All categories

4 years ago

Which shows one way to determine the factors of x^3+11x^2-3x-33 by grouping?

Mathematics

2 answers:

AnnyKZ [126]4 years ago

7 0

Answer:

The correct answer is D. $x^2\left(x+11\right)-3\left(x+11\right)$ .

Step-by-step explanation:

To factor the expression $x^3+11x^2-3x-33$ by grouping you must:

First, notice that there is no factor common to all terms. However, if we group the first two terms together and the last two terms together, each group has its own greatest common factor.

$\left(x^3+11x^2\right)+\left(-3x-33\right)$

Second, factor out $x^2$ from $x^3+11x^2$ :

$x^2\left(x+11\right)$

Next, factor out $-3$ from $-3x-33$ :

$-3\left(x+11\right)$

Therefore,

$x^3+11x^2-3x-33=x^2\left(x+11\right)-3\left(x+11\right)$

wlad13 [49]4 years ago

5 0

That would be Option D

The last step would be
= (x^2 - 3)(x + 11)

You might be interested in

1/6+1/2 simplest form

vlada-n [284]

Lowest common denominator-> 6
1/2 multiply both top and bottom by 3 to get it in sixths
1/6+3/6=4/6
4/6 simplifies to 2/3.
2/3 is the simplest form to your question.

7 0

3 years ago

Find the 52nd term in the following sequence.<br><br> -31, -41, -51, -61,

katrin [286]

Answer:

$a_{52}$ = -541

Step-by-step explanation:

4 0

3 years ago

Please help

Vlada [557]

Answer is AAS hope this helps

3 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago

A piece of machinery depreciates $5000 the first year,

eimsori [14]

Answer:

A- $3500

Step-by-step explanation:

The depreciation rate is $300

$5000, $4700, $4400

$300 x 3 = 900

$4400 - 900 =

$3500.

4 0

3 years ago