Solution :
a). [tex]$\text{Var} (X|Y) =E ((X-E(X|Y))^2 |Y)$[/tex]
Now, if X = Y, then :
[tex]P(X|Y)=\left\{\begin{matrix} 1,& \text{if } x=y \\ 0, & \text{otherwise }\end{matrix}\right.[/tex]
Then, E[X|Y] = x = y
So, [tex]$\text{Var} (X|Y) =E((X-X)^2 |Y)$[/tex]
[tex]$=E(0|Y)$[/tex]
= 0
Therefore, this statement is TRUE.
b). If X = Y , then Var (X) = Var (Y)
And as Var (X|Y) = 0, so Var (X|Y) ≠ Var (X), except when all the elements of Y are same.
So this statement is FALSE.
c). As defined earlier,
[tex]$\text{Var} (X|Y) =E ((X-E(X|Y))^2 |Y=y)$[/tex]
So, this statement is also TRUE.
d). The statement is TRUE because [tex]$\text{Var} (X|Y) =E ((X-E(X|Y))^2 |Y=y)$[/tex].
e). FALSE
Because, [tex]$\text{Var} (X|Y) =E ((X-E(X|Y=y))^2 |Y=y)$[/tex]
A rectangle's length is three times as long as it is wide. Which expression represents the change in area if the width of the rectangle is increased by 1?
1. 3x^2
2. 3x
3. 3x^2+3x
4. the area increases by 3
Step-by-step explanation:
Let's say the rectangle's width is equal to y. We know that the length is three times the width, so the length = 3 * y. We also know that the area for a rectangle is equal to length * width, so the area, z, is equal to
(3*y) * y = z
3 * y² = z
Now, let's increase the width of the rectangle by 1. We can replace y with y+1 (as y+1 is 1 greater than y), and 3 * y with 3 * (y+1) to get
3*(y+1) * (y+1) = new area
(3y+3)*(y+1) = new area
3y²+3y +3 y + 3 = new area
3y² + 6y + 3 = new area
The difference in area is equal to the new area subtracted by the old area, or
3y²+6y+3 - 3y² = 6y +3. The variable for x is not given, so if x = (2y+1), the answer would be the second choice. However, solely using the information given, it is impossible to determine a solution outside of saying that it is not option 4, as 6y + 3 ≠ 3
RESUELVE USANDO LAS PROPIEDADES DE LA POTENCIA
PLISSSSSSSSS CON PROCEDIMIENTOOOOOOO
Answer:
Tenemos dos propiedades de la potencia en este caso:
Para un numero real A:
[tex]A^0 = 1[/tex]
[tex](A^n)^m = A^{n*m}[/tex]
En este caso nuestra ecuación es:
[tex][ [(\frac{0.1234}{-3.2098})^4]^3]^0[/tex]
usando la segunda propiedad, podemos reescribir como:
[tex][ [(\frac{0.1234}{-3.2098})^4]^3]^0 = (\frac{0.1234}{-3.2098})^{4*3*0} = (\frac{0.1234}{-3.2098})^0[/tex]
Y acá tenemos un numero real a la potencia 0, sabemos que esto es igual a 1, entonces:
[tex](\frac{0.1234}{-3.2098})^0 = 1[/tex]
how do we get 24 using 3,3,7 and7
Answer:
2 Answers. #1. +11. [3+(3/7)] times 7 is 24. DarkBlaze347 May 1, 2015. +5. Good job, DB! civonamzuk May 1, 2015.
35 Online Users.
Step-by-step explanation:
brainliest please and follow:D
The probability that an individual has 20-20 vision is 0.18. In a class of 12 students, what is the probability of finding five people with 20-20 vision?
0.417 or 0.185 or 0.18 or 0.037
Answer:
0.417
Step-by-step explanation:
Just divide 12/5 and the answer is 0.416666666...
Round up and you get 0.417.
