Ask question

Ask question

All categories

3 years ago

6

refer back to the first problem on Tuesday,A scale of a map is 1 in:125ft,jon lives 250 feet away from max.how many inches separ

ate Jon's home from Max's on the map?

1 answer:

pishuonlain [190]3 years ago

4 0

Answer:

2 inches

Step-by-step explanation:

The scale means 1 inches on map equal 125 feet in real life

<em>We know in real life Jon lives 250 ft away from Max. 250 is twice of 125, hence in map</em>, it should be 1*2=2 inches.

On the map, it is 2 inches from Jon's home to Max's home.

You might be interested in

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:

$\hat a = max(X_i)$

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.

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

Nine negative third power in standard form

Andre45 [30]

Answer:

9^-3= .0013717421

Step-by-step explanation:

6 0

3 years ago

A dog runs 100 yards from one corner of a rectangular park to the opposite corner. If the length of the park is 28 yards, what i

kap26 [50]

9514 1404 393

Answer:

96 yards

Step-by-step explanation:

The diagonal through the park cuts the rectangle into two right triangles. The Pythagorean theorem tells you the relationship between the sides of a right triangle and its hypotenuse: the sum of the squares of the sides is equal to the square of the hypotenuse.

The diagonal is the hypotenuse, so we have ...

100² = 28² + w²

w = √(10000 -784) = 96

The width of the park is 96 yards.

_____

<em>Additional comment</em>

The integer side lengths of a right triangle form what is called a "Pythagorean triple." One of the most often seen of these is (3, 4, 5). Other commonly seen Pythagorean triples are (5, 12, 13), (7, 24, 25), (8, 15, 17).

You may notice that the numbers here are those of the (7, 24, 25) Pythagorean triple, multiplied by 4. If you recognize the given lengths as having the ratio 28:100 = 7:25, you have the clue you need to determine the answer simply from your knowledge of Pythagorean triples.

8 0

3 years ago

Helpppp me please!!!

Semenov [28]

To find the value of q, we need to find d(-8). Put another way, we need to find the value of d(x) when x = -8

$d(x) = -\sqrt{\frac{1}{2}x+4}$

$d(-8) = -\sqrt{\frac{1}{2}(-8)+4}$

$d(-8) = -\sqrt{-4+4}$

$d(-8) = -\sqrt{0}$

$d(-8) = 0$

So this means q = 0. Note that -0 is just 0.

===========================================

The value of r will be a similar, but now we use f(x) this time.

Plug in x = 0

$f(x) = \sqrt{\frac{1}{2}x+4}$

$f(0) = \sqrt{\frac{1}{2}*0+4}$

$f(0) = \sqrt{0+4}$

$f(0) = \sqrt{4}$

$f(0) = 2$

Therefore, r = 2.

===========================================

For s, we plug x = 10 into f(x)

$f(x) = \sqrt{\frac{1}{2}x+4}$

$f(10) = \sqrt{\frac{1}{2}*10+4}$

$f(10) = \sqrt{5+4}$

$f(10) = \sqrt{9}$

$f(10) = 3$

So s = 3.

===========================================

Finally, plug x = 10 into d(x) to find the value of t

$d(x) = -\sqrt{\frac{1}{2}x+4}$

$d(10) = -\sqrt{\frac{1}{2}(10)+4}$

$d(10) = -\sqrt{5+4}$

$d(10) = -\sqrt{9}$

$d(10) = -3$

A shortcut you could have taken is to note how d(x) = -f(x), so this means

d(10) = -f(10) = -9 since f(10) = 9 was found in the previous section above.

Whichever method you use, you should find that t = -3.

===========================================

<h3>In summary:</h3><h3>q = 0</h3><h3>r = 2</h3><h3>s = 3</h3><h3>t = -3</h3>

8 0

3 years ago

Charlie reads quickly. He reads 1\dfrac371 7 3 1, start fraction, 3, divided by, 7, end fraction pages every \dfrac23 3 2 st

OLEGan [10]

<h2>Answer:</h2>

The number of pages he will read in one minute is:

$2\dfrac{1}{7}\ \text{pages}$

<h2>Step-by-step explanation:</h2>

It is given that:

Charlie reads $1\dfrac{3}{7}\ \text{pages}$ in every $\dfrac{2}{3}\ \text{minutes}$ .

We know that:

$1\dfrac{3}{7}=\dfrac{10}{7}$

This means that:

Charlie reads $\dfrac{10}{7}\ \text{pages}$ in every $\dfrac{2}{3}\ \text{minutes}$ .

This means that:

$\text{In}\ \dfrac{2}{3}\ \text{minutes he reads}\ \dfrac{10}{7}\ \text{pages}\\\\\text{Hence}$

$\text{In}\ 1\ \text{minute he will read}\ \dfrac{\dfrac{10}{7}}{\dfrac{2}{3}}\ \text{pages}$

$\text{In}\ 1\ \text{minute he will read}\ \dfrac{10\times 3}{7\times 2}\ \text{pages}$

Hence,

$\text{In}\ 1\ \text{minute he will read}\ \dfrac{15}{7}\ \text{pages}$

Hence,

$\text{In}\ 1\ \text{minute he will read}\ \dfrac{15}{7}\ \text{pages}$

$\text{In}\ 1\ \text{minute he will read}\ 2\dfrac{1}{7}\ \text{pages}$

4 0

3 years ago

Read 2 more answers