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Vsevolod [243]
2 years ago
6

the total cost of 8 pens and 9 books is $30.05. if the cost of a pen is $a, and a book is .90 MORE expensive. find A. Give your

answer in two decimal places.
Mathematics
2 answers:
AysviL [449]2 years ago
5 0

Answer:

2 Answers By Expert Tutors ... Let's set x = cost of books and y = cost of pens. ... In algebra, this can be written as x + y = 18. ... Using x and y, we get 1 book and 2 pens costs $21, or x + 2y = 21 ... The pen must cost $3 because we are adding $3 to the total when we ... Please provide a valid phone number.

Step-by-step explanation:

Kazeer [188]2 years ago
4 0
A= 2.74375 or just 3
Books= $8.10
Pens= 3 or 2.7475
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PLEASE HELP ME
Zolol [24]
It is 4/3. the quantity of 4 over 3
X+6/3 - x+2/3= x+6-x-2/3 = 4/3.
8 0
2 years ago
Hirva ate 5/8 of a medium pizza Elizabeth ate 1/4 of the pizza how much pizza did they eat all together?
Sliva [168]
I’m pretty sure this is 7/8!
3 0
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)  

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
If you had 38/12 dollars, how much money would that be in dollars and cents (rounded to the nearest penny)?
azamat
$3.17 Just turn the improper fraction into a mixed number which would be 3 1/6. Then, turn THAT into a decimal rounded to the nearest hundredth (pennies) which leads to $3.17.
8 0
2 years ago
Read 2 more answers
Match each graph with its corresponding equation
s344n2d4d5 [400]
The equation of the parabolas given will be found as follows:
a] general form of the parabolas is:
y=k(ax^2+bx+c)
taking to points form the first graph say (2,-2) (3,2), thus
y=k(x-2)(x-3)
y=k(x^2-5x+6)

taking another point (-1,5) 
5=k((-1)^2-5(-1)+6)
5=k(1+5+6)
5=12k
k=5/12
thus the equation will be:
y=5/12(x^2-5x+6)

b] Using the vertex form of the quadratic equations:
y=a(x-h)^2+k
where (h,k) is the vertex
from the graph, the vertex is hence: (-2,1)
thus the equation will be:
y=a(x+2)^2+1
taking the point say (0,3) and solving for a
3=a(0+2)^2+1
3=4a+1
a=1/2
hence the equation will be:
y=1/2(x+2)^2+1


5 0
3 years ago
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