All categories

2 years ago

Is 9/15 irrational??

Mathematics

2 answers:

Mariana [72]2 years ago

4 0

No because irrational cant be a simple fraction and 9/15

is 0.6

so... it is not irrational

ch4aika [34]2 years ago

3 0

Hi Madison!

A rational number is a number that cannot be expressed as a ration between two integers.

So Irrational number are like 3.14.

I searched a chart of numbers and it looks like that fraction will actually be considered an rational number, not irrational.

Hope this helps!

You might be interested in

3) Write the inequality that

masha68 [24]

Answer:

See below

Step-by-step explanation:

Owing represent a negative balance.

So the inequality for this case is:

- 26.00 > - 147.00

3 0

2 years ago

Please help me out!! 24n = 6 n = ?

Travka [436]

Divide 24 from both sides

5 0

2 years ago

Read 2 more answers

A researcher testing seed varieties for tulips got a confidence interval of (14.2%,17.4%) for the percent of seeds that can grow

Alex73 [517]

Answer:

The sample proportion is 0.158.

Step-by-step explanation:

In a confidence interval of proportions, we have the lower bound and the upper bound of the interval. The sample proportion is the halfway points between these bounds, that is, the sum of these bounds divided by 2.

In this problem, we have that:

Lower bound: 0.142

Upper bound: 0.174

Sample proportion: (0.142+0.174)/2 = 0.158

The sample proportion is 0.158.

3 0

3 years ago

Please help.............:)

Alex787 [66]

Answer:

113

Step-by-step explanation:

Multiply

6 0

3 years ago

Transform equation: y = √x from rectangular to polar form.

inessss [21]

$\bf \begin{cases}
x=rcos(\theta )\\
y=rsin(\theta )\\
x^2+y^2=r^2\implies y^2=r^2-x^2
\end{cases}
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\
-------------------------------\\\\
y=\sqrt{x}\implies y^2=x\implies r^2-x^2=x\implies \cfrac{r^2-x^2}{x}=1
\\\\\\
\cfrac{r^2}{x}-\cfrac{x^2}{x}=1\implies \cfrac{r^2}{x}-x=1\implies \cfrac{r^2}{rcos(\theta )}-rcos(\theta )=1$

$\bf \cfrac{r}{cos(\theta )}-rcos(\theta )=1\implies r\left[ \cfrac{1}{cos(\theta )}-cos(\theta ) \right]=1
\\\\\\
r\left[ \cfrac{1-cos^2(\theta )}{cos(\theta )} \right]=1\implies r\left[ \cfrac{sin^2(\theta )}{cos(\theta )} \right]=1\implies r=\cfrac{1}{\frac{sin^2(\theta )}{cos(\theta )}}
\\\\\\
r=\cfrac{cos(\theta )}{sin^2(\theta )}$

3 0

3 years ago