Annual windstorm losses, X and Y, in two different regions are independent, and each is uniformly distributed on the interval [0, 10]. Calculate the covariance of X and Y, given that X+ Y < 10.
Answers
Answer:
[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]
Step-by-step explanation:
Given
[tex]Interval =[0,10][/tex]
[tex]X + Y < 10[/tex]
Required
[tex]Cov(X,Y)[/tex]
First, we calculate the joint distribution of X and Y
Plot [tex]X + Y < 10[/tex]
So, the joint pdf is:
[tex]f(X,Y) = \frac{1}{Area}[/tex] --- i.e. the area of the shaded region
The shaded area is a triangle that has: height = 10; width = 10
So, we have:
[tex]f(X,Y) = \frac{1}{0.5 * 10 * 10}[/tex]
[tex]f(X,Y) = \frac{1}{50}[/tex]
[tex]Cov(X,Y)[/tex] is calculated as:
[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]
Calculate E(XY)
[tex]E(XY) =\int\limits^X_0 {\int\limits^Y_0 {\frac{XY}{50}} \, dY} \, dX[/tex]
[tex]X + Y < 10[/tex]
Make Y the subject
[tex]Y < 10 - X[/tex]
So, we have:
[tex]E(XY) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{XY}{50}} \, dY} \, dX[/tex]
Rewrite as:
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {XY}} \, dY} \, dX[/tex]
Integrate Y
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{XY^2}{2}}} }|\limits^{10 - X}_0 \, dX[/tex]
Expand
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2} - \frac{X(0)^2}{2}}} }\ dX[/tex]
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2}}} }\ dX[/tex]
Rewrite as:
[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X(10 - X)^2\ dX[/tex]
Expand
[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X*(100 - 20X + X^2)\ dX[/tex]
[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 100X - 20X^2 + X^3\ dX[/tex]
Integrate
[tex]E(XY) =\frac{1}{100} [\frac{100X^2}{2} - \frac{20X^3}{3} + \frac{X^4}{4}]|\limits^{10}_0[/tex]
Expand
[tex]E(XY) =\frac{1}{100} ([\frac{100*10^2}{2} - \frac{20*10^3}{3} + \frac{10^4}{4}] - [\frac{100*0^2}{2} - \frac{20*0^3}{3} + \frac{0^4}{4}])[/tex]
[tex]E(XY) =\frac{1}{100} ([\frac{10000}{2} - \frac{20000}{3} + \frac{10000}{4}] - 0)[/tex]
[tex]E(XY) =\frac{1}{100} ([5000 - \frac{20000}{3} + 2500])[/tex]
[tex]E(XY) =50 - \frac{200}{3} + 25[/tex]
Take LCM
[tex]E(XY) = \frac{150-200+75}{3}[/tex]
[tex]E(XY) = \frac{25}{3}[/tex]
Calculate E(X)
[tex]E(X) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{X}{50}}} \, dY} \, dX[/tex]
Rewrite as:
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {X}} \, dY} \, dX[/tex]
Integrate Y
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 { (X*Y)|\limits^{10 - X}_0 \, dX[/tex]
Expand
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)] - [X * 0])\ dX[/tex]
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)]\ dX[/tex]
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 10X - X^2\ dX[/tex]
Integrate
[tex]E(X) =\frac{1}{50}(5X^2 - \frac{1}{3}X^3)|\limits^{10}_0[/tex]
Expand
[tex]E(X) =\frac{1}{50}[(5*10^2 - \frac{1}{3}*10^3)-(5*0^2 - \frac{1}{3}*0^3)][/tex]
[tex]E(X) =\frac{1}{50}[5*100 - \frac{1}{3}*10^3][/tex]
[tex]E(X) =\frac{1}{50}[500 - \frac{1000}{3}][/tex]
[tex]E(X) = 10- \frac{20}{3}[/tex]
Take LCM
[tex]E(X) = \frac{30-20}{3}[/tex]
[tex]E(X) = \frac{10}{3}[/tex]
Calculate E(Y)
[tex]E(Y) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{Y}{50}}} \, dY} \, dX[/tex]
Rewrite as:
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {Y}} \, dY} \, dX[/tex]
Integrate Y
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 { (\frac{Y^2}{2})|\limits^{10 - X}_0 \, dX[/tex]
