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3 years ago

If i randomly pick a polynomial, is it more likely to have real roots or complex roots?

1 answer:

6 0

I would say you are more likely to have complex roots. If you think of graphs of polynomials, if you draw a function that doesn't pass through the x-axis, all roots are complex.

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Richard s kitchen floor measures 22 ft by 16ft. He plans on laying title down and each tile piece measures 16 inches by 16 inche

olya-2409 [2.1K]

We calculate the floor area of the kitchen:
A = (22) * (16)
A = 352 feet ^ 2
We consider the following conversion:
1 foot = 12 inches.
Applying the conversion we have:
A = 352 * (12/1) ^ 2
A = 50688 in ^ 2
We calculate the area of each tile:
A = (16) * (16)
A = 256 in ^ 2
Then, the number of tiles is given by:
N = (50688) / (256)
N = 198
Answer:
He needs to buy 198 tiles

3 0

3 years ago

I wrote it down better

Elenna [48]

number one is 280 because 4/2 is 2 + 2 is 4 then you multiply 7 and that is 28 ^ 2 which is 280

4 0

4 years ago

Read 2 more answers

(4.1.4) Let X and Y be Bernoulli random variables. Let Z = X + Y. a. Show that if X and Y cannot both be equal to 1, then Z is a

Fynjy0 [20]

Step-by-step explanation:

Given that,

X ~ Bernoulli $(p_x)$ and Y ~ Bernoulli $(y_x)$

X + Y = Z

The possible value for Z are Z = 0 when X = 0 and Y = 0

and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0

If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,

Therefore Z is a Bernoulli random variable

If X and Y can not be both equal to 1

then,

$p_z = P(X=1$ or $Y=1)\\$

$p_z = P(X=1)+P(Y=1)-P(=1$ and $Y =1)$

$p_z = P(x=1)+P(Y=1)\\\\p_z=p_x+p_y$

If both X = 1 and Y = 1 then Z = 2

The possible values of the random variable Z are 0, 1 and 2.

since a Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution

7 0

3 years ago

Find the product <br> -1/2 y(2y3-8)

MissTica

$0-(( \frac{1}{2} *y)*2y^3-8) \\ \\ then \ we \ would \ simplify \ 1/2 \\ \\ we \ factor y^3 - 4 \\ \\ 4 \ would \ not \ be \ a \ perfect \ cube. \\ \\ therefore, \ your \ answer \ would \ then \ be \ the \ following: \\ \\ \boxed{ -y * (y^3 - 4)}$

8 0

3 years ago

Say that a supplier claims they are 99% confident that their products will be in the interval of 50.02 to 50.38. You take sample

nydimaria [60]

Answer:

The supplier becomes less accurate than they otherwise would have tried to claim. A further explanation is below.

Step-by-step explanation:

According to the provider, this same width of that same confidence interval would be as follows:

= $50.38 - 50.02$

= $0.36$

Depending on the input observed, the width including its confidence interval would be as follows:

= $50.39 - 50.01$

= $0.38$

As even the width of that interval again for survey asserted > the width including its confidence interval according to the provider's statement, we could conclude that such is the appropriate reaction.

3 0

3 years ago