Hope this helped!
please simplify this one. I need answers fast as possible .(chapter name : surds )
[tex] \sqrt[5]{32} \times 2 \sqrt[3]{81} \\ = {32}^{ \frac{1}{5} } \times 2 {(81)}^{ \frac{1}{3} } \\ = ({{2}^{5}})^{ \frac{1}{5} } \times {2({3}^{3})}^{ \frac{1}{3} } \\ = {2}^{1} \times 2({3}^{1}) \\ = 2 \times 2 \times 3 \\ = 4 \times 3 \\ = 12[/tex]
This is the answer.
Hope it helps!!
Pls solve the above question
Kindly don't spam+_+
Answer:
Step-by-step explanation:
Given expressions are,
[tex]p=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}+\sqrt{5}}[/tex] and [tex]q=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]
Remove the radicals from the denominator from both the expressions.
[tex]p=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}+\sqrt{5}} \times \frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]
[tex]=\frac{(\sqrt{10}-\sqrt{5})^2}{(\sqrt{10})^2-(\sqrt{5})^2}[/tex]
[tex]=\frac{(\sqrt{10}-\sqrt{5})^2}{5}[/tex]
[tex]\sqrt{p}=\sqrt{\frac{(\sqrt{10}-\sqrt{5})^2}{5}}[/tex]
[tex]=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{5}}[/tex]
[tex]q=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]
[tex]=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}\times \frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}+\sqrt{5}}[/tex]
[tex]=\frac{(\sqrt{10}+\sqrt{5})^2}{(\sqrt{10})^2-(\sqrt{5})^2}[/tex]
[tex]=\frac{(\sqrt{10}+\sqrt{5})^2}{5}[/tex]
[tex]\sqrt{q}=\sqrt{\frac{(\sqrt{10}+\sqrt{5})^2}{5}}[/tex]
[tex]=\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}}[/tex]
[tex]\sqrt{q}-\sqrt{p}-2\sqrt{pq}=\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}}-\frac{(\sqrt{10}-\sqrt{5})}{\sqrt{5}}-2(\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}})(\frac{(\sqrt{10}-\sqrt{5})}{\sqrt{5}})[/tex]
[tex]=\frac{1}{\sqrt{5}}(\sqrt{10}+\sqrt{5}-\sqrt{10}+\sqrt{5})-\frac{2}{5}[(\sqrt{10})^2-(\sqrt{5})^2)][/tex]
[tex]=\frac{1}{\sqrt{5}}(2\sqrt{5})-\frac{2}{5}(10-5)[/tex]
[tex]=2-2[/tex]
[tex]=0[/tex]
Suppose U1 and U2 are i.i.d. Unif(0,1) withU1=0.1 and U2=0.8. Use the "cosine" version of Box-Muller to generate a single Nor(-1,4) random variate. Don't forget to use radians instead of degrees.
a. 0.326
b. 0.326
c. 0.663
d. 1.96
Answer:
0.663 ( c )
Step-by-step explanation:
U1 = 0.1 , U2 = 0.8
using the "cosine" version of Box-Muller to generate a single Nor(-1,4) random variable
first step : generate single obsⁿ from N ( -1,4 )
attached below is the detailed solution
Help! please don't just steal my pointss
Answer:
hi, option C is correct because it has a right angel. please give brainliest
one strip is cut into 9 equal bars shade 1/3:of strip
hiiksbsjxbxjsoahwjsissnsks
Drag the tiles to the correct boxes to complete the pairs.
Match each division of rational expressions with its quotient.
Answer:
Step-by-step explanation:
Um where is the diagrahm
What is the product (4.42 x 103)(5 x 10^) written in
scientific notation?
Answer:
2.2763 x 10 to the power of 4
for some reason it doesn't let me put in the explanation
The distribution of the number of children for families in the United States has mean 0.9 and standard deviation 1.1. Suppose a television network selects a random sample of 1000 families in the United States for a survey on TV viewing habits.
Required:
a. Describe (as shape, center and spread) the sampling distribution of the possible values of the average number of children per family.
b. What average numbers of children are reasonably likely in the sample?
c. What is the probability that the average number of children per family in the sample will be 0.8 or less?
d. What is the probability that the average number of children per family in the sample will be between 0.8 and 1.0?
