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3 years ago

8

200 people whee asked how often they go to the movies and whether they prefer action or drama. If 144 people went to the movies

what is the percentage?

1 answer:

Alex Ar [27]3 years ago

6 0

Answer:

It would be 72% trust me it is correct because this level of questions seems a little to young for me lol

Step-by-step explanation:

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Work out the unknown angles. Give reasons for each answer.

zalisa [80]

A)

Because the sum of angles in a triangle must equal 180° we can say:

62+α+90=180

α+152=180

α=28°

b)

Since all sides have equal length, all angles must have equal measure. (This is an equilateral triangle). And again, because the sum of the angle must be 180° we can say:

b+b+b=180

3b=180

b=60°

8 0

3 years ago

Combine like terms 4a2-3b2+ 4b + 10a2

jolli1 [7]

Answer:

14a² - 3b² + 4b

Step-by-step explanation:

Like terms are the terms that have similar stuff on it, like same variables, no variables, and stuff like that.

4a2-3b2+ 4b + 10a2

4a² and 10a² is similar. 4a² + 10a² = 14a²

Now, the equation is 14a² - 3b² + 4b. There is no more like terms left, so that is the answer.

Hope this helps!!

-Ketifa

4 0

3 years ago

Mario wrote the inequality below to compare two numbers.

Firlakuza [10]

Answer:

58 is greater than 51

Step-by-step explanation:

> is greater than and < is less than

8 0

3 years ago

Read 2 more answers

The product of which expression contains four decimal places?

Naya [18.7K]

Answer:

D.) 14.2*0.784

Step-by-step explanation:when you calculate it, there is 4 numbers behind the decimal point.

3 0

3 years ago

A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is

koban [17]

The expected length of code for one encoded symbol is

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha$

where $p_\alpha$ is the probability of picking the letter $\alpha$ , and $\ell_\alpha$ is the length of code needed to encode $\alpha$ . $p_\alpha$ is given to us, and we have

$\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}$

so that we expect a contribution of

$\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375$

bits to the code per encoded letter. For a string of length $n$ , we would then expect $E[L]=1.375n$ .

By definition of variance, we have

$\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2$

For a string consisting of one letter, we have

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4$

so that the variance for the length such a string is

$\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859$

"squared" bits per encoded letter. For a string of length $n$ , we would get $\mathrm{Var}[L]=1.859n$ .

5 0

3 years ago