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2 years ago

Let $x$ be a value such that $8x^2 + 7x - 1 = 0$ and $24x^2+53x-7 = 0.$ What is the value of $x$? Express your answer as a simpl

ified common fraction.

Mathematics

1 answer:

Fittoniya [83]2 years ago

6 0

$8x^2+7x-1=0\ \wedge\ 24x^2+53x-7=0\\\\\text{The equation:}\\\\24x^2+53x-7=8x^2+7x-1\qquad\text{subtract}\ 8x^2\ \text{and}\ 7x\ \text{from both sides}\\\\16x^2+46x-7=-1\qquad\text{add 1 to both sides}\\\\16x^2+46x-6=0\qquad\text{divide both sides by 2}\\\\8x^2+23x-3=0\\\\8x^2+24x-x-3=0\\\\8x(x+3)-1(x+3)=0\\\\(x+3)(8x-1)=0\iff x+3=0\ \vee\ 8x-1=0\\\\x+3=0\qquad\text{subtract 3 from both sides}\\\boxed{x=-3}\\\\8x-1=0\qquad\text{add 1 to both sides}\\8x=1\qquad\text{divide both sides by 8}\\\boxed{x=\dfrac{1}{8}}$

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NISA [10]

Answer:

$50

Step-by-step explanation:

12 x 30 = 360

960 - 360 = 600

600/12 = 50

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2 years ago

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What is the product of (6)(-10)?<br> 70<br> 60<br> -60

iogann1982 [59]

Answer:

-60

Step-by-step explanation:

6 * (-10) = -60

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2 years ago

Find the average rate of change over the interval [-1,2] (from x = -1 to x = 2) for the graph below

Lyrx [107]

Hello!

Find the coordinate where x = -1:

At x = -1, y = 1, so the coordinate is (-1, 1).

Find the coordinate where x = 2:

At x = 2, y = -2, so the coordinate is (2, -2).

Find the rate of change between the points using the slope formula:

$slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Plug in the coordinates above:

$slope = \frac{-2-1}{2-(-1)}$

Simplify:

$slope = \frac{-3}{3} = -1$

Thus, the rate of change between the points is -1.

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2 years ago

Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $

Solnce55 [7]

Answer:

a) $\hat a = max(X_i)$

For this case the value for $\hat a$ is always smaller than the value of a, assuming $X_i \sim Unif[0,a]$ So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

$E(a) - a= 0$ and that's not our case.

b) $E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}$

Since is a negative value we can conclude that underestimate the real value a.

$\lim_{ n \to\infty} -\frac{1}{n+1}= 0$

c) $P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)$

And assuming independence we have this:

$P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n$

$f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]$

e) On this case we see that the estimator $\hat a_1$ is better than $\hat a_2$ and the reason why is because:

$V(\hat a_1) > V(\hat a_2)$

$\frac{a^2}{3n}> \frac{a^2}{n(n+2)}$

$n(n+2) = n^2 + 2n > n +2n = 3n$ and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming $X_1, X_2 , ..., X_n \sim U(0,a)$

And we are are ssuming the following estimator:

$\hat a = max(X_i)$

$E(a) - a= 0$ and that's not our case.

Part b

For this case we assume that the estimator is given by:

$E(\hat a) = \frac{na}{n+1}$

And using the definition of bias we have this:

$E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}$

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

$\lim_{ n \to\infty} -\frac{1}{n+1}= 0$

Part c

For this case we the followng random variable $Y = max (X_i)$ and we can find the cumulative distribution function like this:

$P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)$

And assuming independence we have this:

$P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n$

Since all the random variables have the same distribution.

Now we can find the density function derivating the distribution function like this:

$f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]$

Now we can find the expected value for the random variable Y and we got this:

$E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}$

And the bias is given by:

$E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}$

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

$V(\hat a_1) = \frac{a^2}{3n}$

$V(\hat a_2) = \frac{a^2}{n(n+2)}$

On this case we see that the estimator $\hat a_1$ is better than $\hat a_2$ and the reason why is because:

$V(\hat a_1) > V(\hat a_2)$

$\frac{a^2}{3n}> \frac{a^2}{n(n+2)}$

$n(n+2) = n^2 + 2n > n +2n = 3n$ and that's satisfied for n>1.

8 0

3 years ago

Pleaseeeeeee help me 7th grade math

vampirchik [111]

Answer:

Step-by-step explanation:

Np bro, I've dealt with these all through 7th XD

8 0

3 years ago

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