Answer:
75.935
Step-by-step explanation:
If V = Pe^rt and we are given P (the principle investment) =4630 ,
a rate of 5.7 % which as a decimal = 0.057 (% divided by 100 = decimal)
we can substitute this information in and get V=4630 e^(.057t).
then it says what will it be in 9 years? so we put 9 in for "t" and get
v = 4630 e^(0.057 * 9) simplify using a calculator and we will get our answer.
Mrs. Canales has 1,248 student pictures to displayaround the school. She plans to put them on 24 poster boards with the same amountof pictures on each poster board. How many student pictures will Mrs.Canales place on each poster board?
Answer:
52
Step-by-step explanation:
Its simple, its just division because she is asking how many for EACH(key word) so 1,248/24 equals 52 giving your answer.
mark brainliesttt??
A kindergarten teacher asked the students' parents to send their child to school with two fruits (apples and/or oranges). Below are the combinations of fruits brought to class the next day.
Fruit
2 Apples : 9 students
1 Apple and 1 Orange : 4 students
2 Oranges : 8 students
What is the frequency of oranges in the classroom? Round your answer to 4 decimal places.
The frequency of oranges in the classroom can be calculated as follows:Frequency of oranges in the classroom = Total number of oranges / Total number of fruits= 16/42 = 0.3809 (rounded to 4 decimal places)Thus, the frequency of oranges in the classroom is approximately 0.3809.
To determine the frequency of oranges in the classroom, we need to calculate the proportion of students who brought oranges out of the total number of students.
First, let's calculate the total number of students:
Total students = Number of students with 2 apples + Number of students with 1 apple and 1 orange + Number of students with 2 oranges
Total students = 9 + 4 + 8 = 21
Next, let's calculate the number of students who brought oranges:
Number of students with oranges = Number of students with 1 apple and 1 orange + Number of students with 2 oranges
Number of students with oranges = 4 + 8 = 12
Finally, we can calculate the frequency of oranges:
Frequency of oranges = Number of students with oranges / Total students
Frequency of oranges = 12 / 21 ≈ 0.5714 (rounded to 4 decimal places)
Therefore, the frequency of oranges in the classroom is approximately 0.5714.
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To determine the frequency of oranges in the classroom, it is important to add up the total number of fruits brought to the classroom. The given information is as follows:
2 Apples: 9 students
1 Apple and 1 Orange: 4 students
2 Oranges: 8 students
Thus, the frequency of oranges in the classroom is approximately 0.2105.
Total number of fruits = 2(9) + 1(4) + 2(8)
= 18 + 4 + 16
= 38 fruit in total
So, total number fruits are 38.
Frequency of oranges = Number of oranges / Total number of fruits
There are 8 oranges brought to class, so the frequency of oranges is:
8 / 38 ≈ 0.2105 (rounded to 4 decimal places)
Hence, the frequency of oranges in the classroom is approximately 0.2105.
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If the mode of the data 2,3,3,5,7,7,6,5, x and 8 is 3. Then what is the value of 'x'.
Answer:
x = 3
Step-by-step explanation:
Given that,
The mode of the data 2,3,3,5,7,7,6,5, x and 8 is 3.
We need to find the value of x.
We know that, Mode is the number in a data with Max frequency. x can be 3 or 7. If x = 7, mode becomes 7 and if x = 3, mode equals 3.
Hence, the value of x is equal to 3.
You measure 40 textbooks' weights, and find they have a mean weight of 42 ounces. Assume the population standard deviation is 3.8 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.
Give your answers as decimals, to two places
__________<μ<__________
The 90% confidence interval for the true population mean textbook weight, based on the given data, is approximately 41.44 ounces to 42.56 ounces.
To construct a confidence interval for the population mean textbook weight, we can use the formula:
Confidence Interval = sample mean ± (critical value * standard error)
Given that the sample mean is 42 ounces and the population standard deviation is 3.8 ounces, we need to determine the critical value and the standard error.
For a 90% confidence interval, the critical value corresponds to a two-tailed z-score of 1.645 (from the standard normal distribution).
The standard error can be calculated as the population standard deviation divided by the square root of the sample size. Since the sample size is not provided, we cannot calculate the exact standard error. However, if we assume a large sample size (usually considered to be greater than 30), we can use the formula for the standard error.
