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3 years ago

Is the first quartile 12

Mathematics

1 answer:

likoan [24]3 years ago

5 0

No it is not because 12 is the median

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I haven’t really understood this yet

kakasveta [241]

A, F, and G.

The others have breaks in the range or don’t include all real numbers.

3 0

3 years ago

-12x - 12y = 4 3x + 3y = 0

Svetradugi [14.3K]

Answer:

no Solution

Step-by-step explanation:

-12x-12y=4\\ 3x+3y=0

12x-12y=4

add 12y to both sides

12x-12y+12y=4+12y

divid both sides by -12

\frac{-12x}{-12}=\frac{4}{-12}+\frac{12y}{-12}

simplfy

x=-\frac{1+3y}{3}

\mathrm{Substitute\:}x=-\frac{1+3y}{3}

\begin{bmatrix}3\left(-\frac{1+3y}{3}\right)+3y=0\end{bmatrix}

\begin{bmatrix}-1=0\end{bmatrix}

7 0

3 years ago

Factor the following expression: 16z+20

Evgen [1.6K]

Answer:

4z + 5

Step-by-step explanation:

16 and 20 are both dividable by 4, so you divide them both by four and then you have your answer

3 0

2 years ago

Read 2 more answers

What are the exact solutions of x2 − 5x − 7 = 0, where x equals negative b plus or minus the square root of b squared minus 4 ti

Marat540 [252]

Answer:

$x = \frac{5 + \sqrt{53}}{2}$ or $x = \frac{5 - \sqrt{53}}{2}$

Step-by-step explanation:

Given

$x^2 - 5x - 7 = 0$

Required

Solve for x using:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

First, we need to identify a, b and c

The general form of a quadratic equation is:

$ax^2 + bx + c = 0$

So, by comparison with $x^2 - 5x - 7 = 0$

$a = 1$ $b = -5$ $c = -7$

Substitute these values of a, b and c in

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-(-5) \± \sqrt{(-5)^2 - 4 * 1 * -7}}{2 * 1}$

$x = \frac{5 \± \sqrt{25 +28}}{2}$

$x = \frac{5 \± \sqrt{53}}{2}$

Split the expression to two

$x = \frac{5 + \sqrt{53}}{2}$ or $x = \frac{5 - \sqrt{53}}{2}$

To solve further in decimal form, we have

$x = \frac{5 + 7.28}{2}$ or $x = \frac{5 - 7.28}{2}$

$x = \frac{12.28}{2}$ or $x = \frac{-2.28}{2}$

$x = 6.14$ or $x = -1.14$

4 0

3 years ago

In 1985, there were 285 cell phone subscribers in Arlington. The number of subscribers increased by 75% per year since 1985. Fin

ANTONII [103]

Answer:

The number of subscribers in 2008 is 110,845,988

Step-by-step explanation:

The number of cell phone subscribers in Arlington, in t years after 1985, can be modeled by the following equation.

$N(t) = N(0)(1+r)^{t}$

And which N(0) is the number of subscribers in 1985 and r is the rate it increases per year, as a decimal.

In 1985, there were 285 cell phone subscribers in Arlington.

This means that $N(0) = 285$

The number of subscribers increased by 75% per year since 1985.

This means that $r = 0.75$

$N(t) = 285(1.75)^{t}$

Find the number of subscribers in 2008.

2008 is 2008-1985 = 23 years after 1985. So we have to find N(23).

$N(23) = 285(1.75)^{23} = 110,845,988$

The number of subscribers in 2008 is 110,845,988

8 0

3 years ago