Expand
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] - [\frac{(0)^2}{2}])\ dX[/tex]
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] )\ dX[/tex]
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 [\frac{100 - 20X + X^2}{2}] \ dX[/tex]
Rewrite as:
[tex]E(Y) =\frac{1}{100}\int\limits^{10}_0 [100 - 20X + X^2] \ dX[/tex]
Integrate
[tex]E(Y) =\frac{1}{100}( [100X - 10X^2 + \frac{1}{3}X^3]|\limits^{10}_0)[/tex]
Expand
[tex]E(Y) =\frac{1}{100}( [100*10 - 10*10^2 + \frac{1}{3}*10^3] -[100*0 - 10*0^2 + \frac{1}{3}*0^3] )[/tex]
[tex]E(Y) =\frac{1}{100}[100*10 - 10*10^2 + \frac{1}{3}*10^3][/tex]
[tex]E(Y) =10 - 10 + \frac{1}{3}*10[/tex]
[tex]E(Y) =\frac{10}{3}[/tex]
Recall that:
[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]
[tex]Cov(X,Y) = \frac{25}{3} - \frac{10}{3}*\frac{10}{3}[/tex]
[tex]Cov(X,Y) = \frac{25}{3} - \frac{100}{9}[/tex]
Take LCM
[tex]Cov(X,Y) = \frac{75- 100}{9}[/tex]
[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]
Related Questions
How do you work out the total surface area of a cuboid (equation)
Answers
Answer:
A cuboid has a total of 6 rectangular sides. If you calculate the area of each of the 6 rectangular sides and add them up, you will get the surface area of the cuboid.
Surface Area of a Rectangle = Width x Height
please answer i will mark brainliest :)
Answers
Answer:
c I think that's the right answer
Select all the expressions that are equivalent to 13 - x *
Answers
You gave no options but the simplest way to solve this is to set it equal to all of the answers to see if it is correct.
HELP ANYONE OUT THERE WITH MY MATH
Answers
Answer:
x is 37.,.,.,.,.,.,.,.,.,.,.,.,.,.,.,.
Answer:
x = 37
Step-by-step explanation:
Interior angles of a square is 90 each.
∠JKL = 90°
Given ∠JKL = 3x - 21
Therefore, 3x - 21 = 90
3x = 90 + 21
3x = 111
x = 37
find the value of tan 45° + sin 30°
Answers
Answer:
I think it's approximately 44°
Will mark Brainlest hellpppp
Answers
[tex]h(-3) - h( - 2) = - \frac{5}{ 8} \\ \\ [/tex]
Step-by-step explanation:
[tex]\boxed{h(x) = \frac{2 {x}^{2} - x + 1}{3x - 2}}[/tex]
[tex]h(2) = \frac{2 {(2)}^{2} - (2) + 1}{3(2) - 2} [/tex]
[tex]h(2) = \frac{2 {(4)} - 2 + 1}{6 - 2} [/tex]
[tex]h(2) = \frac{ 8 - 1}{4} [/tex]
[tex]h(2) = \frac{7}{4}[/tex]
[tex]h( - 3) = \frac{2 {( - 3)}^{2} - ( - 3) + 1}{3( - 3) - 2} \\ h( - 3) = \frac{2 (9) + 3 + 1}{ - 9- 2} \\ h( - 3) = \frac{18 + 4}{ - 11} \\ h( - 3) = \frac{22}{ - 11} \\ h(-3) = -2[/tex]
[tex]h( - 2) = \frac{2 {( - 2)}^{2} - ( - 2) + 1}{3( - 2) - 2} \\ h( - 2) = \frac{2 (4) + 2+ 1}{- 6 - 2} \\ h( - 2) = \frac{8 + 3}{- 8} \\ h( - 2) = \frac{11}{- 8} [/tex]
[tex]h(3) - h( - 2) = -2 - \frac{11}{- 8} \\ h(3) - h( - 2) =-2 + \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-2}{1}+ \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-16}{8} + \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-16+11}{8} \\ h(3) - h( - 2) = \frac{-5}{8} \\ - \frac{5}{ 8} [/tex]
22 hours. A
Silas ran 100 m race at a speed of 8 m/s. How long did it take him to com-
plete the race?
Answers
Answer:
100/8 = 12.5 (s)
Step-by-step explanation:
hmm bro
Answer:
[tex]{ \boxed{ \tt{formular : { \bf{ \green{time = \frac{distance}{speed} }}}}}} \\ time = \frac{100}{8} \\ { \boxed{time = 12.5 \: seconds}} \\ \\ { \underline{ \blue{ \tt{becker \: jnr}}}}[/tex]
HELP ME CANT FAIL!! Which parent function is represented by the graph?
