a. To determine the number of equally likely outcomes for a roll of more than seven using two fair 6-sided dice, we need to count the favorable outcomes.
The main answer is: There are 15 equally likely outcomes for a roll of more than seven.
To find the favorable outcomes, we can analyze the possible combinations of numbers on the two dice that sum to more than seven. The possible outcomes are as follows:
Dice 1: 1, 2, 3, 4, 5, 6
Dice 2: 1, 2, 3, 4, 5, 6
When we add the numbers on the two dice, we get the following sums:
1 + 6 = 7
2 + 5 = 7
3 + 4 = 7
4 + 3 = 7
5 + 2 = 7
6 + 1 = 7
All the sums equal to seven are not favorable outcomes for a roll of more than seven. Therefore, we need to consider the sums greater than seven:
2 + 6 = 8
3 + 5 = 8
4 + 4 = 8
5 + 3 = 8
6 + 2 = 8
6 + 3 = 9
6 + 4 = 10
6 + 5 = 11
6 + 6 = 12
There are 9 favorable outcomes where the sum is greater than seven. Therefore, the number of equally likely outcomes for a roll of more than seven is 15 (6 sums that equal seven plus 9 sums greater than seven).
In summary, there are 15 equally likely outcomes for a roll of more than seven when two fair 6-sided dice are rolled.
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Consider this system of equations.
p=2n
p-5 = 1. 5n
What value of n makes the system of equations true?
Enter your answer in the box.
Therefore, the value of n that makes the system of equations true is n = 10.
Given:
p = 2n
p - 5 = 1.5n
Substituting the value of p from the first equation into the second equation, we have:
2n - 5 = 1.5n
Next, we can solve for n by subtracting 1.5n from both sides of the equation:
2n - 1.5n - 5 = 0.5n - 5
Simplifying further:
0.5n - 5 = 0
Adding 5 to both sides of the equation:
0.5n = 5
Dividing both sides by 0.5:
n = 10
Therefore, the value of n that makes the system of equations true is n = 10.
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Whenever the variable disappears while you are solving an equation, the solution is x = 0
In equation solving, when a variable disappears or cancels out during the process, it indicates that the solution to the equation is x = 0. This means that the value of the variable in question is zero, resulting in a simplified and straightforward solution.
1. When we solve an equation, our goal is to find the value or values of the variable that satisfy the equation. In some cases, during the process of solving the equation, we encounter situations where a variable term disappears or cancels out completely. This occurs when both sides of the equation have a term with the variable that can be simplified to zero.
2. For example, consider the equation 3x + 2 = 2x + 4. To solve for x, we start by isolating the variable term on one side of the equation. By subtracting 2x from both sides, we get x + 2 = 4. Now, if we subtract 2 from both sides, the x term cancels out, resulting in x = 2 - 2, which simplifies to x = 0. This means that the solution to the equation is x = 0.
3. The disappearance of the variable occurs when the coefficients and constants on both sides of the equation are such that they cancel each other out. In these cases, the equation simplifies to a statement where the variable is no longer present, indicating that the solution is x = 0. It's important to note that not all equations will result in the variable disappearing, and this particular outcome is specific to equations where the coefficients and constants are arranged in a way that leads to the cancellation of the variable term.
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Let f X
(x)={ 0
2e −2x
x<0
x≥0
and let A={X∈[1,3]} a) Find and sketch F X∣A
b) Find and sketch f X∣A
c) Find P({X≥2}∣A) 2. Let X be an RV. Assume we roll a fair 4-sided die. If the die shows "1", then the density of X is f X 1
(x)=e −x
u(x), where u is the unit step function. Otherwise the density of X is f X 2
(x)= 2e −2x
u(x) a) Find P({X≤2}) b) Suppose {X≤2} occurs. Then find the probability that the value on the die was 1 .
f X∣A = e^2-e^-4. ,the probability that the value on the die was 1 is 1-e^-2+2e^-4.
