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marta [7]
3 years ago
14

To calculate the area of a rectangle we multiply length by width the length of this rectangle below is 9 inches and the area is

36x-27y square inches what is the width of the rectangle
Mathematics
1 answer:
charle [14.2K]3 years ago
8 0

Answer:

547

Step-by-step explanation:

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Step-by-step explanation:

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What is a prime and composite number?
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Read 2 more answers
Kelsie works at a bicycle shop as a salesperson. She records the number of bicycles she sells daily. Here is the probability dis
Andru [333]

Answer:

a) E(B)= \sum_{i=1}^n B_i P(B_i) =0*0.3+1*0.5+2*0.15+ 3*0.05=0.95

b) E(T)= \sum_{i=1}^n T_i P(T_i) =10*0.3+20*0.5+30*0.15+ 40*0.05=19.5

c) E(B^2)= \sum_{i=1}^n B^2_i P(B_i) =0^2 *0.3+1^2 *0.5+2^2 *0.15+ 3^2 *0.05=1.55

And the variance is given by:

Var(B) = E(B^2) -[E(B)]^2 = 1.55- [0.95]^2 =0.6475

And the deviation would be Sd(B) = \sqrt{0.6475}=0.8047

E(T^2)= \sum_{i=1}^n T^2_i P(T_i) =10^2 *0.3+20^2 *0.5+30^2 *0.15+ 40^2 *0.05=445

And the variance is given by:

Var(T) = E(T^2) -[E(T)]^2 = 445- [19.5]^2 =64.75

And the deviation would be Sd(T) = \sqrt{64.75}=8.047

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".  

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).  

And the standard deviation of a random variable X is just the square root of the variance.

Solution to the problem  

Part a

For this case we have the following info:

B            0            1             2              3

____________________________________

T           10           20          30            40

____________________________________

P           0.3          0.5        0.15          0.05

____________________________________

And we can calculate the expected value for the random variable B like this:

E(B)= \sum_{i=1}^n B_i P(B_i) =0*0.3+1*0.5+2*0.15+ 3*0.05=0.95

Part b

Similar to part a we can find the expected value for the random variable T like this:

E(T)= \sum_{i=1}^n T_i P(T_i) =10*0.3+20*0.5+30*0.15+ 40*0.05=19.5

Part c

In order to find the variance for B we need to calculate the second moment given by:

E(B^2)= \sum_{i=1}^n B^2_i P(B_i) =0^2 *0.3+1^2 *0.5+2^2 *0.15+ 3^2 *0.05=1.55

And the variance is given by:

Var(B) = E(B^2) -[E(B)]^2 = 1.55- [0.95]^2 =0.6475

And the deviation would be Sd(B) = \sqrt{0.6475}=0.8047

Similar for the random variable T we have:

E(T^2)= \sum_{i=1}^n T^2_i P(T_i) =10^2 *0.3+20^2 *0.5+30^2 *0.15+ 40^2 *0.05=445

And the variance is given by:

Var(T) = E(T^2) -[E(T)]^2 = 445- [19.5]^2 =64.75

And the deviation would be Sd(B) = \sqrt{64.75}=8.047

7 0
4 years ago
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