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3 years ago

PLZ HELP! Will give BRAINLIEST!

Mathematics

2 answers:

Arisa [49]3 years ago

4 0

Answer:

D -26

Step-by-step explanation:

Vladimir79 [104]3 years ago

4 0

Answer:

A: 8

Step-by-step explanation:

$y = f(g(x)) = ( \sqrt{7x + b)} \\ y= \sqrt{7x + b} \\ at \: (4 \: \: 6) \\ \\ i.e. \: plugging \: x = 4 \: and \: y = 6 \\ \\ 6 = \sqrt{7 \times 4 + b} \\ 36 = 28 + b \\ b = 36 - 28 \\ b = 8 \\$

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What is the quadratic equation if the solutions are -9+ √65 /2, -9- √65 /2

liberstina [14]

Answer:

$4x^2+72x+259=0$

Step-by-step explanation:

If $x_1$ and $x_2$ are the solutions to the quadratic equation, then this equation can be written as

$(x-x_1)(x-x_2)=0$

In your case,

$x_1=-9+\dfrac{\sqrt{65}}{2}\\ \\x_1=-9-\dfrac{\sqrt{65}}{2}$

Then the equation is

$\left(x-\left(-9+\dfrac{\sqrt{65}}{2}\right)\right) \left(x-\left(-9-\dfrac{\sqrt{65}}{2}\right)\right)=0\\ \\\left(x+9-\dfrac{\sqrt{65}}{2}\right)\left(x+9+\dfrac{\sqrt{65}}{2}\right)=0\\ \\x^2+9x+\dfrac{\sqrt{65}}{2}x+9x+81+\dfrac{9\sqrt{65}}{2}-\dfrac{\sqrt{65}}{2}x-\dfrac{9\sqrt{65}}{2}-\dfrac{65}{4}=0\\ \\x^2+18x+\dfrac{259}{4}=0\\ \\4x^2+72x+259=0$

4 0

3 years ago

If a circle has a circumference of 4.6pi what is the area?

Alla [95]

Answer:

r = 0.732 d = 1.46 C = 4.6

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

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

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