A) An exponential parent function
B) Liner parent function
C)Absolute value parent function
D)Quadratic parent function
Answers
Explanation:
A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds. Find the test statistic to decide whether the mean transaction time exceeds 60 seconds.
a. 1.457
b. 2.333
c. 1.848
d. 2.037
Answers
Answer:
b. 2.333
Step-by-step explanation:
Test if the mean transaction time exceeds 60 seconds.
At the null hypothesis, we test if the mean transaction time is of 60 seconds, that is:
[tex]H_0: \mu = 60[/tex]
At the alternate hypothesis, we test if it exceeds, that is:
[tex]H_1: \mu > 60[/tex]
The test statistic is:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.
60 is tested at the null hypothesis:
This means that [tex]\mu = 60[/tex]
A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds.
This means that [tex]n = 16, X = 67, s = 12[/tex]
Value of the test statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{67 - 60}{\frac{12}{\sqrt{16}}}[/tex]
[tex]t = \frac{7}{3}[/tex]
[tex]t = 2.333[/tex]
Thus, the correct answer is given by option b.
Un recipiente cilíndrico tiene un diámetro de 160cm. y la misma altura. ¿Cuánto litros de agua puede caber?
Answers
Respuesta:
3215360 cm³
Explicación paso a paso:
Dado que :
Radio = 160/2 = 80 cm
Altura, h = 160 cm
Volumen de un cilindro = πr²h
Volumen del cilindro = π * 80² * 160
Volumen = 3215360 cm³
Please help
A. B.C. D
Answers
9x5
pls help meeeeeeeeee
Answers
Answer:
45
hope this helps
Answer:
45
Step-by-step explanation:
9x5=45
Somebody help me with this
Answers
Multiply(distribute) (x2-x+7) with (3x2+4x-2) and combine like terms.
Answer:
3x4 +x3+ 15x2+30x-14
Step-by-step explanation:
..........................................................
Answers
Helpppppppp thank u so much
Answers
Answer:
acute triangle
Step-by-step explanation:
First find the other angle
The sum of the angles is 180
x+65+60 = 180
x + 125 = 180
x = 180-125
x = 55
All three angles are different so all three side lengths are different
All the angles are less than 90 so all the angles are acute
This is an acute scalene triangle
Answer:
im is 55 so the only correct option is c
Step-by-step explanation:
hope this helps! have a great day!!
how do I solve this equation?
[tex]x + 8y = 20 \\ 2x + 4y = 4[/tex]
Answers
Answer:
Subtract x x from both sides of the equation. 8y=20−x 8 y = 20 - x. Divide each term by 8 8 and simplify.
Solve for x 2x-4y=4. 2x−4y=4 2 x - 4 y = 4. Add 4y 4 y to both sides of the equation. 2x=4+4y 2 x = 4 + 4 y. Divide each term by 2 2 and simplify.
2(-8y + 20) +4y = 4
-16y + 40 + 4y = 4
-12y +40=4
Subtract 40 from each side
-12y= -36
Divide both by -12
y = 3
Plug Y=3 in to find x
X + 8(3) = 20
X + 24 = 20
Subtract 24
x = -4
Differentiate the function. y = (2x - 5)^2 (5 - x)?