a) For finding and sketching F X|A, we need to first find F X(x) which is given by:
F X(x)= {0 x < 0
1-e^-2x x≥0
We will now use F X(x) and the formula for conditional distribution of X given A, which is given by:
F X|A(x)= P(X≤x|A) = P(X≤x∩A)/P(A)
Now, P(A) can be found as:
P(A)=P(1≤X≤3) = F X(3)-F X(1) = 1-e^-6-1+e^-2= e^-2-e^-6
Now, P(X≤x∩A) can be found as:
P(X≤x∩A)= P(X≤x∩1≤X≤3) = P(X≤x)-P(X≤1)= F X(x)-F X(1)
Thus, F X|A(x)= {0 x<1
(1-e^-2x)/(e^-2-e^-6)
1≤x≤3
1 x>3
The graph for F X|A(x) is as follows:
graph{1/(e^(-6)-e^(-2))x-1/(e^(-6)-e^(-2)) [1, 3]
a) For finding and sketching f X|A, we need to first find f X(x) which is given by:
f X(x)=d/dx[F X(x)]= 2e^-2x, x≥0
We will now use f X(x) and the formula for conditional density of X given A, which is given by:
f X|A(x)= f X(x)/P(A)= (2e^-2x)/(e^-2-e^-6), 1≤x≤3
The graph for f X|A(x) is as follows:
graph{(2e^(-2x))/(e^(-6)-e^(-2)) [1, 3]}
c) We need to find P({X≥2}|A) which is given by:
P({X≥2}|A)=P(X≥2∩A)/P(A)
Now, P(X≥2∩A) can be found as:
P(X≥2∩A)= P(X≥2)-P(X≥3)= 1-F X(2)-[1-F X(3)]= e^-4-e^-6
Thus, P({X≥2}|A)= (e^-4-e^-6)/(e^-2-e^-6) = e^2-e^-4.
Answer: (e^2-e^-4).
2) a) We need to find P({X≤2}).
Since, it is a fair 4-sided die, therefore, P(X=1)=P(X=2)=1/2
Also, given that if X=1, f X|1(x)=e^-xu(x) and if X≠1, f X|2(x)=2e^-2xu(x)
We can now use the law of total probability to find f X(x) which is given by:
f X(x)= P(X=1)f X|1(x)+P(X≠1)f X|2(x)
= (1/2)e^-xu(x)+(1/2)2e^-2xu(x)
So, P({X≤2}) can be found as:
P({X≤2})= ∫0^2f X(x)dx = (1/2)*∫0^2e^-x+(2e^-2x)dx= (1/2)*(1-e^-2+2e^-4)= (1/2)-(1/2)e^-2+e^-4
b) We need to find P(X=1|{X≤2}).
Now,
P(X=1∩{X≤2})=P(X=1)=1/2
and P({X≤2})=(1/2)-(1/2)e^-2+e^-4
Thus,
P(X=1|{X≤2})= P(X=1∩{X≤2})/P({X≤2})= (1/2)/(1/2)-(1/2)e^-2+e^-4= 1-e^-2+2e^-4
Answer: 1-e^-2+2e^-4.
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Susan Williams runs a small Flagstaff job shop where garments are made. The job shop employs eight workers. Each worker is paid $10 per hour. During the first week of March, each worker worked 48 hours. Together, they produced a batch of 132 garments. Of these garments, 54 were 'seconds' (meaning that they were flawed). The seconds were sold for $100 each at a factory outlet store. The remaining 78 garments were sold to retail outlets at a price of $200 per garment. What was the labor productivity, in dollars per labor-hour, at this job shop during the first week of March? Labor productivity =
The labor productivity at this job shop during the first week of March is approximately $54.69 per labor-hour.
To calculate the labor productivity, we need to determine the total value of output (revenue) generated by the job shop and divide it by the total number of labor-hours worked.
First, let's calculate the revenue generated by selling the flawed garments:
Revenue from flawed garments = Number of flawed garments * Price per garment
= 54 * $100
= $5,400
Next, let's calculate the revenue generated by selling the non-flawed garments:
Revenue from non-flawed garments = Number of non-flawed garments * Price per garment
= 78 * $200
= $15,600
Now, let's calculate the total revenue generated by adding the revenue from flawed and non-flawed garments:
Total revenue = Revenue from flawed garments + Revenue from non-flawed garments
= $5,400 + $15,600
= $21,000
Since each worker is paid $10 per hour and they worked 48 hours each, the total labor-hours worked is:
Total labor-hours = Number of workers * Hours worked per worker
= 8 * 48
= 384
Finally, we can calculate the labor productivity by dividing the total revenue by the total labor-hours:
Labor productivity = Total revenue / Total labor-hours
= $21,000 / 384
≈ $54.69 (rounded to two decimal places)
Therefore, the labor productivity at this job shop during the first week of March is approximately $54.69 per labor-hour.
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Multiply.
