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3 years ago

Which polynomial is in standard form

Mathematics

1 answer:

sukhopar [10]3 years ago

7 0

Answer:

The third one

not sure bu i am working on this to so might be right

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What is the value of c in the equation below? 2^-4(2^2)=a^b=c

Triss [41]

Answer:

The answer is option 4.

Step-by-step explanation:

You have to use Indices Law,

${a}^{m} \times {a}^{n} = {a}^{m + n}$

${a}^{ - n} = \frac{1}{ {a}^{n} }$

So for this question :

${2}^{ - 4} \times {2}^{2}$

$= {2}^{ - 4 + 2}$

$= {2}^{ - 2}$

${a}^{b} = c \\ c = {2}^{ - 2}$

$c = \frac{1}{ {2}^{2} }$

$c = \frac{1}{4}$

5 0

3 years ago

5. How many cubic feet are in 3 cubic yards?

Sergeu [11.5K]

Answer:81

Step-by-step explanation: multiply the volume value by 27

8 0

3 years ago

Read 2 more answers

What is the easiest way to solve a quadratic equation?

algol [13]

Answer: This is an opinion, look below! :)

Step-by-step explanation:

Factor the expression. To factor the expression, you have to use the factors of the {\displaystyle x^{2}}x^{2} term (3), and the factors of the constant term (-4), to make them multiply and then add up to the middle term, (-11). Here's how you do it:

Since {\displaystyle 3x^{2}}3x^{2} only has one set of possible factors, {\displaystyle 3x}3x and {\displaystyle x}x, you can write those in the parenthesis: {\displaystyle (3x\pm ?)(x\pm ?)=0}(3x\pm ?)(x\pm ?)=0.

Then, use process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4.[3]

By trial and error, try out this combination of factors {\displaystyle (3x+1)(x-4)}(3x+1)(x-4). When you multiply them out, you get {\displaystyle 3x^{2}-12x+x-4}3x^{2}-12x+x-4. If you combine the terms {\displaystyle -12x}-12x and {\displaystyle x}x, you get {\displaystyle -11x}-11x, which is the middle term you were aiming for. You have just factored the quadratic equation.

As an example of trial and error, let's try checking a factoring combination for {\displaystyle 3x^{2}-11x-4=0}3x^{2}-11x-4=0 that is an error (does not work): {\displaystyle (3x-2)(x+2)}(3x-2)(x+2) = {\displaystyle 3x^{2}+6x-2x-4}3x^{2}+6x-2x-4. If you combine those terms, you get {\displaystyle 3x^{2}-4x-4}3x^{2}-4x-4. Though the factors -2 and 2 do multiply to make -4, the middle term does not work, because you needed to get {\displaystyle -11x}-11x, not {\displaystyle -4x}-4x.

7 0

3 years ago

Help!!!! I have rsm tomowrrow. can u also show me how to do it? thanks!

Tpy6a [65]

Check the picture below.

6 0

2 years ago

There is a 70 percent chance that an airline passenger will check bags. In the next 16 passengers that check in for their flight

GenaCL600 [577]

Answer:

a) $P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332$

b) $P(X$

And we can find the individual probabilities like this:

$P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649$

$P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099$

$P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040$

$P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465$

$P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732$

$P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228$

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033$

And replacing we got:

$P(X$

c) $P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)$

And replacing we got 0.825

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:

$X \sim Binom(n=17, p=0.7)$

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

And we want to find this probability:

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332$

Part b fewer than 10 will check bags

We want this probability:

$P(X$

We can use the complement rule and we have:

$P(X$

And we can find the individual probabilities like this:

$P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649$

$P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099$

$P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040$

$P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465$

$P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732$

$P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228$

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033$

And replacing we got:

$P(X$

Part c at least 10 bags

We can find this probability like this:

$P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)$

And replacing we got 0.825

4 0

3 years ago