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3 years ago

Can anyone simplify 24q2/8q-3

1 answer:

4 0

24q^2 / 8q^-3

lets break this down...
24/8 = 3

q^2 / q^-3....when dividing exponents with the same base, keep the base and subtract the exponents. q^2 / q^-3 = q^(2 -(-3) = q^(2 + 3) = q^5

put them together and u get : 3q^5

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: Sam is 12 years older than John. For the last four years, Sam and John have been friends. Five years ago, Sam was 5 times as o

Firlakuza [10]

Answer:

The present age of Sam is 20 years .

Step-by-step explanation:

Given as :

Let The age of John = J years

Let The age of Sam = S years

∵ Sam is 12 years older than John

so, The age of Sam = 12 + The age of John

i.e S = 12 + J .....1

Again

Before 5 years ago

The age of Sam = 5 times age of John

So, (S - 5) = 5 × (J - 5)

Or, S - 5 = 5 J - 25

Or, 5 J - S = 25 - 5

Or, 5 J - S = 20 .........2

Solving eq 1 and eq 2

[5 J - (12 + J)] = 20

Or, 5 J - 12 - J = 20

Or, 4 J = 20 + 12

Or, 4 J = 32

∴ J = $\frac{32}{4}$

i.e J = 8 years

So, The age of John = J = 8 years

Put The value of J into eq 1

∵ S = 12 + J

So, S = 12 + 8

i.e S = 20 years

So, The present age of Sam = S = 20 years .

Hence, The present age of Sam is 20 years . Answer

7 0

3 years ago

Read 2 more answers

Rewrite the expression as an equivalent expression that does not contain powers of trigonometric functions greater than 1.

Gnesinka [82]

Cos^4(x) =
(cos²x)² =
(1-sen²x)² =
1² - 2*1*sen²x + sen^4x =
1 - 2sen²x + sen^4x

6 0

4 years ago

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

B Find the class that contains the median. c Find the modal class.

Alex_Xolod [135]

Answer:

can you add a paper or more info to answer

Step-by-step explanation:

5 0

3 years ago

Anyone know how to do this

svlad2 [7]

Answer:

The area of the rectangle on the left side is

$9cm \: \times 4cm = 36 {cm}^{2}$

The area of the bottom rectangle is

$6cm \times 2cm = 12 {cm}^{2}$

The total area of the composite figure will be

$36 {cm}^{2} + 12 {cm}^{2} = 48 {cm}^{2}$

Step-by-step explanation:

The area of any given rectangle can be found by multiplying the length of that rectangle by its width. The rectangle on the left side has a length of 9cm but the width is unknown. To find the width, we subtract 6cm from the width of the bottom rectangle: 10cm. And that gives us 4cm.

Therefore, we can now calculate the area to be: length × width = 9cm × 4cm = 36cm²//

The area of the bottom rectangle can be found similarly by multiplying the length: 2cm by the width: 6cm of that rectangle. And the result gives us: 2cm × 6cm = 12cm²//

The total area of the composite figure is calculated by adding the results from the left and bottom rectangles together. And that gives us: 36cm² + 12cm² = 48cm²//

8 0

3 years ago