All categories

2 years ago

what is the domain of f(x) = ³√x ? a). all real numbers b). positive numbers and zero c). all integers d). whole numbers

Mathematics

2 answers:

Mila [183]2 years ago

7 0

Answer:

Step-by-step explanation:

Domain is the ALLOWED $x$ values in a function.

Does plugging in any positive number pose a problem with getting an answer?

No.

Does plugging in 0 pose any problem?

No.

Does plugging in negative numbers pose any problem?

No. Square root of negative numbers are not possible BUT cube root of negative numbers ARE possible.

Also, from the attached snapshot of the graph of $f(x)=\sqrt[3]{x}$ , we can see that ALL real numbers are allowed for $x$

pav-90 [236]2 years ago

7 0

Answer:

Step-by-step explanation:

all real numbers

You might be interested in

Which inequality is equivalent to \-41 <9?

suter [353]

Answer: it’s the letter b ( option 2 )

5 0

3 years ago

Could someone give me a hand

Minchanka [31]

The answer is thirty two

4 0

2 years ago

Please help! I do not know how to find the unknown variables using the Pythagorean theorem. Any help will be greatly appreciated

Aliun [14]

Answer:

pythagorean theorem=a squared + b squared =c squared

where c is the hypotenus(the longest part)

Step-by-step explanation:

16 squared = 8 squared +....

16 ×16=256=8×8=64 + y

256_64=192

sq. root of 192=13.85

y=180_90=90

90÷2=45

y=45

z=45

x=13.85cm

hope you understand

3 0

2 years ago

What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success p? C

charle [14.2K]

Answer:

$(D)E[ X ] =np.$

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success p,

$f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq x\leq n$

$E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}$

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

$E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}$

Now,

$x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)$

Substituting,

$E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}$

Factoring out the n and one p from the above expression:

$E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}$

Representing k=x-1 in the above gives us:

$E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}$

This can then be written by the Binomial Formula as:

$E[ X ] = (np) (p +(1 - p))^{n -1 }= np.$

5 0

3 years ago

Can 1 be expressed as a fraction

elixir [45]

Any integer can be written as a fraction! Just put that integer as the numerator of a fraction with a denominator of 1.

<h2>hope it helps!</h2><h2>#carry on learning</h2>

6 0

2 years ago

Read 2 more answers