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Aliun [14]
3 years ago
9

How would you do this last part?

Mathematics
1 answer:
alina1380 [7]3 years ago
8 0

Answer:

broken heart thats the last one lol

Step-by-step explanation:

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. (16/4) + (10-3)<br> Plzz I need an answer quickly. I really would appreciate it
musickatia [10]

Answer:

11

Explanation

(16/4)+(10-3)

       4+(10-3)

        4+7

          11

         

4 0
3 years ago
Madison is baking party favors she wants to make enough favor so each guest gets the same number of favors she knows there will
iris [78.8K]

Answer:

It would be 24

Step-by-step explanation:

6: 6, 12, 18, 24

8: 8, 16, 24

The least common number is 24, so she should make at least 24.

4 0
3 years ago
Help please help me with this. ​
boyakko [2]

\pi r^2h

\pi \times 7^2 \times 7

\pi \times 7^3

\displaystyle 343\pi

3 0
3 years ago
Read 2 more answers
Solve for x. <br><br> 8x/7+9=30
kodGreya [7K]
X=18 (3/8)
(8x/7) +9 = 30
subtract 9 from both sides
8x/7=21
multiply 7 from both sides
8x=147
divide 8 from both sides and you get
x=18 (3/8)
6 0
3 years ago
Suppose that a college determines the following distribution for X = number of courses taken by a full-time student this semeste
lidiya [134]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

Step-by-step explanation:

Previous concepts

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).  

Solution to the problem

For this case we have the following distribution given:

X          3      4       5        6

P(X)   0.07  0.4  0.25  0.28

We can calculate the mean with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74

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

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36

And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

3 0
3 years ago
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