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3 years ago

What do variables and functions have in common

Mathematics

2 answers:

klemol [59]3 years ago

8 0

Answer:

A function is a mathematical relationship in which the values of a single dependent variable are determined by the values of one or more independent variables. Function means the dependent variable is determined by the independent variable(s).

Step-by-step explanation:

Step2247 [10]3 years ago

6 0

Answer:

Step-by-step explanation:

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Snowcat [4.5K]

306% of 25

= (306/100) * 25

= 3.06 * 25

= 76.5

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3 years ago

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valentinak56 [21]

Answer:

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Step-by-step explanation:

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3 years ago

Write an equation in standard form that is parallel to the line y = 2x - 1 that passes through the point (1, 2). 2x - y = 0 O x

timofeeve [1]

Answer:

2x - y = 0

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 2x - 1 ← is in slope- intercept form

with slope m = 2

Parallel lines have equal slopes, thus

y = 2x + c ← is the partial equation

To find c substitute (1, 2) into the partial equation

2 = 2 + c ⇒ c = 2 - 2 = 0

y = 2x ← equation in slope- intercept form

Subtract y from both sides

0 = 2x - y , that is

2x - y = 0 ← equation in standard form

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3 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