The value of the line integral ∫c f · dr is -cos(1) i - sin(1) j + 1/6 k.
Evaluate the integral?
To evaluate the line integral ∫c f · dr, we need to substitute the given values of f(x, y, z) and r(t) into the integral expression.
[tex]f(x, y, z) = sin(x) i cos(y) j\ x(z) k[/tex]
[tex]r(t) = t^4 i - t^3 j + t k[/tex] , 0 ≤ t ≤ 1
The line integral becomes:
[tex]\int c f * dr = \int c (sin(x) i cos(y) j x(z) k) * (dx i + dy j + dz k)[/tex]
Substituting [tex]x = t^4,\ y = -t^3, and\ z = t:[/tex]
[tex]\int c f * dr = \int c (sin(t^4) i cos(-t^3) j (t^4)(t) k) * (4t^3 dt i - 3t^2 dt j + dt k)[/tex]
Simplifying the expression:
[tex]\int c f * dr = \int c (4t^3 sin(t^4) dt i - 3t^2 cos(t^3) dt j + t^5 dt k)[/tex]
Integrating each component separately:
[tex]\int c f * dr = (\int 0^1 4t^3 sin(t^4) dt) i - (\int 0^1 3t^2 cos(t^3) dt) j + (\int 0^1 t^5 dt) k[/tex]
Evaluating each integral:
[tex]\int c f * dr = [-(cos(t^4))][/tex] evaluated from 0 to [tex]1 i - [sin(t^3)][/tex] evaluated from 0 to [tex]1 j + [t^6/6][/tex] evaluated from 0 to 1 k
Simplifying the expression:
[tex]\int c f * dr = -cos(1) i - sin(1) j + 1/6 k[/tex]
Therefore, the value of the line integral [tex]\int c f * dr\[/tex] is [tex]-cos(1) i - sin(1) j + 1/6 k.[/tex]
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Select the correct answer.
If the parent function RX) = x2 is modified to g(x) = 2x + 1, which statement is true about g(x)?
ΟΑ. .
It is an even function.
ОВ.
It is an odd function,
OC. It is both an even and an odd function.
OD.
It is neither an even nor an odd function.
In a large population, 92% of the households have cable tv. A simple random sample of 225 households is to be contacted and the sample proportion computed. What is the mean and standard deviation of the sampling distribution of the sample proportions
Answer:
the mean and the standard deviation is 0.92 and 0.01808 respectively
Step-by-step explanation:
The computation of the mean and the standard deviation is shown below:
The mean is 0.92
And, the standard deviation is
= √0.92 × (1 - 0.92) ÷ √225
= √0.92 × 0.08 ÷ √225
= √0.0736 ÷ √225
= √3.27
= 0.01808
Hence, the mean and the standard deviation is 0.92 and 0.01808 respectively
What is the value of x? sin(x+37)°=cos(2x+8)° Enter your answer in the box. x =
The answer is x = 15.
15 or 20.33 are the possible values of the x.
What is algebraic Expression?Any mathematical statement that includes numbers, variables, and an arithmetic operation between them is known as an expression or algebraic expression. In the phrase 4m + 5, for instance, the terms 4m and 5 are separated from the variable m by the arithmetic sign +.
We know that sin(x+37)°=cos(90°-(x+37)°) and cos(2x+8)°=sin(90°-(2x+8)°)
So we have sin(x+37)°=cos(2x+8)° becomes sin(x+37)°=sin(82°-2x)
For the above equation to be true, either of the following must hold:
x+37 = 82 - 2x (since the sin function is periodic)
x+37 = 180 - (82-2x)
Solving the first equation for x gives:
3x = 45
x = 15
Solving the second equation for x gives:
3x = 61
x = 20.33 (rounded to two decimal places)
Therefore, the possible values of x are 15 or 20.33 (rounded to two decimal places).
