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3 years ago

Which function is a quadratic function? u(x) = –x + 3x2 – 8 v(x) = 2x2 + 8x3 + 9x y(x) = x2 + 3x5 + 4 z(x) = 7x2 + 2x3 – 3

Mathematics

2 answers:

hichkok12 [17]3 years ago

6 0

Answer:

u(x) = –x + 3x2 – 8

Step-by-step explanation:

The functions are;

u(x) = –x + 3x2 – 8

v(x) = 2x2 + 8x3 + 9x

y(x) = x2 + 3x5 + 4

z(x) = 7x2 + 2x3 – 3

A quadratic function is a function that has the form f(x) = ax2 + bx + c where a, band c are numbers with a not equal to zero. The maximum degree must be 2.

Among all the options, the one that matches the form is;

u(x) = –x + 3x2 – 8

Kisachek [45]3 years ago

5 0

Answer:

u(x)=-x+3x2-8

Step-by-step explanation:

just did the test edg 2021

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4 years ago

Read 2 more answers

Three times a number plus three times a second number is fifteen. Four times the first plus twice the second number is fourteen.

Alexxx [7]

Answer:

x = 2 | y = 3. ( 2,3 )

Step-by-step explanation:

2(3x + 3y = 15)

6x + 6y = 30

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3 years ago

3/4 (m - 500) = 8000 + 4000

miv72 [106K]

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Step-by-step explanation:

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4 years ago

Mathematical Statistics with Applications Homework Help

photoshop1234 [79]

7.37:

a. W follows a chi-squared distribution with 5 degrees of freedom. See theorem 7.2 from the same chapter, which says

$\displaystyle \sum_{i=1}^n\left(\frac{Y_i-\mu}{\sigma}\right)^2$

is chi-squared distributed with n d.f.. Here we have $\mu=0$ and $\sigma=1$ .

b. U follows a chi-squared distribution with 4 degrees of freedom. See theorem 7.3:

$\displaystyle \frac1{\sigma^2}\sum_{i=1}^n (Y_i-\overline Y)^2$

is chi-squared distributed with n - 1 d.f..

c. Y₆² is chi-square distributed for the same reason as W, but with d.f. = 1. The sum of chi-squared distributed random variables is itself chi-squared distributed, with d.f. equal to the sum of the individual random variables' d.f.s. Then U + Y₆² is chi-squared distributed with 5 + 1 = 6 degrees of freedom.

7.38:

a. Notice that

$\dfrac{\sqrt 5 Y_6}{\sqrt W} = \dfrac{Y_6}{\sqrt{\frac W5}}$

and see definition 7.2 for the t distribution. Since Y₆ is normally distributed with mean 0 and s.d. 1, it follows that this random variable is t distributed with 5 degrees of freedom.

b. Similar manipulation gives

$\dfrac{2Y_6}{\sqrt U} = \dfrac{\sqrt4 Y_6}{\sqrt U} = \dfrac{Y_6}{\sqrt{\frac U4}}$

so this r.v. is t distributed with 4 degrees of freedom.

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3 years ago