All categories

3 years ago

Which statement is NOT true of line plots

Mathematics

2 answers:

marysya [2.9K]3 years ago

6 0

Answer:

ehh 2nd

Step-by-step explanation:

Brums [2.3K]3 years ago

3 0

It’s The first one an outlier will show as a gap in the data

You might be interested in

Which triangle is similar to triangle PQR using the pieces of Right Triangles Similarity Theorem?

Dovator [93]

sorry po sorry popopojhbvb

6 0

2 years ago

Which is the equation of a hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0)? y squared over 40 minus x square

N76 [4]

The equation of the hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0) is $\frac{x^2}{10} + \frac{y^2}{15} = 1$

<h3>How to determine the equation of the hyperbola?</h3>

The given parameters are:

Directrices at x = ±2
Foci at (5, 0) and (−5, 0)

The foci of a hyperbola are represented as:

Foci = (k ± c, h)

The center is:

Center = (h,k)

And the directrix is:

Directrix, x = h ± a²/c

By comparison, we have:

k ± c = ±5

h = 0

h ± a²/c = ±2

Substitute h = 0 in h ± a²/c = ±2

0 ± a²/c = ±2

This gives

a²/c = 2

Multiply both sides by c

a² = 2c

k ± c = ±5 means that:

k ± c = 0 ± 5

By comparison, we have:

k = 0 and c = 5

Substitute c = 5 in a² = 2c

a² = 2 * 5

a² = 10

Next, we calculate b using:

b² = c² - a²

This gives

b² = 5² - 10

Evaluate

b² = 15

The hyperbola is represented as:

$\frac{(x - k)^2}{a^2} + \frac{(y - h)^2}{b^2} = 1$

So, we have:

$\frac{(x - 0)^2}{10} + \frac{(y - 0)^2}{15} = 1$

Evaluate

$\frac{x^2}{10} + \frac{y^2}{15} = 1$

Hence, the equation of the hyperbola is $\frac{x^2}{10} + \frac{y^2}{15} = 1$

Read more about hyperbola at:

brainly.com/question/3405939

#SPJ1

6 0

2 years ago

Identifying Rigid Transformations

Alex Ar [27]

Answer:

Rotation and Translation

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v squared w

Sati [7]

Answer:

Step-by-step explanation:

The model fo the shell is given by the following equation of equilibrium:

$\Sigma F = -b\cdot v^{2} - m\cdot g = m\cdot \frac{dv}{dt}$

This first-order differential equation has separable variables, which are cleared herein:

$\int\limits^t_{0\,s} \, dt = -\frac{m}{b} \int\limits^{0\,\frac{m}{s} }_{600\,\frac{m}{s} } {\frac{1}{ v^{2}+\frac{m}{b}\cdot g } } \, dv$

The solution of this integral is:

$t = -\frac{m}{2b}\cdot \left[\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } }\right) - \tan^{-1} \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }\right)\right]$

$\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } } \right)=-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)$

$\frac{v}{\sqrt{\frac{m\cdot g}{b} } }=\tan \left[-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\right]$

$v = \sqrt{\frac{m\cdot g}{b} } \left [\frac{\tan \left(-\frac{2\cdot b \cdot t}{m} \right)+ \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)}{1 - \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\cdot \tan \left(-\frac{2\cdot b \cdot t}{m} \right) }\right]$

4 0

2 years ago

Is that correct ????

Travka [436]

Yes I believe it is :)

7 0

3 years ago