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Current Attempt in Progress The following table lists the ages (in years) and the prices (in thousands of dollars) by a sample of six houses. Age Price 27 165 15 182 3 205 35 177 9 180 18 161 The null hypothesis is that the slope of the population regression line of price on age is zero and the alternative hypothesis is that the slope of this population regression line is less than zero. The significance level is 5%. What is the value of the test statistic, t. rounded to three decimal places?
Whwn the null hypothesis is that the slope of the population regression line of price on age is zero, the test statistic is -2.43.
How to calculate the valueThe test statistic is calculated as follows:
t = (b - 0) / SE(b)
In this case, the slope of the regression line is estimated to be -11.50, the standard error of the slope is 4.77, and the hypothesized value of the slope is 0.
t = (-11.50 - 0) / 4.77
= -2.43
The test statistic is -2.43, The p-value for this test statistic can be calculated using a t-table. The degrees of freedom for this test are 5 - 2 = 3. The p-value for a t-statistic of -2.43 with 3 degrees of freedom is 0.048.
Since the p-value is less than the significance level of 0.05, we can reject the null hypothesis and conclude that there is sufficient evidence to support the alternative hypothesis.
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please help I will give brainliest
I don't want to see any link
if I do you will be reported
Answer:
24
Brainliest?
Answer:
nxbxbxxbznznxnxnxnxnxnxnxbxbbxbnxxnxnx
Step-by-step explanation:
xnncxbbxbxbxbxbxbx hahahahahahahahahahahah
Solve the system of inequalities graphically:
x−2y≤3,3x+4y≥12,x≥0,y≥1
The solution to the system of inequalities is the region above the x-axis and to the right of the line y = 1, shaded in green.
Given system of inequalities is: x - 2y ≤ 3 ...(1)3x + 4y ≥ 12 ...(2)x ≥ 0 ...(3)y ≥ 1 ...(4)
We graph the lines x - 2y = 3 and 3x + 4y = 12 and shade the appropriate regions.
Let's start with the line x - 2y = 3.
We rewrite this as y = (1/2)x - 3/2 and plot the line as shown below: graph{(1/2)x - 3/2 [-10, 10, -5, 5]}
Now we determine which side of the line we want to shade.
Since the inequality is of the form ≤, we shade below the line y = (1/2)x - 3/2 (including the line itself) as shown below: graph {(1/2)x - 3/2 [-10, 10, -5, 5](-10,-5)--(10,0)}
Next, we graph the line 3x + 4y = 12. We rewrite this as y = (-3/4)x + 3 and plot the line as shown below: graph{(-3/4)x + 3 [-10, 10, -5, 5]}
We determine which side of the line we want to shade. Since the inequality is of the form ≥, we shade above the line y = (-3/4)x + 3 (including the line itself) as shown below: graph{(-3/4)x + 3 [-10, 10, -5, 5](-10,4)--(10,0)}
Finally, we shade the region that satisfies x ≥ 0 and y ≥ 1.
This is the region above the x-axis and to the right of the line y = 1 as shown below: graph{(-3/4)x + 3 [-10, 10, -5, 5](-10,4)--(10,0)(0,1)--(10,1)[above]}
The shaded region is the region that satisfies all three inequalities.
Thus, the solution to the system of inequalities is the region above the x-axis and to the right of the line y = 1, shaded in green.
We graph the lines x - 2y = 3 and 3x + 4y = 12 and shade the appropriate regions.
Let's start with the line x - 2y = 3. We rewrite this as y = (1/2)x - 3/2 and plot the line.
Now we determine which side of the line we want to shade. Since the inequality is of the form ≤, we shade below the line y = (1/2)x - 3/2 (including the line itself).
Next, we graph the line 3x + 4y = 12. We rewrite this as y = (-3/4)x + 3 and plot the line. We determine which side of the line we want to shade.
Since the inequality is of the form ≥, we shade above the line y = (-3/4)x + 3 (including the line itself).
Finally, we shade the region that satisfies x ≥ 0 and y ≥ 1.
This is the region above the x-axis and to the right of the line y = 1. The shaded region is the region that satisfies all three inequalities.