Answers
Answer:
[tex]\displaystyle y' = -(2x - 5)(6x - 25)[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightDistributive Property
Algebra I
Terms/CoefficientsFactoringCalculus
Derivatives
Derivative Notation
Derivative of a constant is 0
Basic Power Rule:
f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹Derivative Rule [Product Rule]: [tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
Step 1: Define
Identify
y = (2x - 5)²(5 - x)
Step 2: Differentiate
Derivative Rule [Product Rule]: [tex]\displaystyle y' = \frac{d}{dx}[(2x - 5)^2](5 - x) + (2x - 5)^2\frac{d}{dx}[(5 - x)][/tex]Chain Rule [Basic Power Rule]: [tex]\displaystyle y' = [2(2x - 5)^{2 - 1} \cdot \frac{d}{dx}[2x]](5 - x) + (2x - 5)^2\frac{d}{dx}[(5 - x)][/tex]Simplify: [tex]\displaystyle y' = [2(2x - 5) \cdot \frac{d}{dx}[2x]](5 - x) + (2x - 5)^2\frac{d}{dx}[(5 - x)][/tex]Basic Power Rule: [tex]\displaystyle y' = [2(2x - 5) \cdot 1 \cdot 2x^{1 - 1}](5 - x) + (2x - 5)^2(1 \cdot -x^{1 - 1})][/tex]Simplify: [tex]\displaystyle y' = [2(2x - 5) \cdot 2](5 - x) + (2x - 5)^2(-1)[/tex]Multiply: [tex]\displaystyle y' = 4(2x - 5)(5 - x) - (2x - 5)^2[/tex]Factor: [tex]\displaystyle y' = (2x - 5)[4(5 - x) - (2x - 5)][/tex][Distributive Property] Distribute 4: [tex]\displaystyle y' = (2x - 5)[20 - 4x - (2x - 5)][/tex][Distributive Property] Distribute negative: [tex]\displaystyle y' = (2x - 5)[20 - 4x - 2x + 5][/tex][Subtraction] Combine like terms (x): [tex]\displaystyle y' = (2x - 5)[20 - 6x + 5][/tex][Addition] Combine like terms: [tex]\displaystyle y' = (2x - 5)(25 - 6x)[/tex]Factor: [tex]\displaystyle y' = -(2x - 5)(6x - 25)[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Derivatives
Book: College Calculus 10e
Peggy constructed the 95 percent confidence interval (4.8,5.2) to estimate the slope of a regression model for a set of bivariate data with 24 data values. Peggy claims that the width of the confidence interval will increase if a sample size of 30 is used, all other things remaining the same. Quincy claims that the width of the confidence interval will decrease if a sample size of 30 is used. Which statement is true about the claims made by Peggy and Quincy?
А. Peggy's claim is correct.
B. Quincy's claim is correct.
C. Both Peggy's claim and Quincy's claim are correct
D. Neither Peggy's claim nor Quincy's claim is correct.
E. There is not enough information to determine whether the claims are correct.
Answers
Answer:
B. Quincy's claim is correct.
Step-by-step explanation:
Margin of error of a confidence interval:
The margin of error of a confidence interval has the following format:
[tex]M = z\frac{s}{\sqrt{n}}[/tex]
In which z is related to the confidence level, s is the standard error and n is the size of the sample.
From this interval, we have that the margin of error and the sample size are inversely proportional, that is, if we increase the sample size, the margin of error decreases, and so does the width of the confidence interval.
Peggy claims that the width of the confidence interval will increase if a sample size of 30 is used, all other things remaining the same.
Peggy is wrong, as the increase of the sample size results on the decrease of the margin of error, and a decrease of the width.
Quincy claims that the width of the confidence interval will decrease if a sample size of 30 is used.
Margin of error decreases, and so does the width of the interval, thus, Quincy's claim is correct, and the correct answer is given by option b.
Right triangle ABC is shown.
Triangle A B C is shown. Angle A C B is a right angle and angle C B A is 50 degrees. The length of A C is 3 meters, the length of C B is a, and the length of hypotenuse A B is c.
Which equation can be used to solve for c?
sin(50o) = StartFraction 3 Over c EndFraction
sin(50o) = StartFraction c Over 3 EndFraction
cos(50o) = StartFraction c Over 3 EndFraction
cos(50o) =
Answers
Answer:
A and A.
Step-by-step explanation:
Got it correct on edge 2022.
A company makes plastic baseballs. They put 6 baseballs nackage. Which teple shows the values for Op, the number of baseballs in p packages? Number of packages. P 7 Tamber of baseballs. 5p42 47 52 number of cackages out of basebis 50
Answers
Answer:
Table C
Step-by-step explanation:
Given
[tex]1\ package = 6\ balls[/tex]
Required
Which table is correct
We have:
[tex]1\ package = 6\ balls[/tex]
For 7 packages, it will be:
[tex]7\ packages = 42\ balls[/tex] --- i.e. 6 * 7
For 8:
[tex]8\ packages = 48\ balls[/tex] --- i.e. 8 * 7
For p packages
[tex]f(p) = 6p[/tex]
Using the above formula, we can conclude that table (C) is correct
If x = 82°, find the measures of angles 1, 2, and 3
Answers
Answer:
98
Step-by-step explanation:
x(o) + 1(o) = 180(o)
∠2° = ∠x°
∠1° = ∠3°
Answer:
angle 2 is 82 also and angle 3 and 4 is 180 - 82 = 98
Dertemine a área total at
Answers
Square: A = a^2
25. Approximate the sample variance and standard deviation given the following frequency distribution: Class Frequency 0–9 13 10–19 7 20–29 10 30–39 9 40–49 11
Answers
Sample variance = 228.408
Standard deviation = 15.113
Step-by-step explanation:The well formatted frequency table has been attached to this response.