Assuming a large sample size, the standard error would be 3.8 ounces divided by the square root of the sample size.
Using the formula for the confidence interval, we can now calculate the range:
Confidence Interval = 42 ± (1.645 * standard error)
Substituting the values, we get:
Confidence Interval = 42 ± (1.645 * 3.8 / sqrt(sample size))
Since we do not know the sample size, we cannot calculate the exact confidence interval. However, based on the given data, we can conclude that the true population mean textbook weight falls between approximately 41.44 ounces and 42.56 ounces with 90% confidence.
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Evaluate the algebraic expression5m + 4n – 3 whenm=3 and n=4. Show your work.
Answer:
Given, m = 3 and n = 4
5×3 + 4×4 – 3
15 + 16 - 3
31 - 3
28
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From a survey taken by Survey 'R Us, 243 respondents out of the 1523 cat owners surveyed claim that their cats speak to them.
A.) With an 85% confidence level, provide the confidence interval that could be used to estimate the proportion of the population that hears their cats talking to them: use all three notations Set notation, interval notation, +/- notation
B.) Do the same as you did for 1a, but use a 95% confidence level instead Set Notation, Interval Notation, +/- notation
C.) Describe the differences between the ranges given for 1a and 1b. Why are the ranges different D.) Provide an interpretation for the interval given in 1b.
The interpretation of the interval in 1b (95% confidence level) is that we can be 95% confident that the true proportion of the population that hears their cats talking to them falls within the range of 0.1241 to 0.2137.
A.) With an 85% confidence level, the confidence interval that could be used to estimate the proportion of the population that hears their cats talking to them is [0.1459, 0.1919] in set notation, (0.1459, 0.1919) in interval notation, and +/- 0.023 in +/- notation.
B.) With a 95% confidence level, the confidence interval that could be used to estimate the proportion of the population that hears their cats talking to them is [0.1241, 0.2137] in set notation, (0.1241, 0.2137) in interval notation, and +/- 0.045 in +/- notation.
C.) The ranges for 1a and 1b are different because the confidence level affects the width of the interval. A higher confidence level requires a wider interval to provide a more reliable estimate. In this case, the 95% confidence level has a wider range compared to the 85% confidence level.
This means that if we were to repeat the survey multiple times, approximately 95% of the intervals calculated would contain the true proportion.
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a) The 85% confidence interval is given as follows: (0.146, 0.174).
b) The 95% confidence interval is given as follows: (0.142, 0.178).
c) The interval for item a is narrower than the interval for item b, as the lower confidence level leads to a lower critical value and a lower margin of error.
d) We are 95% sure that the true population proportion is between the two bounds found in item b.
What is a confidence interval of proportions?A confidence interval of proportions has the bounds given by the rule presented as follows:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which the variables used to calculated these bounds are listed as follows:
[tex]\pi[/tex] is the sample proportion, which is also the estimate of the parameter.z is the critical value.n is the sample size.For the confidence level of 85%, the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.85}{2} = 0.925[/tex], so the critical value is z = 1.44.
For the confidence level of 95%, the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.
The parameters for the confidence interval are given as follows:
[tex]n = 1523, \pi = \frac{243}{1523} = 0.16[/tex]
Hence the bounds of the 85% confidence interval are given as follows:
[tex]0.16 - 1.44\sqrt{\frac{0.16(0.84)}{1523}} = 0.146[/tex][tex]0.16 + 1.44\sqrt{\frac{0.16(0.84)}{1523}} = 0.174[/tex]The bounds of the 95% confidence interval are given as follows:
[tex]0.16 - 1.96\sqrt{\frac{0.16(0.84)}{1523}} = 0.142[/tex][tex]0.16 + 1.96\sqrt{\frac{0.16(0.84)}{1523}} = 0.178[/tex]More can be learned about the z-distribution at https://brainly.com/question/25890103
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A rotating lawn sprinkler sprays water in a circular area of grass, as shown in the
picture. The diameter of the circular area of grass is 16 ft. what is the closest measurement to the area in square feet ?