Thus, the solution to the system of inequalities is the region above the x-axis and to the right of the line y = 1, shaded in green.
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se the confidence interval to find the margin of error and the sample mean. question content area bottom part 1 the margin of error is . 069. part 2 the sample mean is . 381.
The "margin-of-error" is 3.6 and the "sample-mean" is 18.7 based on the given confidence interval (15.1, 22.3).
To find the "margin-of-error" and the "sample-mean" from the given confidence interval, we use the formula:
Confidence-Interval = Sample mean ± Margin of error,
In this case, the given confidence-interval is (15.1, 22.3),
To find the margin-of-error, we need to consider the range between the upper and lower bounds of the confidence interval and divide it by 2,
The "Margin-of-error" is = (Upper bound - Lower bound)/2,
"Margin-of-error" is = (22.3 - 15.1)/2 = 3.6,
So, the margin of error is 3.6.
To find "sample-mean", we calculate average of the upper and lower bounds of the confidence-interval,
The "Sample-Mean" is = (Upper bound + Lower bound)/2,
"Sample-Mean" is = (22.3 + 15.1)/2 = 18.7,
Therefore, the "sample-mean" is 18.7.
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The given question is incomplete, the complete question is
Use the confidence interval to find the margin of error and the sample mean. (15.1, 22.3).
Suppose a and n are relatively prime such that g.c.da, n=1, prove that \/ b 1 b) If n = 1, we cannot conclude that x=a (mod n) has solutions.
If a and n are relatively prime (gcd(a, n) = 1), it does not guarantee that the equation x ≡ a (mod n) has solutions.
If a and n are relatively prime, denoted by gcd(a, n) = 1, it means that a and n do not have any common factors other than 1. However, this does not guarantee that the equation x ≡ a (mod n) has solutions.
The equation x ≡ a (mod n) represents a congruence relation, where x is congruent to a modulo n. This equation implies that x and a have the same remainder when divided by n.
To have solutions for this congruence equation, it is necessary for a to be congruent to some number modulo n. In other words, a must lie in the residue classes modulo n. However, the fact that gcd(a, n) = 1 does not ensure that a is congruent to any residue modulo n, hence not guaranteeing the existence of solutions for the equation.
Therefore, when n = 1, we cannot conclude that the equation x ≡ a (mod n) has solutions.
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Select all that are equivalent to sin GFH
Answer:
FGH DEJ .................
Consider the following IVP: u"(t) + u'(t) - 12u (t)=0 (1) u (0) = 60 and u'(0) = 56. Show that u(t)=c₁₁e² + c ₂² -4 satisifes ODE (1) and find the values of c ER and C₂ ER such that the solution satisfies the given initial values.
The values of c₁ and c₂ that satisfy the initial values u(0) = 60 and u'(0) = 56 are:
c₁ = 148 / (3e²)
c₂ = (20 - 148/9)e⁴
The given solution, u(t) = c₁e² + c₂e⁻⁴, indeed satisfies the given ordinary differential equation (ODE) u"(t) + u'(t) - 12u(t) = 0. To find the values of c₁ and c₂ such that the solution satisfies the initial values u(0) = 60 and u'(0) = 56, we substitute these values into the solution.
First, let's find u(0) by substituting t = 0 into the solution:
u(0) = c₁e² + c₂e⁻⁴
Since u(0) = 60, we have:
60 = c₁e² + c₂e⁻⁴ (Equation 2)
Next, let's find u'(0) by differentiating the solution with respect to t and substituting t = 0:
u'(t) = 2c₁e² - 4c₂e⁻⁴
u'(0) = 2c₁e² - 4c₂e⁻⁴
Since u'(0) = 56, we have:
56 = 2c₁e² - 4c₂e⁻⁴ (Equation 3)
Now we have a system of two equations (Equations 2 and 3) with two unknowns (c₁ and c₂). We can solve this system to find the values of c₁ and c₂.
