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2 years ago

5

A bag contains 3 red pens 5 blue pens and 2 black pens a pen is selected at random and is not replaced what is the probability o

f selecting a red pen then a blue pen

2 answers:

MAVERICK [17]2 years ago

6 0

Answer: 1/6

Step-by-step explanation:

3 red

5 blue

2 black

10 total

pick a red pen first: 3/10

pick a blue pen next: 5/9

3/10*5/9 = 15/90 or 1/6

LUCKY_DIMON [66]2 years ago

4 0

Answer: 1/6

Step-by-step explanation:

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3 years ago

Which of the values shown are potential roots of f(x) = 3x3 – 13x2 – 3x + 45? Select all that apply.

galina1969 [7]

Answer:

All potential roots are 3,3 and $-\frac{5}{3}$ .

Step-by-step explanation:

Potential roots of the polynomial is all possible roots of f(x).

$f(x)=3x^3-13x^2-3x+45$

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

p=All the positive/negative factors of 45

q=All the positive/negative factors of 3

$p=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45$

$q=\pm 1,\pm 3$

All possible roots

$\frac{p}{q}=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45,\pm \frac{1}{3},\pm \frac{5}{3}$

Now we check each rational root and see which are possible roots for given function.

$f(1)= 3\times 1^3-13\times 1^2-3\times 1+45\Rightarrow 32\neq 0$

$f(-1)= 3\times (-1)^3-13\times (-1)^2-3\times (-1)+45\Rightarrow \neq 32$

$f(-3)= 3\times (-3)^3-13\times (-3)^2-3\times (-3)+45\Rightarrow \neq -144$

$f(3)= 3\times (3)^3-13\times (3)^2-3\times (3)+45\Rightarrow =0\\\\ \therefore x=3\text{ Potential roots of function}$

Similarly, we will check for all value of p/q and we get

$f(-5/3)=0$

Thus, All potential roots are 3,3 and $-\frac{5}{3}$ .

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