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1 year ago

Indira makes a box-and-whisker plot of her data. She finds that the distance from the minimum value to the first

Mathematics

2 answers:

Kipish [7]1 year ago

8 0

Answer: D: The mean is less than the median because the data is skewed to the left.

Step-by-step explanation:

elena-14-01-66 [18.8K]1 year ago

4 0

Indira makes a box-and-whisker plot of her data. She finds that the distance from the minimum value to the first quartile is greater than the distance between the third quartile and the maximum value. Which is most likely true?

The mean is less than the median because the data is skewed to the left.

Did this on quizlet, hope this helps!

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Two solutions of different concentrations of acid are mixed creating 40 mL of a solution that is 32% acid. One-quarter of the so

Georgia [21]

In order to construct this equation, we will use the variables:
V to represent mixture volume (40 ml)
C to represent mixture concentration (0.32)
v₁ to represent volume of first solution (40 / 4 = 10 ml)
c₁ to represent concentration of first solution (0.2)
v₂ to represent the volume of the second solution (40 * 3/4 = 30 ml)
c₂ to represent the concentration of the second solution

We know that the total amount of substance, product of the volume and concentration, in the final solution is equal to the individual amounts in the two given solutions. Thus:
VC = v₁c₁ + v₂c₂
40(0.32) = 10(0.2) + 30c

6 0

2 years ago

Read 2 more answers

Performance task: A parade route must start And and at the intersections shown on the map. The city requires that the total dist

GaryK [48]

Answer:

Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B: For the total distance is as close to 3 miles as possible, the start point of the parade should be at the point on Broadway with coordinates (9.941, 4.970)

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue; (4, 2)

For Broadway; (7.97, 2.49)

Step-by-step explanation:

Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

$l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}$

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

$l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}$

Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit

The scale of the drawing is 1 unit = 0.25 miles

Therefore;

The actual length of the initial parade =0.25×12.265 unit = 3.09 miles

The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B:

For an actual length of 3 miles, the length on the scale drawing should be given as follows;

1 unit = 0.25 miles

0.25 miles = 1 unit

1 mile = 1 unit/(0.25) = 4 units

3 miles = 3 × 4 units = 12 units

With the same end point and route, we have;

$l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}$

y² + (6 - x)² = 176 - 96·√2

y² = 176 - 96·√2 - (6 - x)²............(1)

Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2

Which gives;

y/x = 1/2

y = x/2 ..............................(2)

Equating equation (1) to (2) gives;

176 - 96·√2 - (6 - x)² = (x/2)²

176 - 96·√2 - (6 - x)² - (x/2)²= 0

176 - 96·√2 - (1.25·x²- 12·x+36) = 0

Solving using a graphing calculator, gives;

(x - 9.941)(x + 0.341) = 0

Therefore;

x ≈ 9.941 or x = -0.341

Since l₁ is required to be 12 - 4·√2, we have and positive, we have;

x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97

Therefore, the start point of the parade should be the point (9.941, 4.970) on Broadway so that the total distance is as close to 3 miles as possible

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue;

Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)

For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

5 0

2 years ago

8) Write an absolute value equation that has 5 and 15 as its solutions?

marissa [1.9K]

(I'm going to use brackets as my absolute value bars lol)
[5 x -3]
[-15]
=15

7 0

2 years ago

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$