Answer:
Area of the lawn = 64π square feet
Step-by-step explanation:
Area of the circular lawn = πd²/4
d is the diameter of the lawn = 16ft
Substitute the given value into the formula
Area = π(16)²/4
Arrea of the lawn = 256π/4
Area of the lawn = 64π square feet
8. find h[n], the unit impulse response of the system described by the following equation y[ n+2] +2y[ n+1] + y[n] = 2x[n+ 2] − x[n+ 1]
The unit impulse response of the system described by the equation y[n+2] + 2y[n+1] + y[n] = 2x[n+2] − x[n+1] is h[n] = 2δ[n+2] − δ[n+1] + δ[n], where δ[n] represents the unit impulse function.
To find the unit impulse response, we need to determine the output of the system when an impulse is applied at the input, i.e., x[n] = δ[n].
Substituting x[n] = δ[n] into the given equation, we have:
y[n+2] + 2y[n+1] + y[n] = 2δ[n+2] − δ[n+1].
Since δ[n] = 0 for n ≠ 0 and δ[0] = 1, we can simplify the equation:
y[n+2] + 2y[n+1] + y[n] = 2δ[n+2] − δ[n+1] + δ[n] = 2δ[n+2] − δ[n+1] + δ[n]δ[n].
Now, comparing the equation with the standard form of the unit impulse response:
h[n] = 2δ[n+2] − δ[n+1] + δ[n],
we can conclude that h[n] is the unit impulse response of the given system.
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was 10 years and 6 months old when I moved to Mumbai. have been living in Mumbai for the past 13
years 11 months. What is my age now?
24 years and 5 months
add 13 years to ten get 23 add 6 to 11 to get 1 year and 5 months
23 plus 1 years and 5 months = 24 years an 5months
have a nice day please mark brainliest
5. Calculate the area.
9ft
4
11 ft.
ООО
40
44
о
396
[tex]area = b \times h \\ = 11 \times 4 \\ = 44[/tex]
at traget a 5 pack of gaterade cost 8.78 how much would 21 packs of gatorade cost
Answer:
21 packs of gatorade would cost $36.88.
Step-by-step explanation:
Mathematically:
8.78 / 5 = 1.756
1.756 * 21 = 36.876 ~= $36.88
plssssssss help solve
Answer:
cosine = adjacent/hypotenuse
cos A = 20/29 (choice: yellow)
Step-by-step explanation:
Answer:
yellow
Step-by-step explanation:
If x=(y+2)^2 and y= -7 then what is the value of x
Answer:
25
Step-by-step explanation:
(y+2)^2=x
(-7+2)^2=x
(-5)^2=x
(-5)(-5)=x
25=x
Hope that helps :)
Answer:
x=25
Step-by-step explanation:
Plug it innnn plug it innnn
The company ALTA Ltd issued a bank accepted bill to fund its working capital requirement. The bill is issued for 60 days, with a face value of $150,000 and a yield of 2.5% per annum to the original discounter. After 25 days, the bank bill is sold by the wwwwww original discounter into the secondary market for $138,222. The purchaser holds the bill to maturity. What is the yield received by the holder of the bill at the date of maturity?
the yield received by the holder of the bill at the date of maturity is approximately 10.15%.
To calculate the yield received by the holder of the bill at the date of maturity, we need to use the formula for yield to maturity. The formula is:
Yield to Maturity = (Face Value - Purchase Price) / Purchase Price * (365 / Days to Maturity)
In this case:
Face Value = $150,000
Purchase Price = $138,222
Days to Maturity = 60 - 25 = 35
these values in the formula, we can calculate the yield to maturity:
Yield to Maturity = ($150,000 - $138,222) / $138,222 * (365 / 35)
Yield to Maturity ≈ 0.1015 or 10.15%
Therefore, the yield received by the holder of the bill at the date of maturity is approximately 10.15%.
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1.13 UNIT TEST GRAPH OF SINUSOIDAL FUNCTION PART 1
What is the equation of the midline for the function f(x)?
f(x) =1/2 sin(x)+6
Answer:
The equation of the midline for the function [tex]f(x)[/tex] is [tex]y = 6[/tex].
Step-by-step explanation:
The sinusoidal function of the form [tex]y = A_{o}+A\cdot \sin x[/tex] is a periodic function whose range is bounded between [tex]A_{o}-A[/tex] (minimum) and [tex]A_{o}+A[/tex] (maximum). The equation of the midline is a line paralel to the x-axis, that is:
[tex]y = c,\forall\, c\in \mathbb{R}[/tex] (1)
Where [tex]c[/tex] is mean of the upper and lower bounds of the sinusoidal function, that is:
[tex]c = \frac{(A_{o}+A+A_{o}-A)}{2}[/tex]
[tex]c = A_{o}[/tex] (2)
If we know that [tex]y = \frac{1}{2}\cdot \sin x + 6[/tex], then the equation of the midline for the function [tex]y[/tex] is:
[tex]c = A_{0} = 6[/tex]
[tex]y = 6[/tex]
The equation of the midline for the function [tex]f(x)[/tex] is [tex]y = 6[/tex].
