Factorise completely

1 answer:

4 0

1. Number that adds up to -12 and multiplied by the same # that gives u -13. Set it up like this because both are negative so, u need 1 negative and 1 positive.
(x-13)(x+1)

2. Same rule applies. GCF
8y(4y+1)

3. (x+3)(x-3)

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Sylvie is going on vacation. She has already driven 58 miles in one hour. If she then travels for 6 hours at an average speed of

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The answer would be 360 miles

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3 years ago

Identify the constant of proportionality from the graph. 11 10 9 00 7 6 5 1 2 3 4 5 6 7 8 9 10 11 O A. 4 B. 8 O c. D. 3

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Answer:

A. 4

Step-by-step explanation:

Constant of proportionality (k) = y/x

We can use the coordinates of any point on the line to find k.

Let's use (2, 8)

Constant of proportionality (k) = 8/2

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2 years ago

The variance can never bea.zerob.larger than the standard deviationc.negatived.smaller than the standard deviation

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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