All categories

3 years ago

Which of the following statements is true about a distribution that appears to have a gap when displayed as a histogram?

Mathematics

1 answer:

makkiz [27]3 years ago

5 0

Using the concept of histogram, it is found that the correct option is given by:

B. The distribution has a region between two data values where no data were observed.

-----------------------

When a data-set is divided into intervals, the value of the histogram for each interval represents the amount of times the interval was visited, that is, the number of observations in the data-set that assume values in that interval.
If there is a gap on the histogram, it means that the value at that position is 0, that is, there were no observations in that interval, which is what is described in option B.

A similar problem is given at brainly.com/question/14552629

You might be interested in

This is my last question please help

Nina [5.8K]

Answer:

Step-by-step explanation:

I hope this helps!!

6 0

3 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago

-3s = -24 what does S equal

qaws [65]

-3s=-24 to find s you will divide both sides by -3
S=8

8 0

3 years ago

Read 2 more answers

Which line is parallel to 4x + 2y = 0? A. y=2x+3 B. -1/2×+3 C. y=-2x+3 D. y=1/2x+3

egoroff_w [7]

Line A is parallel to line B if line A and B have the same slope and have the same or different y-intercept.

Convert the standard form equation to slope-intercept form as it will be easier to see the slope.

Solve for y.

4x + 2y = 0
2y = -4x <-- Subtract 4x from each side
y = -2x <-- Divide both sides by 2

In the slope-intercept form y = -2x, a y-intercept of 0 is implied

Now find a slope-intercept form equation that has the same slope as y = -2x

So, the answer is C y = -2x + 3.

7 0

3 years ago

Read 2 more answers

Please help thank you!

ser-zykov [4K]

Answer:

a) 91.4%

b) 37.6%

c) 26.8%

Step-by-step explanation:

For each of these, we'll get a z-score, then use our calculators to find the area using normalcdf (on TI-84, 2nd->VARS->2)

a) z = (9-6.4) / 1.9 = 1.368

normalcdf(-999,1.368) = .914 = 91.4%

b) z = (7-6.4) / 1.9 = .316

normalcdf(.316,999) = .376 = 37.6%

c) normalcdf(.316,1.368) = .268 = 26.8%

3 0

3 years ago