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2 years ago

How to write eight ten-thousandths of thirty-five million in scientific notation?

1 answer:

5 0

Eight ten-thousandths = 8*(1/10,000) = 8/10⁴ = 8 x 10⁻⁴.
Thirty-five million = 35*(1,000,000) = 35 x 10⁶

Therefore eight ten-thousandths of thirty-five million is
(8 x 10⁻⁴) * (35 x 10⁶)
= 280 x 10⁶⁻⁴
= 280 x 10²
= (2.8 x 10²) * 10²
= 2.8 x 10⁴

Answer: 2.8 x 10⁴

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2 years ago

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ABC Fence Company charges $35 more per foot of fencing.

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ABC charges $35 more per foot of fencing.; The slope of each function represents the rate per foot of fencing.

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1 year ago

From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def

jasenka [17]

Answer:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

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

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

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

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X 0 1 2 3

P(X) 0.92 0.03 0.03 0.02

We can calculate the expected value with the following formula:

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

And replacing we got:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

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

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

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2 years ago