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OLEGan [10]
2 years ago
7

A card is drawn from a deck of cards, Then the card is replaced, the deck is reshuffled, second card is drawn. What is the proba

bility of getting.
1. a 3 on the first draw and a red card on the second draw?
2. an ace on the first draw and a heart on the second draw?
3. a queen on the first draw and a king on the second draw
4. a jack on the first draw and multiple of 5 on the second draw
Mathematics
1 answer:
LenaWriter [7]2 years ago
3 0

Answer:

https://socratic.org/questions/a-card-is-drawn-from-a-standard-deck-a-second-card-is-drawn-without-replacing-th-1#131182

Step-by-step explanation:

:)

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Which of the following correctly uses absolute value to show the distance between −60 and 13? (5 points)
ioda

A.)|−60 − 13| = |−73| = 73 units is the answer

7 0
3 years ago
Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat
drek231 [11]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

Step-by-step explanation:

For this case we have the following probability distribution given:

X          0            1        2         3        4         5

P(X)   0.031   0.156  0.313  0.313  0.156  0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

We can verify that:

\sum_{i=1}^n P(X_i) = 1

And P(X_i) \geq 0, \forall x_i

So then we have a probability distribution

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

6 0
3 years ago
What is the range of this graph
Arlecino [84]

Answer:

[0,12]

Step-by-step explanation:

7 0
3 years ago
ASAP! Guaranteed Brainliest for first decent answer
Gelneren [198K]

Answer:

First differences means take the ordered pairs in increasing order for their x coordinates (assuming the x's go by ones), then subtract each y (except the last) from the y that comes after it.  Keep doing that and you get the first differences.  (The second differences would be what you get if you then subtract each of the first differences from the one that came after it (TMI probably but ...)

Say the points were (1,3), (2,7), (3,11) and (4,15).  The first differences are

7-3 = 4

11-7 = 4

15 - 11 = 4  all 4.  This would also be the slope.  So that is the relationship - they are the same.

6 0
3 years ago
7 rectangles in every 140 cm.how many rectangles will be presented in 220 cm
Monica [59]
140/7=20
220/20=11
the answer is 11
5 0
3 years ago
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