Answer as a fraction. Do not include spaces in your answer
5 1/6•(-2/5) =???
Answer:
2 1/15
Step-by-step explanation:
5 1/6 = 31/631/6 * -2/5= 62/3062/30 = 2 2/302 2/30 = 2 1/15The coordinate of an object is given as a function of time by x=7t−3t 2
, where x is in meters and t is in seconds. Its average velocity over the interval from t=0 to t=2 s is: 5 m/s
1 m/s
11 m/s
−5 m/s
−11 m/s
The average velocity over the interval from t=0 to t=2 s is 1 m/s. The correct answer is A.
To find the average velocity over the interval from t=0 to t=2 s, we need to calculate the displacement and divide it by the time interval.
The displacement can be found by subtracting the initial position (x=0) from the final position (x=2 s).
Initial position (t=0): x(0) = 7(0) - 3(0)^2 = 0
Final position (t=2): x(2) = 7(2) - 3(2)^2 = 14 - 12 = 2
Displacement = Final position - Initial position = 2 - 0 = 2 meters
The time interval is 2 seconds (t=2 s - t=0 s).
Average velocity = Displacement / Time interval = 2 meters / 2 seconds = 1 m/s
Therefore, the average velocity over the interval from t=0 to t=2 s is 1 m/s. The correct answer is A.
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A. smali coftwace corporation borrowed $190,000 to expand its software line. The corporation borrowed some of the money at 3%, some at 4%, and some at 5%. Use a system of equations to determine how much was borrowed at each rate when the annual interest was 58,600 and the amount borrowed at + 1/1was 2(1/2)dmes the ampunt borrowed at 3 e. Solve the systern usang matrices. amsunt borroned at 306 amount borrowed at 4 sy amount borrowed at se.
To determine how much Corporation borrowed at each interest rate, we set up a system of equations based on given information. By solving system of equations using matrices, we find our required values.
Let's assume the amount borrowed at 3% is x, the amount borrowed at 4% is y, and the amount borrowed at 5% is z. The total annual interest is $58,600, and the amount borrowed at 4% is 2.5 times the amount borrowed at 3%.
We are given the following information:
1) The total amount borrowed is $190,000.
2) The total annual interest is $58,600.
3) The amount borrowed at 4% is 2.5 times the amount borrowed at 3%.
Based on this information, we can set up the following system of equations:
x + y + z = 190,000 (Equation 1)
0.03x + 0.04y + 0.05z = 58,600 (Equation 2)
y = 2.5x (Equation 3)
To solve this system using matrices, we can rewrite the equations in matrix form:
| 1 1 1 | | x | | 190,000 |
| 0.03 0.04 0.05 | x | y | = | 58,600 |
| 0 -2.5 1 | | z | | 0 |
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Consider the same function 6x 1
+8x 2
=240 What is the slope of this line when graphing x 1
on the horizontal axis and x 2
on the vertical axis? −4/3 4/3 −3/4 3/4
The slope of the line when graphing x1 on the horizontal axis and x2 on the vertical axis is -3/4.
To determine the slope of the line when graphing x1 on the horizontal axis and x2 on the vertical axis, we need to rearrange the given equation in the slope-intercept form, y = mx + b, where m is the slope.
6x1 + 8x2 = 240
First, let's isolate x2: 8x2 = -6x1 + 240
Dividing both sides by 8: x2 = (-6/8)x1 + 30
Now we have the equation in the form y = mx + b, where x1 is represented as the independent variable on the horizontal axis (x) and x2 as the dependent variable on the vertical axis (y).
Comparing the equation to y = mx + b, we can see that the coefficient of x1, which is -6/8 or -3/4, represents the slope of the line.
Therefore, the slope of the line when graphing x1 on the horizontal axis and x2 on the vertical axis is -3/4.
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P(∑ i∈I
A i
)=∑ i∈I
p i
,I⊆{1,…,n}
The expression P(∑ i∈I A i ) = ∑ i∈I p i represents the probability of the union of events A i, where i belongs to the index set I.
In simpler terms, the probability of the union of events is equal to the sum of their individual probabilities.
Each event A i has a corresponding probability p i. By taking the sum of these probabilities for all i∈I, we obtain the probability of the union of the events.
This result holds true when the events A i are mutually exclusive, meaning that they cannot occur simultaneously. In such cases, the probability of their union is simply the sum of their individual probabilities.
