A histogram can only display what data?

Mathematics

2 answers:

Andrej [43]3 years ago

8 0

I agree with her answer, plot frequency. It may determine a frequency of a variable and equality of class interval.

zlopas [31]3 years ago

6 0

The answer is plot frequency.

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Based on the side lengths given (a, b, and c), which triangles are right triangles?

ASHA 777 [7]

Answer:

It's the third option.

Step-by-step explanation:

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3 years ago

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Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab

Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

$\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

<u>Right part:</u>

$\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

Since simplified left and right parts are the same, then the equality is true.

3 0

3 years ago

Eva wants to make two pieces of pottery.She needs 3/5 pound of clay for one piece and 7/10 pound of clay for the other piece.She

marta [7]

Answer:

a) Eva will need 2 bags of clay to make both pieces of pottery.

b) She will have leftover $\frac{11}{10}$ pounds of clay.

Step-by-step explanation:

a) We have to find the total amount of clay, as she needs $\frac{3}{5}pound=P_{1}$ and $\frac{7}{10}pound=P_{2}$ , so $P_{1}+P_{2}= \frac{3}{5} +\frac{7}{10} =\frac{6+7}{10}=\frac{13}{10}pounds$ , is the total amount of clay. Then as we Know that one bag of clay has $\frac{4}{5} =\frac{8}{10} pounds$ , It isn´t enough, but with two bags $2*\frac{8}{10}=\frac{16}{10} pounds$ , It is enough clay.

b) She will have leftover $\frac{16}{10} -\frac{13}{10} =\frac{3}{10}pounds$ of the second bag, then we add the third one, so $\frac{3}{10} +\frac{8}{10} =\frac{11}{10}pounds$ are leftover.