To do that, let's first divide Equation 3 by 2:
28 = c₁e² - 2c₂e⁻⁴
Next, let's multiply Equation 2 by 2:
120 = 2c₁e² + 2c₂e⁻⁴
Adding the two equations, we get:
148 = 3c₁e²
Dividing both sides by 3e², we find:
c₁ = 148 / (3e²)
Substituting this value of c₁ back into Equation 2, we can solve for c₂:
60 = (148 / (3e²))e² + c₂e⁻⁴
60 = 148/3 + c₂e⁻⁴
60 - 148/3 = c₂e⁻⁴
20 - 148/9 = c₂e⁻⁴
c₂ = (20 - 148/9)e⁴
Therefore, the values of c₁ and c₂ that satisfy the initial values u(0) = 60 and u'(0) = 56 are:
c₁ = 148 / (3e²)
c₂ = (20 - 148/9)e⁴
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Which expression is equivalent to 1.5a + 2.4 (a + 0.5b) - 0.2b?
a. 3.9a + b
b. 2.5a + 0.7b
c. 3.9a + 0.3b
d. 2.5a + 0.3b + 2.4
The answer is A, 3.9a + b
Step-by-step explanation:
youre welcome
Survey / Statistical Question:
How many times has each person moved states?
Answer:
a regular person would gave moved 4 to 5 times
Use digits to write the value of the 4 in this number.
842,963
Answer:
40,000
Step by step explanation:
842,963
840,000
40,000
A stick of length 10 is broken at a point X which is uniformy distributed on (0,10). Given X = 1, another breakpoint Y is chosen uniformly on (0,3). The joint part(X,Y) is given by f(x,y) for 0 <<<10 10s The marginal pdf of Y is given by fY() = 0.1 for Oy10
The joint probability density function (pdf) is f(x,y) = 0.03 for 0 < x < 1 and 0 < y < 3.
The joint pdf, f(x,y), represents the probability density function for the random variables X and Y. Given that X is uniformly distributed on (0,10), we have fX(x) = 0.1 for 0 < x < 10. The probability of X being less than 1 is 1/10, so the conditional pdf f(x|X<1) = 0.1 for 0 < x < 1.
Furthermore, Y is uniformly distributed on (0,3), so fY(y) = 0.1 for 0 < y < 3. To find the joint pdf, we multiply the conditional pdf of X with the marginal pdf of Y: f(x,y) = f(x|X<1) * fY(y) = 0.1 * 0.1 = 0.01 for 0 < x < 1 and 0 < y < 3. Therefore, the joint pdf is f(x,y) = 0.01 for 0 < x < 1 and 0 < y < 3.
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Please help me with this, please
Answer:
Hopefully it makes sense
Step-by-step explanation:
Good luck
(22) + (3x) = 4
solve for x
Answer:
-6
Step-by-step explanation:
Step One: The goal is to isolate the x. So first, we would do 4-22, which is -18. The equation is: 3x=-18.
Step Two: Lastly, we need to divide by three to completely isolate the x. This is -6.
A cylinder has a volume of 2,309.07 cubic cm and a height of 15 cm. What is the
radius of the cylinder? Use 3.14 for st in your calculations and round to the
nearest whole number.
cm
Answer:
7
Step-by-step explanation:
the answer I got was 7 hope this helped
the probability that an at home pregnancy test will correctly identify a pregnancy is 0.98. suppose 11 randomly selected pregnant women with typical hormone levels are each given the test. rounding your answer to four decimal places, find the probability that all 11 tests will be positive at least one test will be negative
The probability that all 11 tests will be positive is 0.8007
The probability that at least one test will be negative is 0.1993
Finding the probability that all 11 tests will be positiveFrom the question, we have the following parameters that can be used in our computation:
p = 0.98
n = 11
The probability that all 11 tests will be positive is
P = pⁿ
So, we have
P = 0.98¹¹
Evaluate
P = 0.8007
Finding the probability that at least one test will be negativeHere, we use
P = 1 - P(No negative)
So, we have
P = 1 - 0.8007
Evaluate
P = 0.1993
Hence, the probability is 0.1993
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You have 1/8 gallon of melted crayon wax. You pour the wax equally into 8 different molds to make new crayons. What fraction of a cup of melted wax is in each mold? Think: 1 gallon is 16 cups.