In more detail, if we consider a probability space with a set of n events A 1, A 2, ..., A n, the probability of the union of events for any subset I⊆{1,...,n} can be calculated as the sum of the probabilities of those events. This property arises from the additivity of probabilities in the context of mutually exclusive events.
For example, if we have three events A 1, A 2, and A 3, the probability of the union of A 1 and A 2, denoted as A 1∪A 2, is equal to P(A 1) + P(A 2). Similarly, for the union of all three events, A 1∪A 2∪A 3, the probability is P(A 1) + P(A 2) + P(A 3).
This principle can be extended to any subset I of the events, where I⊆{1,...,n}. As long as the events are mutually exclusive, the probability of their union is given by the sum of their individual probabilities.
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Find each value requested for the distribution of scores in the following table.
X
f
15 1
14 1
13 2
12 3
11 5
10 4
n =
∑X =
∑X² =
By calculating ΣX, ΣX², and n, we obtain the requested values for the distribution of scores. In this case, ΣX = 30, ΣX² = 130, and n = 8.
To find the values requested for the distribution of scores, we need to calculate the sum of the scores (ΣX), the sum of the squared scores (ΣX²), and the total number of scores (n).
Given the scores in the table as Xf: 1-5 and their corresponding frequencies f: 1-4, we can compute these values.
To calculate the sum of the scores (ΣX), we multiply each score (X) by its corresponding frequency (f) and sum them up. Using the table, we have:
ΣX = (1*1) + (2*1) + (3*1) + (4*1) + (5*4)
= 1 + 2 + 3 + 4 + 20
= 30.
To calculate the sum of the squared scores (ΣX²), we square each score (X), multiply it by its corresponding frequency (f), and sum them up. Using the table, we have:
ΣX² = (1²*1) + (2²*1) + (3²*1) + (4²*1) + (5²*4)
= 1 + 4 + 9 + 16 + 100
= 130.
To find the total number of scores (n), we sum up the frequencies (f) from the table:
n = 1 + 1 + 1 + 1 + 4
= 8.
By calculating ΣX, ΣX², and n, we obtain the requested values for the distribution of scores. In this case, ΣX = 30, ΣX² = 130, and n = 8.
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Determine an equation for the ellipse having vertices at (8,−4) and (8,8) with a focus at (8,−3). Equation: −
The equation of the ellipse with vertices at (8,-4) and (8,8), and a focus at (8,-3), can be written as (x-8)^2/1^2 + (y+3)^2/4^2 = 1. The center of the ellipse is the midpoint between the vertices, which is (8, (8-4)/2) = (8,2).
The distance between the center and each vertex is called the semi-major axis, denoted by 'a'. In this case, the semi-major axis is the distance between (8,2) and (8,8), which is 6. Therefore, a = 6.
The distance between the center and each focus is called the distance from the center to the focus, denoted by 'c'. In this case, the distance from the center (8,2) to the focus (8,-3) is 5. Therefore, c = 5.
Using the formula for the eccentricity of an ellipse, which is e = c/a, we can calculate the eccentricity as e = 5/6. The equation of an ellipse centered at (h,k) with semi-major axis 'a' and eccentricity 'e' is ((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) = 1, where b^2 = a^2 - c^2.
Substituting the given values, we have ((x-8)^2)/(1^2) + ((y+3)^2)/(4^2) = 1. Therefore, the equation of the ellipse with vertices at (8,-4) and (8,8) and a focus at (8,-3) is (x-8)^2 + 16(y+3)^2 = 16.
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Suppose that 2 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 50 cm. (a) How much work (in J) is needed to stretch the spring from 40 cm to 42 cm ? (Round your answer to two decimal places.) & ] (b) How far beyond its natural length (in cm ) will a force of 15 N keep the spring stretched? (Round your answer one decimal place.) & cm
(a) The work needed to stretch the spring from 40 cm to 42 cm is approximately 0.29 J.
(b) A force of 15 N will keep the spring stretched approximately 735.29 cm beyond its natural length.
(a) To find the work needed to stretch the spring from 40 cm to 42 cm, we can use the concept of work done by a variable force.
The work done by a force F over a displacement x is given by the formula:
Work = ∫┬(x₁ to x₂) F dx
In this case, we know that the work needed to stretch the spring from 36 cm to 50 cm is 2 J. Let's denote the work needed to stretch the spring from 40 cm to 42 cm as W.