To calculate the sample variance and standard deviation of the given grouped data, follow these steps:
i. Find the midpoint (m) of the class interval.
This is done by adding the lower bounds and upper bounds of the class intervals and dividing the result by 2. i.e
For class 0 - 9, we have
m = (0 + 9) / 2 = 4.5
For class 10 - 19, we have
m = (10 + 19) / 2 = 14.5
For class 20 - 29, we have
m = (20 + 29) / 2 = 24.5
For class 30 - 39, we have
m = (30 + 39) / 2 = 34.5
For class 40 - 49, we have
m = (40 + 49) / 2 = 44.5
This is shown in the third column of the attached table.
ii. Find the product of each of the frequencies of the class intervals and their corresponding midpoints. i.e
For class 0 - 9, we have
frequency (f) = 13
midpoint (m) = 4.5
=> f x m = 13 x 4.5 = 58.5
For class 10 - 19, we have
frequency (f) = 7
midpoint (m) = 14.5
=> f x m = 7 x 14.5 = 101.5
For class 20 - 29, we have
frequency (f) = 10
midpoint (m) = 24.5
=> f x m = 10 x 24.5 = 245
For class 30 - 39, we have
frequency (f) = 9
midpoint (m) = 34.5
=> f x m = 9 x 34.5 = 310.5
For class 40 - 49, we have
frequency (f) = 11
midpoint (m) = 44.5
=> f x m = 11 x 44.5 = 489.5
This is shown in the fourth column of the attached table.
iii. Calculate the mean (x) of the distribution i.e
This is done by finding the sum of all the results in (ii) above and dividing the outcome by the sum of the frequencies. i.e
x = ∑(f x m) ÷ ∑f
Where;
∑(f x m) = 58.5 + 101.5 + 245 + 310.5 + 489.5 = 1205
∑f = 13 + 7 + 10 + 9 + 11 = 50
=> x = 1205 ÷ 50
=> x = 24.1
Therefore, the mean is 24.1
This is shown on the fifth column of the attached table.
iv. Calculate the deviation of the midpoints from the mean.
This is done by finding the difference between the midpoints and the mean. i.e m - x where x = mean = 24.1 and m = midpoint
For class 0 - 9, we have
midpoint (m) = 4.5
=> m - x = 4.5 - 24.1 = -19.6
For class 10 - 19, we have
midpoint (m) = 14.5
=> m - x = 14.5 - 24.1 = -9.6
For class 20 - 29, we have
midpoint (m) = 24.5
=> m - x = 24.5 - 24.1 = 0.4
For class 30 - 39, we have
midpoint (m) = 34.5
=> m - x = 34.5 - 24.1 = 10.4
For class 40 - 49, we have
midpoint (m) = 44.5
=> m - x = 44.5 - 24.1 = 20.4
This is shown on the sixth column of the attached table.
v. Find the square of each of the results in (iv) above.
This is done by finding (m-x)²
For class 0 - 9, we have
=> (m - x)² = (-19.6)² = 384.16
For class 10 - 19, we have
=> (m - x)² = (-9.6)² = 92.16
For class 20 - 29, we have
=> (m - x)² = (0.4)² = 0.16
For class 30 - 39, we have
=> (m - x)² = (10.4)² = 108.16
For class 40 - 49, we have
=> (m - x)² = (20.4)² = 416.16
This is shown on the seventh column of the attached table.
vi. Multiply each of the results in (v) above by their corresponding frequencies.
This is done by finding f(m-x)²
For class 0 - 9, we have
=> f(m - x)² = 13 x 384.16 = 4994.08
For class 10 - 19, we have
=> f(m - x)² = 7 x 92.16 = 645.12
For class 20 - 29, we have
=> f(m - x)² = 10 x 0.16 = 1.6
For class 30 - 39, we have
=> f(m - x)² = 9 x 108.16 = 973.44
For class 40 - 49, we have
=> f(m - x)² = 11 x 416.16 = 4577.76
This is shown on the eighth column of the attached table.
vi. Calculate the sample variance.
Variance σ², is calculated by using the following relation;
σ² = ∑f(m-x)² ÷ (∑f - 1)
This means the variance is found by finding the sum of the results in (vi) above and then dividing the result by one less than the sum of all the frequencies.
