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svlad2 [7]
3 years ago
12

What is the quadralatic formula

Mathematics
2 answers:
Lyrx [107]3 years ago
7 0

Answer:

Step-by-step explanation:

The quadratic formula is as follows:

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

It is used for factoring second degree polynomials ONLY!!!  That's why it's called the quadratic formula...quadratics are second degree polynomials.  The a, b and c are found in the standard form of a quadratic equation:

y=ax^2+bx+c

Just plug them into the formula, but be careful with the "-4ac".  Sometimes that mistakenly leads to an unwanted negative sign under you radical.  Unless you are dealing with imaginary numbers, your solutions should be real.

Sholpan [36]3 years ago
4 0

X equals negative B and plus and minus the square root of b squared minus 4ac all over 2a

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sammy [17]

Answer:

\large\boxed{\bigg(4\times10^8\bigg)^2=1.6\times10^{17}}

Step-by-step explanation:

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Apply the distributive property to create an equivalent expression.<br> −9⋅(5j+k)
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Step-by-step explanation:

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Vikki [24]
Here’s your answer. Hope this helped!!

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Suppose in the University of Manitoba, 40% of the students live in apartments. If 600 students are randomly selected, then the a
just olya [345]

Answer:

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So then the correct answer would be:

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Step-by-step explanation:

The exact way to solve this problem is using the binomial distribution, assuming that our random variable of interest is "number of students living in apartments" represented by X and X \sim Bin(n=600, p =0.4)

And we want this probability:

P(200 \leq X \leq 400) [tex]In order to find this probability we can use the foloowing excel code:"=BINOM.DIST(400,600,0.4,TRUE)-BINOM.DIST(199,600,0.4,TRUE)"And we got:[tex] P(200 \leq X \leq 400) =0.999675[tex]But for this case the problem says that we need to approximate, so then we can use the normal approximation to the normal distribution. We need to check the conditions in order to use the normal approximation.&#10;[tex]np=600*0.4=240\geq 10

n(1-p)=600*(1-0.4)=360 \geq 10

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

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\sigma=\sqrt{np(1-p)}=\sqrt{600*0.4(1-0.4)}=12

And then X \sim N (\mu = 240, \sigma= 12)

And we are interested on the following probability:

P(200\leq X\leq 400)=P(\frac{200-240}{12}\leq \frac{X-\mu}{\sigma}\leq \frac{400-240}{12})=P(-3.33\leq Z \leq 13.33)

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So then the correct answer would be:

B) .9996

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3 years ago
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