PLS HELP
Answer:
2 cups
Step-by-step explanation:
16/8=2 so it is 2 cups
If f(x) is not defined at c, then f(x) cannot be continuous on any interval. True False
Answer:
True
Step-by-step explanation:
The wording of the question is a little tricky, but here's what I think.
If a function f(x) is not defined at a point c, then the function has a discontinuity at that point. In order for a function to be continuous on an interval, it must be defined and have no abrupt changes or jumps within that interval. Since f(x) is not defined at c, it violates the condition of continuity, and therefore f(x) cannot be continuous on any interval that includes c.
The given statement "If f(x) is not defined at c, then f(x) cannot be continuous on any interval." is false because it does not automatically mean that f(x) cannot be continuous on any interval.
Continuity of a function depends on the behavior of the function around the point of interest, rather than just the absence of a definition at a single point. A function can still be continuous on an interval except at the specific point where it is not defined.
For example, consider the function f(x) = 1/x. This function is not defined at x = 0, but it is continuous on any interval that does not include x = 0. This is because f(x) approaches positive or negative infinity as x approaches 0 from the left or right side, respectively, indicating that there is no abrupt jump or discontinuity.
In general, the continuity of a function is determined by its behavior around a point, including its limit as x approaches that point. The absence of a definition at a single point does not automatically imply that the function cannot be continuous on any interval.
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Part A
An economist has measured people's annual salary (in thousands of dollars) and their years of relevant job experience, thinking that a linear relationship between them might exist.
Let the proposed regression relationship between Salary and experience be as follows: E(Salary) = beta subscript 0 space plus space beta subscript 1 space cross times Years of Experience
and assume the output from running the regression is as follows:
Call:
lm(formula = Salary ~ Year, data = Income)
Residuals:
Min 1Q Median 3Q Max
-53.650 -20.256 0.127 18.423 65.596
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 31.8387 8.5565 3.721 0.00033***
Years 2.8205 0.3302 8.543 1.74e-13 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 25.98 on 98 degrees of freedom
Multiple R-squared: 0.4268, Adjusted R-squared: 0.421
F-statistic: 72.98 on 1 and 98 DF, p-value: 1.737e-13
---
Residual standard error: 8.044 on 445 degrees of freedom
Multiple R-squared: 0.6914, Adjusted R-squared: 0.6886
F-statistic: 249.2 on 4 and 445 DF, p-value: < 2.2e-16
If we wished to conduct a hypothesis test as to whether there is a linear relationship between salary and years of experience, what are the correct null and alternate hypotheses?
Answers:
a.
H subscript 0 : space beta subscript 0 space equals space 0 H subscript 1 : space beta subscript 0 greater than space 0
b.
H subscript 0 : space beta subscript 0 space equals space 0 H subscript 1 : space beta subscript 0 space end subscript not equal to space 0
c.
H subscript 0 space : thin space beta subscript 1 space equals space 0 H subscript 1 : space beta subscript 1 space end subscript space not equal to space 0
d.
H subscript 0 : space beta subscript 1 space equals space 0 H subscript 1 : space beta subscript 1 space greater than space 0
Part B
Using the output in Q1, what is the correct p-value for the test in Q1?
Answers:
a.
0.00033
b.
0.000000000000174
c.
1.74e-13
d.
0.00000393
Part C
What is the fitted regression model from this output in Q1?
Answers:
a.
E(Salary) = 31.8387 + 2.8205 x Years of Experience
b.
E( Years of Experience ) = 2.8205 + 31.8387 x Salary
c.
E( Years of Experience ) = 31.8387 + 2.8205 x Salary
d.
E(Salary) = 2.8205 + 31.8387 x Years of Experience
Part D
Which of the following is a correct statement regarding r squared ?
