All categories

3 years ago

Which statement about a histogram is true?

2 answers:

6 0

The correct answer is b) Histograms can be used to exhibit the shape of distributions. This is because histograms count the frequency of each possible answer (not each individual value), so one is able to see which results are more common and which are less common.

BigorU [14]3 years ago

5 0

le answer to le question is b

You might be interested in

Karen, Mary and Amy sold candy bars as a fundraiser. Karen sold three more than twice the number of bars Mary sold. Amy sold fiv

Pepsi [2]

karen 3 more than 2 times of mary amy sold 5 more than karen

(3x)2+5

4 0

3 years ago

Read 2 more answers

Two-thirds of a number minus six is 2

Vlada [557]

(2/3)n - 6 = 2

Add 6 to both sides.
(2/3)n = 8

Divide by 2/3.
n = 9 / (2/3)
n = 27/2, or 13.5

4 0

4 years ago

Slope is 3,and (2,5) is on the line.

bazaltina [42]

Standard form of line:

Ax + By = C

where A, B, and C are integers, and A is positive

slope-point form of a line:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is a point on the line. Replacing with m = 3 and point (2, 5):

y - 5 = 3(x - 2)

y - 5 = 3x - 3(2) distributing

y - 5 = 3x - 6

y - 5 - y = 3x - 6 - y subtracting y at both sides

- 5 = 3x - 6 - y

- 5 + 6 = 3x - 6 - y + 6 adding 6 at both sides

1 = 3x - y which has the standard form

4 0

2 years ago

The number of people and dogs taking in obedience class increased from 1800 students to 1600 student what is the percent increas

kolbaska11 [484]

Increased or decreased?

I attached an image above of how to solve this equation.

7 0

3 years ago

Find the exact value of the trigonometric expression.

Dmitrij [34]

$\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
-------------------------------\\\\
\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}$

$\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}$

$\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}$

which simplifies thus far to

$\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}$

$\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}$

7 0

4 years ago