Simplify the expression using PEMDAS<br> "15+3-2(2-1) <br> Show me the steps not just the answer!

Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function.

marishachu [46]

Answer:

$\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18$

Step-by-step explanation:

We have been given the function

$f(x)=-2x^2+4x^3+3x^2+18$

From the rational zeros theorem, we have

$\text{Possible rational zeros}=\pm\frac{\text{Factors of constant term}}{\text{Factors of leading coefficient}}$

From the given function,

Leading coefficient = 2

Factors of 2 are 1,2

Constant term = 18

Factors of constant term = 1, 2, 3, 6, 9, 18

Hence, we have

$\text{Possible rational zeros}=\pm\frac{1,2,3,6,9,18}{1,2}\\\\\text{Possible rational zeros}=\pm1,\pm\frac{1}{2},\pm2,\pm3,\pm\frac{3}{2},\pm6,\pm9,\pm\frac{9}{2},\pm18$

6 0

3 years ago

When a certain number is divided by 25 the result is 32 reminder 17 . What is the number

hoa [83]

Multiply 25 by 32 then add 17:

25 x 32 = 800

800 + 17 = 817

The number is 817

4 0

3 years ago

Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p

Ymorist [56]

Answer:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

$A\sim Uniform(1 ,7)$

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

$B\sim Exp(\lambda=\frac{1}{30 min})$

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

$E(A +B)$

Since the two random variables are assumed independent, then we have this

$E(A+B) = E(A)+E(B)$

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by $E(X) =\frac{a+b}{2}$ where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got: