1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lesantik [10]
3 years ago
8

3/5 of your class earned an A or a B last week. 1/4 of those teammates earned an A. what fraction of you class earned an A? Poss

ible answers could be 4/9, 7/20, 2/1, 3/20 but which one is it?
Mathematics
1 answer:
Keith_Richards [23]3 years ago
6 0

Answer:

3/20

Step-by-step explanation:

  • 3/5 earned an A or B = <em>"3/5 of the class"</em>
  • 1/4<em> </em>of 3/5 earned an A = <em>"3/5 of class </em><u><em>divided by 4</em></u><em>"</em>

\frac{3}{5}  ÷  4    =     \frac{3}{5} *   \frac{1}{4}    =      \frac{3 * 1}{5 * 4}     =     \frac{3}{20}

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
Warm-up
Vitek1552 [10]

Answer:

the percent is 18

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Look at Jerry's monthly budget below. What percentage of Jerry's net income is spent on food?
Mila [183]
450/2,970*100
450*10/297=15.2
7 0
3 years ago
Read 2 more answers
What is 4/5 of 55?<br> Pls help me
rjkz [21]

Answer:

44

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
What is 89+78+90-67÷34
Mariulka [41]
Simplify the expression.
Exact Form:
8671/34

Decimal Form:
255.02941176
…
Mixed Number Form:
255 1/34
5 0
3 years ago
Read 2 more answers
Other questions:
  • Find the missing angles with reasons. plz quick it is due today
    11·2 answers
  • What is the total area of an 8”x12” picture and it’s Frame if the frame is 1.5” thick
    12·1 answer
  • Factor completely 96-6x^2
    11·1 answer
  • Solve for x: <br> 3x+2y=7
    7·1 answer
  • What is the answer to − 4/7 = 5/2t ( its a one step equation)
    14·1 answer
  • I WILL MARK BRAINLIST
    6·1 answer
  • PrettPlease plsssss I beg of you sor
    15·1 answer
  • AYYYY YO rate my fine self
    7·1 answer
  • Identify the statement that uses the verb mood in the interrogative. will give brainliest
    12·2 answers
  • PLSSS HELP IF YOU TURLY KNOW THISS
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!