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3 years ago

The ratio 60:40 in its simplest form

Mathematics

2 answers:

Ksenya-84 [330]3 years ago

5 0

Answer:

30:20

Step-by-step explanation:

vladimir1956 [14]3 years ago

4 0

Answer:

3:2

Step-by-step explanation:

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What is the value of 4 1/3Ã·1/9?Enter your answer as a fraction in simplest form, like this: 3/14

Vlada [557]

Hey there!

$\bold{4\frac{1}{3}\times\frac{1}{9}}$
$\bold{4\frac{1}{3}=\frac{13}{3}}$
$\bold{\frac{13}{3}(\frac{1}{9})}$
$\bold{13\times1=13}$
$\bold{9\times3=27}$
$\boxed{\boxed{\bold{Answer:\frac{13}{27} }}}$

Good luck on your assignment and enjoy your day!

~ $\bold{LoveYourselfFirst:)}$

8 0

3 years ago

How many triangles can be constructed with angles measuring 60 degrees 60 degrees and 40 degrees

erma4kov [3.2K]

None, a triangle need to have angles measuring 180 degrees , its only possible if the triangle has three angles equaling 60

8 0

3 years ago

Read 2 more answers

Suppose that you draw two cards from a deck. After drawing the first card, you do not put the first card back in the deck. What

AVprozaik [17]

Answer:

The correct answer is 0.05882.

Step-by-step explanation:

A deck of cards have 52 cards, 13 cards of each suit.

We are drawing two cards without replacement.

We need to find the probability of getting a diamond as the first card.

Favorable outcomes are 13 and total number of outcomes are 52.

Thus this probability is $\frac{13}{52} = \frac{1}{4}$ .

Now for the next draw we again want to pick a diamond card.

Favorable outcomes are 12 and total number of leftover cards are 51.

Thus this probability is $\frac{12}{51}$ .

Now the probability that both cards are diamonds is $\frac{1}{4}$ × $\frac{12}{51}$ = $\frac{3}{51} = \frac{1}{17}$ = 0.0588235 ≈ 0.05882

4 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

2 years ago

The sum of 7, -8 and 2 is less than zero. true or false

Usimov [2.4K]

False ssssssssssssssssssssssssssssssssssssssssssssss

8 0

3 years ago