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madreJ [45]
3 years ago
9

What is 3.2 × 10−4 m/h in centimeters per second? A. 8.889 × 10−6 cm/s B. 8.889 × 10−5 cm/s C. 1.152 × 10−5 cm/s D. 1.152 × 10−7

cm/s
Mathematics
1 answer:
atroni [7]3 years ago
4 0
3.2 x 10^{-4} x  \frac{100}{3600} = 8.888 x  10^{-6}

therefore ur answer is OPTION A 8.889 x 10-6 cm/s
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Answer:

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2 years ago
What all numbers whose absolute value is 6
GalinKa [24]

Answer:

6, -6

Explanation:

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3 0
3 years ago
Will give brainliest for *BEST* answer!
Nataliya [291]
Use substitution and substitute y into the other equation. One of the equations already gives you y in terms of x, so use that and substitute it into the other equation. y = 3x - 4

Plug into the other equation: -3y = -9x + 12 
-3(3x-4) = -9x + 12
-9x + 12 = -9x + 12
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8 0
3 years ago
What is the maximum volume in cubic inches of an open box to be made from a 12-inch by 16-inch piece of cardboard by cutting out
Romashka-Z-Leto [24]

If you start with a 12x16 rectangle and cut square with side length x, when you bend the sides you'll have an inner rectangle with sides 12-2x and 16-2x, and a height of x.

So, the volume will be given by the product of the dimensions, i.e.

(12-2x)(16-2x)x = 4x^3-56x^2+192x

The derivative of this function is

12x^2-112x+192

and it equals zero if and only if

12x^2-112x+192=0 \iff x = \dfrac{14\pm 2\sqrt{13}}{3}

If we evaluate the volume function at these points, we have

f\left(\dfrac{14-2\sqrt{13}}{3}\right) = \dfrac{64}{27}(35+13\sqrt{13})> f\left(\dfrac{14-2\sqrt{13}}{3}\right) = -\dfrac{64}{27}(13\sqrt{13}-35)

So, the maximum volume is given if you cut a square with side length

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5 0
3 years ago
The variance can never bea.zerob.larger than the standard deviationc.negatived.smaller than the standard deviation
MrRa [10]

Answer:

c.negative

True, by definition since the variance take in count the differences around the mean squared, then we can't have a negative value for a sum of square values.

Step-by-step explanation:

We need to remember that the variance is a measure of dispersion for a dataset respect to a measure of central tendency called the mean. The mean is defined as:

\bar x = \frac{\sum_{i=1}^n X_i}{n}

And the population mean is given by:

\mu= \frac{\sum_{i=1}^n X_i}{N}

The population variance is defined by:

\sigma^2 = \frac{\sum_{i=1}^n (X_i -\mu)^2}{N}

And the sample variance is defined as:

s^2= \frac{\sum_{i=1}^n (X_i -\bar x)}{n-1}

Now if we analyze the options we have this:

a.zero

False, if all the values are the same then the mean would be the same to the values and the difference between each value and the mean would 0 and indeed the variance would be 0.

b.larger than the standard deviation

False, if the population standard deviation is \sigma=2 then the variance would be \sigma^2 = 4 and as we can see is higher than the standard deviation

c.negative

True, by definition since the variance take in count the differences around the mean squared, then we can't have a negative value for a sum of square values.

d.smaller than the standard deviation

False, if the population standard deviation is \sigma=0.1 then the variance would be \sigma^2 = 0.01 and as we can see is lower than the standard deviation

7 0
3 years ago
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