Answer:
a) By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.
b) Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.
c) 0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less
d) 0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean 0.9 and standard deviation 1.1.
This means that [tex]\mu = 0.9, \sigma = 1.1[/tex]
Suppose a television network selects a random sample of 1000 families in the United States for a survey on TV viewing habits.
This means that [tex]n = 1000, s = \frac{1.1}{\sqrt{1000}} = 0.035[/tex]
a. Describe (as shape, center and spread) the sampling distribution of the possible values of the average number of children per family.
By the Central Limit Theorem, it has an approximately normal shape, with mean(center) 0.9 and standard deviation(spread) 0.035.
b. What average numbers of children are reasonably likely in the sample?
By the Empirical Rule, 95% of the sample is within 2 standard deviations of the mean, so:
0.9 - 2*0.035 = 0.83
0.9 + 2*0.035 = 0.97
Average numbers of children between 0.83 and 0.97 are reasonably likely in the sample.
c. What is the probability that the average number of children per family in the sample will be 0.8 or less?
This is the p-value of Z when X = 0.8. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]
[tex]Z = -2.86[/tex]
[tex]Z = -2.86[/tex] has a p-value of 0.0021
0.0021 = 0.21% probability that the average number of children per family in the sample will be 0.8 or less.
d. What is the probability that the average number of children per family in the sample will be between 0.8 and 1.0?
p-value of Z when X = 1 subtracted by the p-value of Z when X = 0.8.
X = 1
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1 - 0.9}{0.035}[/tex]
[tex]Z = 2.86[/tex]
[tex]Z = 2.86[/tex] has a p-value of 0.9979
X = 0.8
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.8 - 0.9}{0.035}[/tex]
[tex]Z = -2.86[/tex]
[tex]Z = -2.86[/tex] has a p-value of 0.0021
0.9979 - 0.0021 = 0.9958
0.9958 = 99.58% probability that the average number of children per family in the sample will be between 0.8 and 1.0
Recall the creative calligraphy case we discussed in class. Suppose you have received a rush order of 55 invitations that have to be created in the next (four hour) work session. What is the most time You (i.e. not Susie) can spend writing on the card and envelope for each invitation and still fill the order
Answer:
About 4 minutes and 3 seconds
Step-by-step explanation:
If I have 55 invitations to be created in four hours.
4 hours × 60 =240 minutes
240/55= 4.36 minutes
So if I spend 4 minutes and 3 seconds on an invitation, I should be able to fill the order.
In the diagram, point D is the center of the medium-sized circle that passes through C and E, and it is also the center of the largest circle that passes through A and G. Each of the diameters of the small circles with centers B and F equals the radius of the medium-sized circle with center D. The shaded area is what fraction of the largest circle?Single choice.
9514 1404 393
Answer:
5/8
Step-by-step explanation:
The area of the smaller circles is proportional to the square of the ratio of their diameters. The two smallest circles have diameters equal to 1/4 the diameter of the largest circle. Hence their areas are (1/4)^2 = 1/16 of that of the largest circle.
Similarly, the medium circle has a diameter half that of the largest circle, so its area is (1/2)^2 = 1/4 of the are of the largest circle.
The smaller circles subtract 2×1/16 +1/4 = 3/8 of the area of the largest circle. Then the shading is 1-3/8 = 5/8 of the area of the largest circle.
Aron flips a penny 9 times. Which expression represents the probability of getting exactly 3 heads? P (k successes) = Subscript n Baseline C Subscript k Baseline p Superscript k Baseline (1 minus p) Superscript n minus k. Subscript n Baseline C Subscript k Baseline = StartFraction n factorial Over (n minus k) factorial times k factorial EndFraction Subscript 9 Baseline C Subscript 3 Baseline (0.5) cubed (0.5) Superscript 6 Subscript 9 Baseline C Subscript 3 Baseline (0.5) cubed Subscript 9 Baseline C Subscript 3 Baseline (0.5) cubed (0.5) Superscript 9 Subscript 9 Baseline C Subscript 6 Baseline (0.5) Superscript 6
Answer:
[tex]P(3) = ^9C_3 * 0.5^3 *0.5^6[/tex]
Step-by-step explanation:
Given
[tex]n = 9[/tex] --- number of flips
Required
[tex]P(x = 3)[/tex]
The probability of getting a head is:
[tex]p = \frac{1}{2}[/tex]
[tex]p = 0.5[/tex]
The distribution follows binomial probability, and it is calculated using:
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]
So, we have:
[tex]P(3) = ^9C_3 * 0.5^3 * (1 - 0.5)^{9-3}[/tex]
[tex]P(3) = ^9C_3 * 0.5^3 *0.5^6[/tex]
Answer:
Aron flips a penny 9 times. Which expression represents the probability of getting exactly 3 heads?
