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3 years ago

What is 2.085×106 in standard form?

Mathematics

2 answers:

Gelneren [198K]3 years ago

8 0

2.085 x 10^6 =

move the decimal 6 places to the right: 2,085,000

svetoff [14.1K]3 years ago

6 0

Sorry my answer was deleted for some reason but the answer was c 2.085 x 106 in standard form is 2,085,000 which is c

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suppose a parabola has a vertex of (-4 7) and also passes through the point (-3,8). what is the equation of the parabola in vert

jasenka [17]

The equation would be y = (x+4)² + 7

Because adding 4 to 'x' creates a transformation 4 units left and adding 7 to the whole function moves the parabola up 7

Hope this helps!

8 0

3 years ago

Read 2 more answers

Please help thanks very much

erik [133]

X/(x+5) = (x-2)/(x+1)
x(x+1) = (x -2) (x + 5)
x^2 + x = x^2 - 2x + 5x - 10
x = 3x - 10
3x - x = 10
2x = 10
x = 5

answer is A
5

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3 years ago

If I'm putting £100 a month into a fund that has 13.03% annual interest annually for 48 years, what's the the total??

Katena32 [7]

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3 years ago

Wilma earns $9 per hour at her job. Then she spent $15 on a birthday gift for her best friend. Write an expression to represent

frozen [14]

Answer:

Wilma has $12.

Step-by-step explanation:

Since we don't know how many hours she worked, we know that Wilma earned 9x dollars at her job. Then she spends $15, so we can just subtract 15 from 9x.

$9x-15$

We can just replace the x with 3 to get:

$27-15\\12$

Hence, Wilma has $12.

3 0

2 years ago

Read 2 more answers

The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly

Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between $\\ (\mu - 6\sigma)$ and $\\ (\mu + 6\sigma)$ . It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

$\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027$

For those with defects operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects in a 5 sigma level, we have:

$\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697$ .

The probability of defects is:

$\\ 1 - p = 1 - 0.999999426697 = 0.000000573303$

Well, the defects present in a six sigma level and a five sigma level are, respectively:

$\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}$

$\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}$

Then, comparing both fractions, we can confirm that a 6 sigma level is markedly different when it comes to the number of defects present:

$\\ {6\sigma} \approx \frac{2}{10^9}$ [1]

$\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9}$ [2]

Comparing [1] and [2], a six sigma process has 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

8 0

3 years ago