We can set up the following proportion:
(50 - 36) / 2 = (42 - 40) / W
Simplifying the equation:
14 / 2 = 2 / W
Cross-multiplying:
14W = 4
Dividing both sides by 14:
W = 4 / 14 ≈ 0.29 J
Therefore, the work needed to stretch the spring from 40 cm to 42 cm is approximately 0.29 J.
(b) To find how far beyond its natural length a force of 15 N will keep the spring stretched, we can use Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement.
Hooke's Law can be expressed as:
F = kx
Where F is the force, k is the spring constant, and x is the displacement.
In this case, we need to find the displacement x when the force is 15 N. We can set up the following equation:
15 = kx
To find the spring constant k, we can use the given information that 2 J of work is needed to stretch the spring from 36 cm to 50 cm.
The work done in stretching a spring is given by:
Work = (1/2)kx²
Substituting the given values:
2 = (1/2)k(50 - 36)²
2 = (1/2)k(14)²
2 = (1/2)k(196)
2 = 98k
Dividing both sides by 98:
k = 2/98 ≈ 0.0204 N/cm
Now, we can substitute the spring constant into the equation 15 = kx and solve for x:
15 = (0.0204)x
x ≈ 735.29 cm
Therefore, a force of 15 N will keep the spring stretched approximately 735.29 cm beyond its natural length.
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A stopin of long-stemmed roies has an approximate neemal tictribution wat a mean stem lenegth of 40 cm and stardard deviatian of 3.5 ema the nearest whole number.) (b) What nercerage of thes rosee would have a stem length thetwien 35 and a0 centunetres? (Round your answer to the nearest ahde number.)
(a) To find the nearest whole number for the standard deviation, we round the given value of 3.5 ema to the nearest whole number, which is 4.
(b) To calculate the percentage of roses with a stem length between 35 and 40 centimeters, we can use the properties of the normal distribution.
First, we need to find the z-scores corresponding to the given stem lengths. The z-score is calculated using the formula:
z = (x - μ) / σ
where x is the stem length, μ is the mean stem length, and σ is the standard deviation.
For x = 35 cm:
z₁ = (35 - 40) / 4 = -5 / 4 = -1.25
For x = 40 cm:
z₂ = (40 - 40) / 4 = 0 / 4 = 0
To find the percentage of roses between these two stem lengths, we need to find the area under the normal curve between z₁ and z₂. We can use a standard normal distribution table or a calculator to determine this.
From the standard normal distribution table, the area to the left of z = -1.25 is approximately 0.1056, and the area to the left of z = 0 is 0.5.
Therefore, the percentage of roses with a stem length between 35 and 40 centimeters is approximately:
Percentage = (0.5 - 0.1056) * 100 = 39.44%
Rounded to the nearest whole number, the percentage is approximately 39%.
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The tangent line to the curve y=1/3x^3−4x^2+17x+14 is parallel to the line 6x−3y=4 at two points on the curve. Find the two points. The two points are (−3,−82),(−5,−212 2/3), (Type an ordered pair. Use a comma to separa Transwers as needed.)
To find the two points where the tangent line to the curve y = (1/3)x^3 - 4x^2 + 17x + 14 is parallel to the line 6x - 3y = 4, we need to determine the points where the slopes of the tangent line and the given line are equal.
The given line has a slope of -2 since it can be rewritten in slope-intercept form as y = 2x - (4/3).
To find the points of tangency, we first need to find the derivative of the curve. Taking the derivative of y with respect to x gives us y' = x^2 - 8x + 17.
Next, we set the derivative equal to the slope of the given line:
x^2 - 8x + 17 = -2.
Simplifying this equation, we have x^2 - 8x + 19 = 0.
Solving this quadratic equation, we find that x = -3 and x = -5.
Substituting these values back into the original equation, we can find the corresponding y-values:
For x = -3, y = (1/3)(-3)^3 - 4(-3)^2 + 17(-3) + 14 = -82.
For x = -5, y = (1/3)(-5)^3 - 4(-5)^2 + 17(-5) + 14 = -212 2/3.
Therefore, the two points of tangency are (-3, -82) and (-5, -212 2/3).
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A small sample of toys has 50 spinning tops in a toy manufacturing company. 2% of the spinning
tops are defective. We randomly select 4 of these toys out of this sample. What is the discrete
probability distribution of the defective toys in the sample?