Answer: A
Step-by-step explanation:
a boat leaves port at 13:52 and arrives at its destination 3 and a half hours later. at what time does the boat arrive
16:52
Hope this helps! :)
If f(x) = 4x ^ 2 - 4x - 8 and g(x) = 2x ^ 2 + 3x - 6 then f(x) - g(x) * i * s
Answer:
[tex]4 {x}^{2} - 4x - 8 - (2 {x}^{2} + 3x - 6) = 4 {x}^{2} - 4x - 8 - 2 {x}^{2} - 3x + 6 = 2 {x}^{2} - 7x - 2[/tex]
find the slope of a line perpendicular to the line below. y=2x+4
A family has inherited $300,000. If they choose to invest the $300,000 at 12\% per year compounded quarterly, how many quarterly withdrawals of $25000 can be made? (Assume that the first withdrawal is three months after the investment is made).
Step-by-step explanation:
ejejejejrjruruehehhr
Help Pleasss I will give brainlyest!!!! :D
Answer: The answer is 2,4. May I have the brainiest? pls I only need one more.
A school band found they could arrange themselves in rows of 6, 7, or 8 with no one left over. What is the minimum number of students in the band?
Answer:
168 is the answer if i m not wrong.I took the LCM.
If school band found they could arrange themselves in rows of 6, 7, or 8 with no one left over, the minimum number of students in the band is 168.
To find the minimum number of students in the band, we need to determine the least common multiple (LCM) of the numbers 6, 7, and 8.
The LCM is the smallest multiple that is divisible by all the given numbers.
Prime factorizing each number, we have:
6 = 2 * 3
7 = 7
8 = 2 * 2 * 2
To find the LCM, we take the highest exponent for each prime factor:
2³ * 3 * 7 = 168
By having 168 students, they can arrange themselves into rows of 6 (28 rows), 7 (24 rows), or 8 (21 rows) without anyone being left over. Any fewer than 168 students would result in at least one row having students left over.
To learn more about LCM click on,
https://brainly.com/question/1771764
#SPJ2
a/b=2/5 and b/c=3/8 find a/c
Answer:
3/20
Step-by-step explanation:
By question it's given that ,
[tex]\implies \dfrac{a}{b}=\dfrac{2}{5}[/tex]
[tex]\implies \dfrac{b}{c}=\dfrac{3}{8}[/tex]
And we need to find out the value of a/c .For that Multiply both of them , we have ;
[tex]\implies \dfrac{a}{b} \times\dfrac{b}{c}=\dfrac{2}{5}\times \dfrac{3}{8} \\\\\implies \dfrac{a}{c}= \dfrac{3}{20}[/tex]
Hence the required answer is 3/20 .
Tim and Al are bricklayers. Tim can construct an outdoor grill in 5 days. If Al helps Tim, they can build it in only 3 days. How long
would it take Al to build the grill alone? Write your answer as an integer, simplified fraction, or mixed number.
It would take Al
days to build the grill alone.
Answer:
It would take Al 7.5 days to build the grill alone.
Step-by-step explanation:
Since Tim and Al are bricklayers, and Tim can construct an outdoor grill in 5 days, and if Al helps Tim, they can build it in only 3 days, to determine how long would it take Al to build the grill alone should be done the following calculation:
1/5 + X = 1/3
0.20 + X = 0.333
X = 0.333 - 0.20
X = 0.1333333
X = 1 / 7.5
Therefore it would take Al 7.5 days to build the grill alone.
