) = 18/√3 units ×√2/1

= 18× √2/ √3 units

= 18 √2/√3 units

Therefore x = Option B. 18 ROOT 2/ROOT 3

3 0
3 years ago
1 point
Firdavs [7]

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Answer:

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Step-by-step explanation:

To determine the value of a digit, set all other digits to zero. Your digit has a value of ...

  0.020  . . .  two hundredths = 2/100

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3 years ago
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6r-2(r+8)=11<br> Solve equation
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4 years ago
Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo
Shkiper50 [21]

Answer:

a) P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288

P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335

b) P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452

c) P(X \geq 8) = 1-P(X

d) P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that X \sim Poisson(\lambda=4)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda=4  , Var(X)=\lambda=2, Sd(X)=2

a. Compute both P(X≤4) and P(X<4).

P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183

P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733

P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465

P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646

P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)

P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692

b. Compute P(4≤X≤ 8).

P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595

P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298

P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452

c. Compute P(8≤ X).

P(X \geq 8) = 1-P(X

P(X \geq 8) = 1-P(X

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559

4 0
3 years ago
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