Please help I’ll give brainliest if you answer all :)
Step-by-step explanation:
a^2 +b^2 = c^2
b^2 = c^2 - a^2
3 - solving for h
A = bh
h = A ÷h
4 - solving for r
I = prt
r = I÷ pt
5- solving for b
A= 1/2 bh
can rewrite as
A = bh ÷2
b = 2A ÷ h
In 2005, there are 705 cable users in the small town of Whoville. The number of users
increases by 56% each year after 2005. Find the number of users to the nearest whole in 2020.
Answer:2008
Step-by-step explanation:
Alex is a faster runner than Carlos. The chart below relates how many laps around the track each runs in the same amount of time.
Number of Laps Run
Alex Carlos
4 3
8 6
12 9
How many laps will Alex have run in the time it take Carlos to run 12 laps?
Answer:
16
Step-by-step explanation:
Alex will run 16 laps in the time it take for Carlos to run 12 laps.
What is Multiplication?Multiplication of two numbers is defined as the addition of one of the number repeatedly until the times of the other number.
a × b means that a is added to itself b times or b is added to itself a times.
Given are the values related to the number of laps run by Alex and Carlos.
The number of laps run by Alex is forming the multiples of 4 or 4x.
Number of laps run by Carlos is forming the multiples of 3 or 3x.
When x = 1,
Alex : 4 × 1 = 4
Carlos : 3 × 1 = 3
When x = 2,
Alex : 4 × 2 = 8
Carlos : 3 × 2 = 6
When x = 3,
Alex : 4 × 3 = 12
Carlos : 3 × 3 = 9
When x = 4,
Alex : 4 × 4 = 16
Carlos : 3 × 4 = 12
Hence Alex ran 16 laps in the same time for which Carlos ran 12 laps.
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2 Which expressions can be used to find the volume of the rectangular prism?
3 ft
5 ft
7 ft
apply.
5
35 + 5
35 x 3
(7 + 5) x3
7+3+5
35 + 35 + 35
Answer:
To find the volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units.
Find the volume of the con round you answer to the nearest tenth
Answer:
16.76 inch³
Step-by-step explanation:
volume of cone is 1/3 πr²h
h=4
r=2
then 1/3* 22/7* 2*2 *4
=352/21
=16.76
evaluating quadratic functions using equations evaluate the function g(x) = –2x2 3x – 5 for the input values –2, 0, and 3. g(–2) = –2(–2)2 3(–2) – 5 g(–2) = –2(4) – 6 – 5 g(–2) = g(0) = g(3) =
Evaluating the quadratic function we will get:
g(-2) = -3
g(0) = -5
g(3) = 31
How to evaluate the quadratic function?Here we need to evaluate the quadratic function:
g(x) = -2x² + 3x - 5
To do so, just replace the value of x by the correspondent number.
For example, if x = -2
g(-2) = 2*(-2)² + 3*(-2) - 5 = -3
if x = 0
g(0) = 2*(0)² + 3*(0) - 5 = -5
if x = 3
g(3) = 2*(3)² + 3*(3) - 5 = 31
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How far apart are - 7 and |-7| on a number
line?
Answer:
the answer is 7. :)
Step-by-step explanation:
Algebra 2 PLEASE HELP
Answer:
[tex]\frac{x+20}{x+4}[/tex]
Step-by-step explanation:
Can't cancel out terms like that
need to factor out the top and bottom
[tex]\frac{x^{2} +16x-80}{x^{2} -16}[/tex] = [tex]\frac{(x+20)(x-4)}{(x-4)(x+4)}[/tex] now cancel out the (x-4) from the top and bottom
= [tex]\frac{x+20}{x+4}[/tex]
Answer & Explanation:
Error: individual terms in an equation in a fraction cannot be directly canceled out.