Can anyone help please?
Answer:
h(t) = -16t(t-6)
h(2) = 128
Step-by-step explanation:
h(t) = -16t² + 96t
h(t) = -16t(t-6)
t = 3
h(2) = -16(2)(2 - 6)
h(2) = 128
Find the 11th term of the sequence
3, -6, 12, -24,...
3072
6144
-6144
-3072
Answer:
-3072
Step-by-step explanation:
The greatest common factor of 45a^2b^3 and 18a^4b
Answer:
9a²b
Step-by-step explanation:
Hi there!
We need to find the greatest common factor out of 45a²b³ and 18[tex]a^{4}[/tex]b
We can split apart the monomials to make it easier
45a²b³ is 45*a²b³
18[tex]a^{4}[/tex]b is 18*[tex]a^{4}[/tex]b
First, let's find the GCF out of 45 and 18 (the number coefficients)
we can find all of the multiples of the 2 numbers:
45 is made up of 9 and 5
9 is made up of 3 and 3
so 3*3*5 is 45
18 is made up of 2 and 9
9 is made up of 3 and 3
so 2*3*3 is 18
3*3 is in both 45 and 18, so 9 is the GCF out of 45 and 18
Now let's find the GCF out of a²b³ and [tex]a^{4}[/tex]b
a²b³ made up of a² and b³
so a²b³ is a*a*b*b*b
[tex]a^{4}[/tex]b is made up of [tex]a^{4}[/tex] and b
so [tex]a^{4}[/tex]b is a*a*a*a*b
a*a*b is in both a²b³ and [tex]a^{4}[/tex]b, so the GCF out of a²b³ and [tex]a^{4}[/tex]b is a²b
Now multiply 9 and a²b together, as they are only the GCF of the parts of the monomials
9*a²b=9a²b
there's the greatest common factor of the 2 monomials
Hope this helps!
A developer wants to purchase a plot of land to build a house. The area of the plot can be described by the following expression: (5x+1)(7x−7) where x is measured in meters. Multiply the binomials to find the area of the plot in standard form
Answer:
35x^2 - 28x - 7
Step-by-step explanation:
Find all the roots of the equation
[tex] {x}^{6} - 64 = 0[/tex]
Answer:
x = 2
x =-2
x =1 +√3i
x =1 -√3i
x =-1 +√3i
x =-1 -√3i
Step-by-step explanation:
[tex]x^{6} -64 = 0\\(x^{3} -8)(x^{3} + 8) = 0\\ \\( x - 2) (x^{2} + 2x + 4)( x + 2) (x^{2} -2x + 4) = 0\\[/tex]
x = 2
x =-2
x =1 +√3i
x =1 -√3i
x =-1 +√3i
x =-1 -√3i
How many cubes with side lengths of 1/2 cm does it take to fill the prism?
Answer:
24
Step-by-step explanation:
You first find out how many cubes can fit into each measurement, then multiply them. (2*4*3=24)
Answer:
It will take 24 cubes to fill the rectangular prism.
Step-by-step explanation:
Find the volume of a cube with side lengths of 1/2 cm:
1/2^3 = 1/8
1/8 cm^3
Find the volume of the whole rectangular prism (lwh):
1 x 3/2 x 2
= 3/2 x 2
= 3
3 cm^3
Divide the volume of the prism by the bolume of one cube:
3 ÷ 1/8 = 24
Therefore it will take 24 cubes to fill the prism. Hope this helps!
Rajah / Diagram 5 (b) Dalam Rajah 6, PQ ialah tangen sepunya dua bulatan. AQ dan BQ ialah tangen bagi bulatan yang masing-masing berpusat E dan F. Cari nilai x dan y. In Diagram 6, PQ is the common tangent of two circles. AQ and BQ is the tangent to the circles with centre E and F respectively. Find the value of x and y. [3 markah.
Answer:
36281629273781646181993836619946527189119292937467482919198$7473828191927364732818919283838292927383883829118661552621718919191019284746617171819001187373765252728
Step-by-step explanation:
173899918377+28910873638282