∑f(m-x)² = sum of the results in (vi)
∑f(m-x)² = 4994.08 + 645.12 + 1.6 + 973.44 + 4577.76 = 11192
∑f - 1 = 50 - 1 = 49 {Remember that ∑f was calculated in (iii) above}
∴ σ² = 11192 ÷ 49 = 228.408
Therefore, the variance is 228.408
vii. Calculate the standard deviation
Standard deviation σ, is calculated by using the following relation;
σ =√ [ ∑f(m-x)² ÷ (∑f - 1) ]
This is done by taking the square root of the variance calculated above.
σ = [tex]\sqrt{228.408}[/tex]
σ = 15.113
Therefore, the standard deviation is 15.113
Illustrative Example 1:
The grades in Statistics of 10 students are 87, 84, 85, 85, 86, 90, 79, 82, 78, 76.
What is the mode?
Answers
Given:
The data set is:
87, 84, 85, 85, 86, 90, 79, 82, 78, 76
To find:
The mode of the given data set.
Solution:
We have,
87, 84, 85, 85, 86, 90, 79, 82, 78, 76
Arrange the data values in the ascending order.
76, 78, 79, 82, 84, 85, 85, 86, 87, 90
We know that the mode of date set is the most frequency value of the data set.
From the above data set it is clear that the number 85 has the highest frequency 2.
Therefore, the mode of the data set is 85.
ASAP PLEASE!!The table and the relative frequency histogram show the distribution of the number of tails and three coins are tossed. Find the probability P(T=1). write your answer as a fraction.
Answers
Explanation:
P(T = 1) is the notation that means "The probability of getting exactly one tail". The table shows 3/8 in the bottom row, under the "1" in the top row. So that's why P(T = 1) = 3/8
Or it might make more sense to say P(one tail) = 3/8 so we don't have too many equal signs going on.
Calculate the unit price. Round to the nearest cent, if necessary.
0.75 lb of salami for $2.65
$4.64 per lb of salami
$2.65 per lb of salami
$3.53 per lb of salami
$1.99 lb of salami
per
Answers
Answer:
$3.53 per lb of salami
Step-by-step explanation:
0.75 lb of salami for $2.65
The price per of salami will be :
0.75 lb = $2.65
1 lb = x
Using cross multiplication :
0.75 * x = $2.65 * 1
0.75x = $2.65
Divide both sides by 0.75
0.75x / 0.75 = $2.65 / 0.75
x = $3.533
This means that :
The price per unit of salami is $3.53
Answer:
0.75 lb of salami for $2.65 is $3.53 per lb of salami.
Perimeter of a 14cm,10cm,6cm and 20cm shape?
Answers
Answer:
The perimeter is 50 cm.
Step-by-step explanation:
P= 14 + 10 + 6 + 20= 50 cm
In function notation, f(x) is another way of saying ______.
1.)y
2.)x
3.)or none of the above
Answers
Answer:
y
Step-by-step explanation:
In function notation, f(x) is another way of saying y. Then the correct option is A.
What is a function?A function is an assertion, concept, or principle that establishes an association between two variables. Functions may be found throughout mathematics and are essential for the development of significant links.
Tables, symbols, and graphs can all be used to represent functions. Every one of these interpretations has benefits. Tables provide the functional values of certain inputs in an explicit manner. How to compute direct proportionality is succinctly stated in symbolic representation.
The function is represented as,
y = f(x)
In function notation, f(x) is another way of saying y. Then the correct option is A.
More about the function link is given below.
https://brainly.com/question/5245372
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i need the answer of this with steps and solutions fast plz
Answers
Answer:
[tex] 1 \dfrac{13}{15} [/tex]
Step-by-step explanation:
[tex] 2 \dfrac{1}{5} - \dfrac{1}{3} = [/tex]
[tex] = \dfrac{11}{5} - \dfrac{1}{3} [/tex]
[tex] = \dfrac{33}{15} - \dfrac{5}{15} [/tex]
[tex] = \dfrac{28}{15} [/tex]
[tex] = 1 \dfrac{13}{15} [/tex]
PLEASE I NEED THE CORRECT ANSWER ILL GIVE BRAINLISET 100 students are interviewed to see which of biology, chemistry or physics they prefer.
22 of the students are girls. 6 of the girls like biology best.
23 of the boys prefer physics.
14 out of the 32 who prefer chemistry are girls.
What percentage of the students prefer biology?