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3 years ago

Use the slope-intercept form to graph the equation

1 answer:

8 0

What is the equation??
But just to help until I get the equation, slope intercept form is y=mx+b
M is the slope
Y is the y-intercept

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Helia invested 4000 AED for 3 years with a compound interest rate of 1.5%, Calculate the worth of her investment at the end of t

sineoko [7]

Answer:

4180 AED

Step-by-step explanation:

Given data

Principal= 4000

Time= 3 years

Rate= 1.5%

The compound interest expression is given as

A= P(1+r)^t

substitute

A=4000(1+0.015)^3

A= 4000(1.015)^3

A= 4000*1.045

A= 4180

Hence the worth will be 4180 AED

6 0

3 years ago

Write an equation for the orbit of Saturn in the form x^2/a^2 + y^2/b^2 = 1

frozen [14]

Answer:

x²/2166784 +y²/2159989 = 1

Step-by-step explanation:

The relationship between the semi-axes and the eccentricity of an ellipse is ...

e = √(1 -b²/a²)

In order to write the desired equation we need to find 'b' from 'e' and 'a'.

<h3>semi-minor axis</h3>

Squaring the equation for eccentricity gives ...

e² = 1 -b²/a²

Solving for b², we have ...

b²/a² = 1 -e²

b² = a²(1 -e²)

<h3>ellipse equation</h3>

Using the given values, we find ...

b² = 1472²(1 -0.056²) = 2166784(0.996864) ≈ 2159989

The desired equation is ...

x²/2166784 +y²/2159989 = 1

7 0

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

4 years ago

Mark ran 875 miles this year in the track club.Mark ran in 52 track meets and ran the same number of miles in each.How many mile

shepuryov [24]

Answer:

The number of miles Mark runs in each track be $16 \frac{43}{52}\ miles$

Step-by-step explanation:

Let us assume that the number of miles Mark run in each track meet be x.

As given

Mark ran 875 miles this year in the track club.

Mark ran in 52 track meets and ran the same number of miles in each.

Than the equation becomes

52 × x = 875

$x = \frac{875}{52}$

$x = 16 \frac{43}{52}\ miles$

Therefore the number of miles Mark runs in each track be $16 \frac{43}{52}\ miles$

7 0

3 years ago

What is the equation for the kind of reflection?

andre [41]

Answer:

A vertical reflection is given by the equation y=−f(x) y = − f ( x ) and results in the curve being “reflected” across the x-axis. A horizontal reflection is given by the equation y=f(−x) y = f ( − x ) and results in the curve being “reflected” across the y-axis.

4 0

3 years ago

Use the​ slope-intercept form to graph the equation

Use the slope-intercept form to graph the equation