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2 years ago

Write a quadratic function whose zeros are 2 and -12. 60 Point question Answer ASAP for brainliest!

Mathematics

1 answer:

ycow [4]2 years ago

5 0

Answer:

x^2 + 10x - 24 = 0

Step-by-step explanation:

Two zeros are 2 and -12:

x - 2 = 0
x + 12 = 0

So the function is (x-2)(x+12) = 0

<=> x^2 + 12x - 2x - 24 = 0

<=> x^2 + 10x - 24 = 0

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f a random sample of 300 adults is taken from the state of Colorado, where the rate of obesity is 19.8%, can we use the Normal a

Troyanec [42]

Answer:

Probability of at least 50 obese individuals in our sample is 0.92364 .

Step-by-step explanation:

We are given that a random sample of 300 adults is taken from the state of Colorado, where the rate of obesity is 19.8% .

Let X = Number of obese individuals

Firstly, X ~ $Binom(n=300,p=0.198)$

For approximating binomial distribution into normal distribution, firstly we have to calculate $\mu$ and $\sigma^{2}$ .

Mean of Normal distribution, $\mu$ = n * p = 300 * 0.198 = 59.4

Variance of Normal distribution, $\sigma^{2}$ = n * p * (1-p) = 300 *0.198 *0.802 = 47.64

So, now X ~ N( $\mu = 59.4 , \sigma^{2} = 47.64$ )

The standard normal z score distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

So, probability of at least 50 obese individuals in our sample = P(X >= 50)

P(X >= 50) = P(X > 49.5) {using continuity correction}

P(X > 49.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{49.5 - 59.4}{6.9}$ ) = P(Z > -1.43) = P(Z < 1.43) = 0.92364

Therefore, required probability is 0.92364 .

8 0

3 years ago

What is the length of AB?

kramer

Answer:

We are given to find the length of side AB. We know that in a triangle, if two angles have equal measures, then the sides opposite to them are equal in length. Thus, the length of side AB is 6 units.

Hope this helps!!

3 0

3 years ago

Solve. 4 - 2x = 10 A)-7 B)-3 C)0 D)3

mars1129 [50]

Answer: The answer should be B) -3. To solve the equation you have to rewrite the equation.

4 0

3 years ago

Express the quotient of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5C

Lesechka [4]

Answer:

Solution : $-\frac{3}{4}-\frac{3}{4}i$

Step-by-step explanation:

$-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]$

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

$\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}$