The line plots shows the number of hours two different age groups of teenagers spent online the previous week.

Mathematics

2 answers:

Serjik [45]3 years ago

4 0

The range is <span>is greater for data set 2.
- This is because it goes from 21-34 in the first set and 12-36 in the second.

The mean is greater for data set 2.
- This is because the first mean is 27.4 and the second is 28.05. </span>

ale4655 [162]3 years ago

3 0

Answer:

is greater for data set 2 ; is greater for data set 2

Step-by-step explanation:

The range is found by subtracting the highest and lowest value. The range of the first plot would be

34-21 = 13.

The range of the second plot would be

36-12 = 24.

The mean is found by adding together all of the data points and dividing by the number of data points. For the first plot, the sum of the data is 274 and there are 10 data points; this makes the mean 274/10 = 27.4.

For the second plot, the sum of the data is 280.5 and there are 10 data points; this makes the mean 280.5/10 = 28.05.

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Pls help sixowiskosixo

erma4kov [3.2K]

The given question is a quadratic equation and we can use several methods to get the solutions to this question. The solution to the equation are 3/4 and -5/6 and the greater of the two solutions is 3/4

<h3>Quadratic Equation</h3>

Quadratic equation are polynomials with a second degree as it's highest power.

An example of a quadratic equation is

$y = ax^2 + bx + c$

The given quadratic equation is $24x^2 + 2x = 15$

Let's rearrange the equation

$24x^2 + 2x = 15\\24x^2 + 2x - 15 = 0$

This implies that

a = 24
b = 2
c = -15

The equation or formula of quadratic formula is given as

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}$

We can substitute the values into the equation and solve

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}\\y = \frac{-2 +- \sqrt{2^2 -4 * 24 * (-15)} }{2*24} \\y = \frac{-2+-\sqrt{4+1440} }{48} \\y = \frac{-2+-\sqrt{1444} }{48} \\y = \frac{-2+- 38}{48} \\y = \frac{-2+38}{48} \\y = \frac{3}{4}\\ \\or\\y = \frac{-2-38}{48} \\y = \frac{-40}{48} \\y = -\frac{5}{6}$