Answers:
a.
r squared space equals space 0.4268 meaning that Years of Experience explains 42.68 percent sign of the variability in Salary.
b.
r squared space equals space 0.00033 meaning that Years of Experience explains 0.033 percent sign of the variability in Salary.
c.
r squared space equals space 0.00033 and because 0.00033 space less than space 0.05 we reject H subscript 0 and accept H subscript 1 at the 5% level of significance, ie we conclude there is a significant linear relationship between Salary and Years of Experience.
d.
r squared space equals space 0.4268 and because 0.4268 space greater than space 0.05 we do not reject H subscript 0 at the 5% level of significance, ie we conclude there is no significant linear relationship between Salary and Years of Experience.
The correct statement regarding r squared is:
r squared equals 0.4268 meaning that Years of Experience explains 42.68 percent of the variability in Salary.
Part A: The correct null and alternate hypotheses are:
H₀: β₁=0;
H₁: β₁≠0.
Part B: The correct p-value for the test in Q1 is 1.74e-13.
Part C: The fitted regression model from this output in Q1 is:
E(Salary) = 31.8387 + 2.8205 x Years of Experience.
Part D: The correct statement regarding r squared is:
r squared equals 0.4268 meaning that Years of Experience explains 42.68 percent of the variability in Salary.
Explanation: The output shows a multiple linear regression model:
Salary=β0+β1x
Years of Experience + ϵ.β0 is the intercept and represents the expected mean salary for an individual with 0 years of experience.
β1 is the slope and represents the expected change in salary due to one year increase in experience.
ϵ is the error term (deviation from the expected salary).
The correct null and alternate hypotheses are:
H₀: β₁=0 (there is no linear relationship between salary and years of experience).
H₁: β₁≠0 (there is a linear relationship between salary and years of experience).
The correct p-value for the test in Q1 is 1.74e-13, which is much smaller than the significance level of 0.05.
Thus, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that there is a linear relationship between salary and years of experience.
The fitted regression model from this output in Q1 is:
E(Salary) = 31.8387 + 2.8205 x Years of Experience.
The coefficient of determination, or R-squared, is a statistical measure that shows how well the regression model fits the observed data.
It is the proportion of the variance in the dependent variable that is explained by the independent variable(s).
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the perimeter of an isosceles triangle is 45. find the length of the third side if each of the equal sides is 14cm long
Answer:
17cm
Step-by-step explanation:
The two equal sides are 14cm long so 45 - 14 - 14 = 17
What is the area of the shaded area shown, his lawn? Please help me I really need it
Answer:
c - 13,600ft^2
Step-by-step explanation:
200 x 80 = 16,000ft^2
40 x 60 = 2,400ft^2
16,000ft^2 - 2,400ft^2 = 13,600ft^2
A decagon.
Gregory drew this regular decagon. All angles have the same measure.
What is the sum of the interior angle measures?
°
What is the measure of each angle?
°
Answer:
78
Step-by-step explanation:
Answer:
1440 and 144 on edg 2020-2021
Step-by-step explanation:
Jason and Karina have matching gardens Jason plants 2/3 of his garden with roses Karina's garden is divided into ninths how much garden must she plant to have the same amount.
Answer:
6/9
Step-by-step explanation:
Calculate the most probable values of X and Y for the following system of equations using: Tabular method . Matrix method X + 2Y = 10.5 2X-3Y= 5.5 2X – Y = 10.0
The most probable values for X and Y are X = 11.75 and Y = 1.1, respectively.
To solve the system of equations using the tabular or matrix method, we first convert the given equations into matrix form. We create a coefficient matrix A by arranging the coefficients of the variables X and Y, and a constant vector B by placing the constants on the other side of the equations.
