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guapka [62]
4 years ago
5

1/8y verbal expression

Mathematics
1 answer:
Brums [2.3K]4 years ago
5 0

The algebraic expression 1/8y verbal expression is: one is divided by the product of 8 and a number.
 

To decipher the mathematical information in a problem, a verbal expression (or verbal model) uses words. From a verbal model,  an equation can often be written.

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If f(x)=8x+5, what is f(x)^-1?
Molodets [167]
F(8x+5)^-1=0.2

(8+5)^-1=0.077
5^-1=0.2
8x^-1=0.125
0.125-0.2=-0.075/0.2-0.125=0.075
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

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choli [55]
Yes, I'm relatively sure any number can be written in scientific notation.
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Five decimals between 0.42 and 0.44
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Answer: 0.421, 0.422, 0.423, 0.43, 0.434

Step-by-step explanation:

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Answer:

1 plane

I hope this helps!

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