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Komok [63]
3 years ago
12

Can someone help again I’m still in class

Mathematics
2 answers:
AnnZ [28]3 years ago
8 0

Answer:

X3

Step-by-step explanation:

6x3=18

5x3=15

4x3=12

malfutka [58]3 years ago
8 0
The answer 3x.
Explanation
6x3=18
5x3=15
4x3=12
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Solve 50r − 6 = 82r − 70
Basile [38]

Answer:

Step-by-step explanation: 50r - 6= 82r-70

Add 50r to 82r which equals to 132 r

Now add the the 70 to the left side so -6+70 should equal 76.

Now it should look like this 76=132r

To get the value of r you have to divide 76/132 on both sides of the equation and your answer should be 0.57=r

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3 years ago
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Flora's Florist charges a standard delivery charge of
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$10.00 = delivery charge

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nadezda [96]

Answer:

Alice wins if a total of 5 is rolled. Finn wins if a total of 9 is rolled.

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SOLVE ATTACHMENT 30 POINTS!!!!!!
Masteriza [31]

Answer:

<h2>B = \frac{6D + 18 {f}^{3} }{M}</h2>

Step-by-step explanation:

- 3 {f}^{3}  +  \frac{1}{6} MB = D

Multiply through by 6 to eliminate the fraction

That's

- 18 {f}^{3}  + MB = 6D

Move - 18f³ to the right side of the equation

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MB = 6D + 18 {f}^{3}

Divide both sides by M to make B stand alone

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\frac{MB}{M}  =  \frac{6D + 18 {f}^{3} }{M}

We have the final answer as

B = \frac{6D + 18 {f}^{3} }{M}

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3 years ago
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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
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