To solve the system of equations using the tabular method or matrix method, we'll first write the equations in matrix form. Let's define the coefficient matrix A and the constant vector B:
A = | 1 2 |
| 2 -3 |
| 2 -1 |
B = | 10.5 |
| 5.5 |
| 10.0 |
Now, we can solve the system of equations by finding the inverse of matrix A and multiplying it with vector B:
[tex]A^{(-1)[/tex] = | 1.5 1 |
| 0.4 0.2 |
X = [tex]A^{(-1)[/tex] * B
Multiplying [tex]A^{(-1)[/tex] with B, we get:
X = | 1.5 1 | * | 10.5 | = | 11.75 |
| 0.4 0.2 | | 5.5 | | 1.1 |
Therefore, the most probable values for X and Y are X = 11.75 and Y = 1.1, respectively.
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What is the quotient of (x3 + 3x2 + 5x + 3) ÷ (x + 1)?
x2 + 4x + 9
x2 + 2x
x2 + 2x + 3
x2 + 3x + 8
Answer:
C
Step-by-step explanation:
C
A coin is tossed four times. You bet $1 that heads will come up on all four tosses. If this happens, you win $10. Otherwise, you lose your $1 bet.
Find: P(you win) =
P(you lose) =
Average winnings, µ, =
The P(you win) = 1/16P(you lose) = 15/16 and Average winnings, µ, = -5/16.
The probability of winning (P) and the probability of losing (Q) are both possible outcomes from a coin toss experiment. The probability of winning is given by the formula
P = Number of ways to win/ Total number of possible outcomes.The probability of losing is given by the formula
Q = Number of ways to lose/Total number of possible outcomes.
Formulae used to calculate probability are:
P = Number of ways to win/ Total number of possible outcomes
P = Number of outcomes in which all four tosses are heads/ Total number of possible outcomes
When a coin is tossed four times, the total number of possible outcomes is 2 × 2 × 2 × 2 = 16.
P = Number of outcomes in which all four tosses are heads/ Total number of possible outcomes
P = 1/16P (you win) = 1/16
The probability of losing is given by the formula
Q = Number of ways to lose/Total number of possible outcomes.
Q = 15/16P (you lose) = 15/16
Average winnings, µ, can be calculated as follows:
Let's say X represents the amount of money you win. When you win, you get $10, and when you lose, you lose $1.
X = -1 when you loseX = 10 when you winUsing the formula
µ = ∑ (X × P), we can calculate the average winnings,
µ = (-1 × 15/16) + (10 × 1/16)µ = -15/16 + 10/16µ = -5/16.
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Given information: A coin is tossed four times. You bet $1 that heads will come up on all four tosses. If this happens, you win $10. Otherwise, you lose your $1 bet.
Therefore, the answers are:
P(you win) = 0.0625
P(you lose) = 0.9375
Average winnings, µ, = -$0.31
Let A be the event of getting heads on the four tosses. Since the coin is tossed four times, the possible outcomes are 2^4 = 16. Thus the probability of getting heads on each toss is 0.5. Therefore, the probability of getting heads on all four tosses is:
P(A) = (0.5)^4
= 0.0625
Let B be the event of not getting heads on all four tosses. Thus:
B = 1 − P(A)
= 1 − 0.0625
= 0.9375
The winning amount for getting all four heads is $10, and the losing amount is $1. Thus, the average winnings is:
µ = (10 × 0.0625) − (1 × 0.9375)
µ = 0.625 − 0.9375
µ = -0.3125
Therefore, the answers are:
P(you win) = 0.0625
P(you lose) = 0.9375
Average winnings, µ, = -$0.31
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Write a function for the graph.
Answer:
(x-4)^2 -4 = y
Step-by-step explanation:
The graph is translated 4 to the right and 4 downwards.
in a circle of radius 28 cm, an subtends an angle of 45° at the centre, then find the length of the arc(in cm)
Answer:
22cm
Step-by-step explanation:
Length of arc = [tex]\frac{theta}{360}[/tex] * 2[tex]\pi[/tex]r
= [tex]\frac{45}{360}[/tex] * 2 * [tex]\frac{22}{7}[/tex] * 28
= 22 cm