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3 years ago

What is the value of c in the equation y = 4x^2 -10x + c if the y-coordinate of the vertex is -7

Mathematics

1 answer:

love history [14]3 years ago

6 0

Answer: $c=-\frac{3}{4}$

Step-by-step explanation:

Find the x-coordinate of the vertex with this formula:

$x=\frac{-b}{2a}$

In this case:

$a=4\\b=-10$

Substituting values, we get:

$x=\frac{-(-10)}{2(4)}=\frac{5}{4}$

Now we can substitute the coordinates of the vertex into the equation $y = 4x^2 -10x + c$ and then solve for "c".

Then:

$-7 = 4(\frac{5}{4})^2 -10(\frac{5}{4}) + c\\\\-7+\frac{25}{4}=c\\\\c=-\frac{3}{4}$

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Answer:

$X_(r) >> X_(n)$

The mean for this case would increase since is defined as:

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

The interquartile range would not change since the definition for the IQR is $IQR =Q_3 -Q_1$ and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

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

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

Step-by-step explanation:

For this case we assume that we have a random sample given $X_(1), X_(2) ,..., X_(n)$ and for each observation $X_i \sim N(\mu, \sigma)$ since the problem states that the data is approximately normal.

Let's assume that the largest value on this sample is $X_(n)$ and for this case we are going to replace this value by another one extremely higher so we satisfy this condition:

$X_(r) >> X_(n)$

The mean for this case would increase since is defined as:

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

The interquartile range would not change since the definition for the IQR is $IQR =Q_3 -Q_1$ and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

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

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

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