Correction:
the easy way (solve the quadratic equation in the numerator and complete the square in the denominator):
(x^2 + 16x - 80) / (x^2 - 16)
(x+20)(x-4) / (x+4)(x-4)
x+20 / x+4
the complicated way (manipulate the exponents and common factors):
(x^2 + 16x - 80) / (x^2 - 16)
(x^2 - 4x + 20x - 80) / x^2 - 2^4
x(x^2-1 - 2^2)+4*5(x - 2^4-2) / (x - 2^2)(x + 2^2)
x(x-4)+20(x-4) / (x-4)(x+4)
x(x-4)+(2^2 (5))(x-4) / (x-4)(x+4)
(x-4)(x + 2^2 (5)) / (x-4)(x+4)
(x-4)(x+20) / (x-4)(x+4)
x+20 / x+4
Write a Matlab code to solve the following problems. 1-use Bisection Method x3 + 4x2 - 10 = 0 for x = [0,5] x3 - 6x2 + 10x - 4 = 0 for xe [0,4] 2-Use Newton Method x3 + 3x + 1 = 0 for x = (-2,0] 3-Use fixed point Method x3 - 2x - 1 = 0 for x E (1.5,2] 4-Use secant Method 1-2e -* - sin(x) = 0 for x € (0,4] 2-x3 + 4x2 - 10 = 0 for x € [0,4]
a) Bisection Method MATLAB code for equation [tex]x^3 + 4x^2 - 10 = 0[/tex] in the interval [0,5]:
function root = bisection_method()
f = [tex]x^3 + 4*x^2 - 10[/tex];
a = 0;
b = 5;
tol = 1e - 6;
while (b - a) > tol
c = (a + b) / 2;
if f(c) == 0
break;
elseif f(a) * f(c) < 0
b = c;
else
a = c;
end
end
root = (a + b) / 2;
end
b) Bisection Method MATLAB code for equation [tex]x^3 - 6x^2 + 10x - 4 = 0[/tex] in the interval [0,4]:
function root = bisection_method()
f = [tex]x^3 - 6*x^2 + 10*x - 4[/tex];
a = 0;
b = 4;
tol = 1e-6;
while (b - a) > tol
c = (a + b) / 2;
if f(c) == 0
break;
elseif f(a) * f(c) < 0
b = c;
else
a = c;
end
end
root = (a + b) / 2;
end
c) Newton's Method MATLAB code for equation [tex]x^3 + 3x + 1 = 0[/tex] in the interval (-2,0]:
function root = newton_method()
f = [tex]x^3 + 3*x + 1[/tex];
df = [tex]3*x^2 + 3[/tex];
[tex]x_0[/tex] = -1;
tol = 1e-6;
while abs(f([tex]x_0[/tex])) > tol
[tex]x_0 = x_0 - f(x_0) / df(x_0)[/tex];
end
root = [tex]x_0[/tex];
end
d) Fixed-Point Method MATLAB code for equation [tex]x^3 - 2x - 1 = 0[/tex] in the interval (1.5,2]:
function root = fixed_point_method()
g = [tex](x^3 - 1) / 2[/tex];
[tex]x_0 = 2[/tex];
tol = 1e-6;
while abs([tex]g(x_0) - x_0[/tex]) > tol
[tex]x_0 = g(x_0)[/tex];
end
root = [tex]x_0[/tex];
end
e) Secant Method MATLAB code for equation 1 - 2*exp(-x) - sin(x) = 0 in the interval (0,4]:
function root = secant_method()
f = 1 - 2*exp(-x) - sin(x);
[tex]x_0[/tex] = 0;
[tex]x_1[/tex] = 1;
tol = 1e-6;
while abs(f([tex]x_1[/tex])) > tol
[tex]x_2 = x_1 - f(x_1) * (x_1 - x_0) / (f(x_1) - f(x_0))[/tex];
[tex]x_0 = x_1[/tex];
[tex]x_1 = x_2[/tex];
end
root = [tex]x_1[/tex];
end
f) Secant Method MATLAB code for equation [tex]2 - x^3 + 4*x^2 - 10 = 0[/tex] in the interval [0,4]:
function root = secant_method()
f = [tex]2 - x^3 + 4*x^2 - 10[/tex];
[tex]x_0 = 0[/tex];
[tex]x_1 = 1[/tex];
tol = 1e-6;
while abs(f([tex]x_1[/tex])) > tol
[tex]x_2 = x_1 - f(x_1) * (x_1 - x_0) / (f(x_1) - f(x_0))[/tex];
[tex]x_0 = x_1[/tex];
[tex]x_1 = x_2[/tex];
end
root = [tex]x_1[/tex];
end
How to find the MATLAB code be used to solve different equations numerically?MATLAB provides several numerical methods for solving equations. In this case, we have used the Bisection Method, Newton's Method, Fixed-Point Method, and Secant Method to solve different equations.
