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3 years ago

Which statement is true about the ranges for the box plots?

Mathematics

2 answers:

dimulka [17.4K]3 years ago

7 0

Answer: A. The range of the Morning box plot is the same as the range of the Afternoon box plot.

agasfer [191]3 years ago

4 0

Answer:

The range of the Morning box plot is the same as the range of the Afternoon box plot.

Step-by-step explanation:

Data provided in the question

Number line = 0 to 16

Morning whiskers range = 3 to 15

Box range = 5 to 12

Line divided = 8

Afternoon whiskers range = 4 to 16

Box range = 8 to 15

Line divided = 14

Based on the above information,

The morning whiskers range difference is

= 12 - 5

= 7

And, the afternoon whiskers range difference is

= 15 - 8

= 7

Therefore the first option is correct

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A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt

Katena32 [7]

Answer:

$2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}$

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be $y$ . The magnitude of $40\text{ m/s}$ will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is $60^{\circ}$ , we have:

$\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}$ (Recall that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ )

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation $v_f^2=v_i^2+2a\Delta y$ to solve this problem, where $v_f/v_i$ is final and initial velocity, respectively, $a$ is acceleration, and $\Delta y$ is distance travelled. The only acceleration is acceleration due to gravity, which is approximately $9.8\:\mathrm{m/s^2}$ . However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.

What we know:

$v_i=20\sqrt{3}\text{ m/s}$
$a=-9.8\:\mathrm{m/s^2}$
$\Delta y =50\text{ m}$

Solving for $v_f$ :

$v_f^2=(20\sqrt{3})^2+2(-9.8)(50),\\v_f^2=1200-980,\\v_f^2=220,\\v_f=\sqrt{220}=\boxed{2\sqrt{55}\text{ m/s}}$

3 0

3 years ago

Point DD is located at (-2,-6)(−2,−6) on the coordinate plane. Point DD is reflected over the x-axis to create point D'D. Point

bearhunter [10]

Point D (-2,-6)>>(x,-y)>> Point D’(-2,6)
Point D’ (-2,6)>> (-x,y)>> Point D’’(2,6)

7 0

2 years ago

Read 2 more answers

Find the indicated limit, if it exists. limit of f of x as x approaches 9 where f of x equals x plus 9 when x is less than 9 and

SVEN [57.7K]

Answer:

B. 18

Step-by-step explanation:

For the function

$f(x)=\left\{\begin{array}{l}x+9,\ \ x$

we can find the value of the function for all x that are very close to 9 but are less than 9 and for all values of x that are very close to 9 but are greater than 9.

1. For $x$