The Bisection Method starts with an interval and iteratively narrows it down until the root is found within a specified tolerance. It relies on the intermediate value theorem.
Newton's Method, also known as Newton-Raphson Method, approximates the root using the tangent line at an initial guess. It iteratively refines the guess until the desired accuracy is achieved.
The Fixed-Point Method transforms the equation into an equivalent fixed-point iteration form. It repeatedly applies a function to an initial guess until convergence.
The Secant Method is a modification of the Newton's Method that uses a numerical approximation of the derivative. It does not require the derivative function explicitly.
By implementing these methods in MATLAB, we can numerically solve various equations and find their roots within specified intervals.
These numerical methods are powerful tools for solving equations when analytical solutions are not feasible or not known.
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Emily asked the players on her volleyball team their height in inches and listed the results below. What is the mean of the data set? Round to the nearest tenth when necessary
Answer:the range is 10 pls gimme brainless plsss I need it
Step-by-step explanation:
¿can you help me, please?
a) give stating reasons five other angles each equal to x
b) prove that AECF is a parallelogram
Simple Proof:
a) In the image, we know that ABCD is a parallelogram and that means opposite angel measures should be the same. We know that angel DCB is made up by angel 1 and 2, and angel DAB and DCB are equal and angel DAB is made up by angel 1 and x. So now we can conclude that angel x is equal to angel 2.
b) According to the definitions of a parallelogram, opposite angel measures have to be the same, while AECF have angle 1 to angel 1 and angel 2 to angel 1. We can conclude that AECF is NOT a parallelogram. (Sorry, you didn't give me the full question so some information remains unclear. )
"Marty purchased a car. The car cost him $16,500 and it depreciates in value at a rate of 4.3% per year. How much will the car be worth in 12 years?"
Answer:
"Marty purchased a car. The car cost him $16,500 and it depreciates in value at a rate of 4.3% per year. How much will the car be worth in 12 years?"
Step-by-step explanation:
Which shows a correct way to determine the volume of the right rectangular prism?
Answer:
the last answer
Step-by-step explanation:
[tex]V=l*w*h\\l=8\\w=9\\h=1\\V=8*9*1\\V=72[/tex]
In a game of chance, a fair die is tossed. If the number is 1 or 2, you will win $3. If the number is 3, you win $5. If the number is 4 or 5, you win nothing, and if the number is 6 you lose S2. Should you play the game, based on the long run expected amount you would win? von $3= / 116 (A) Yes! In the long run, you are expected to win $2.16. (B) Yes! In the long run, you are expected to win $1.00. (C) Yes! You have more opportunities to win money than you have to lose money, (D) No. In the long run, you are expected to lose $0.33 (E) No. Even with the opportunities to win money, it is not worth the risk to lose $2 in the long run
In the long run, you are expected to win $1.50 when playing the game. Therefore, the correct answer is :
(B) Yes! In the long run, you are expected to win $1.00.
To determine whether you should play the game based on the long run expected amount you would win, we need to calculate the expected value.
The probability of winning $3 is 2/6 (numbers 1 and 2), the probability of winning $5 is 1/6 (number 3), the probability of winning nothing is 2/6 (numbers 4 and 5), and the probability of losing $2 is 1/6 (number 6).
Now let's calculate the expected value:
Expected Value = (Probability of winning $3 * $3) + (Probability of winning $5 * $5) + (Probability of winning nothing * $0) + (Probability of losing $2 * -$2)
Expected Value = (2/6 * $3) + (1/6 * $5) + (2/6 * $0) + (1/6 * -$2)
Expected Value = ($6/6) + ($5/6) + ($0) + (-$2/6)
Expected Value = $11/6 - $2/6
Expected Value = $9/6
Expected Value = $1.50
Therefore, in the long run, you are expected to win $1.50.
The correct answer is option (B) Yes! In the long run, you are expected to win $1.00.
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