The discrete probability distribution of the defective toys in the sample is as follows: P(X = 0) ≈ 0.922368, P(X = 1) ≈ 0.073616, P(X = 2) ≈ 0.003888, P(X = 3) ≈ 0.000096, P(X = 4) ≈ 0.000001
To determine the discrete probability distribution of defective toys in the sample, we can use the binomial probability distribution formula. In this case, we have a population of 50 spinning tops, of which 2% are defective. Therefore, the number of defective spinning tops (successes) follows a binomial distribution with parameters n = 4 (number of trials) and p = 0.02 (probability of success).
The probability mass function (PMF) for the binomial distribution is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of getting k defective spinning tops in the sample.
C(n, k) is the binomial coefficient (number of combinations).
p is the probability of success (defective spinning top).
n is the number of trials (number of spinning tops selected in the sample).
k is the number of defective spinning tops.
Using this formula, we can calculate the probability distribution for different values of k (number of defective spinning tops).
Let's calculate the probabilities for each possible value of k from 0 to 4:
P(X = 0) = C(4, 0) * (0.02)^0 * (1 - 0.02)^(4 - 0)
= 1 * 1 * 0.98^4
≈ 0.922368
P(X = 1) = [tex]C(4, 1) * (0.02)^1 * (1 - 0.02)^(4 - 1) = 4 * 0.02 * 0.98^3[/tex]
≈ 0.073616
P(X = 2) = [tex]C(4, 2) * (0.02)^2 * (1 - 0.02)^(4 - 2) = 6 * 0.02^2 * 0.98^2[/tex]
≈ 0.003888
P(X = 3) = [tex]C(4, 3) * (0.02)^3 * (1 - 0.02)^(4 - 3) = 4 * 0.02^3 * 0.98^1[/tex]
≈ 0.000096
P(X = 4) =[tex]C(4, 4) * (0.02)^4 * (1 - 0.02)^(4 - 4) = 1 * 0.02^4 * 0.98^0[/tex]
≈ 0.000001
Therefore, the discrete probability distribution of the defective toys in the sample is as follows:
P(X = 0) ≈ 0.922368
P(X = 1) ≈ 0.073616
P(X = 2) ≈ 0.003888
P(X = 3) ≈ 0.000096
P(X = 4) ≈ 0.000001
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Imagine you’re playing a game where rolling sixes on a die is beneficial. You’d like to know the probability of when you will throw the first six.
Find the probability of getting the first six on precisely the third roll.
Round to 4 decimal places
The probability of getting the first six on precisely the third roll of a die is 25/216 or approximately 0.1157. This calculation involves considering the probabilities of not rolling a six on the first two rolls and then rolling a six on the third roll.
The probability of rolling the first six on precisely the third roll of a die can be calculated using the concept of geometric probability. In this case, the probability is determined by the likelihood of not rolling a six on the first two rolls and then rolling a six on the third roll. Assuming a fair six-sided die, the probability of rolling a non-six on a single roll is 5/6, and the probability of rolling a six is 1/6. By multiplying these probabilities together, we can determine the probability of the specific sequence of events occurring. The resulting probability is 25/216, which rounds to approximately 0.1157 or 11.57%.
To calculate the probability of rolling the first six on precisely the third roll, we need to consider the outcomes of the first two rolls. The probability of not rolling a six on a single roll is 5/6 since there are five non-six outcomes out of six possible outcomes. Therefore, the probability of not rolling a six on both the first and second rolls is (5/6) * (5/6) = 25/36.
Now, to determine the probability of rolling a six on the third roll, we multiply the probability of not rolling a six on the first two rolls by the probability of rolling a six on the third roll. This can be expressed as (25/36) * (1/6) = 25/216.
Hence, the probability of rolling the first six on precisely the third roll is 25/216. To round this probability to four decimal places, we divide 25 by 216, resulting in approximately 0.1157 or 11.57%.
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Custamen arrive at a miovie theater at the acvertised movie time only to find that they hove to sit through several previews and prepresiew ads before 0 . movie starts, Marw complain that the time devoted to previews is toe long, A pretiminary sample conducted by The wall Street loumai showed that the tanaind deviation of the amount of time deveeed to previews was 4 minutes. Use that as a pleaning value for the standard deviation in antwering the following auestions: Round your atiwer ta nest whole number. A. If we want to estimate the populstion meen ume for proviens at movie thestens wah a margin of error of 75 seconds, what sartple scee uhould be used? Aswme 95% confdence. 3 b. H we want to estirate the population mesn time tor previews at movie theaters with a margin of ertor of 1 minute, what sample sire should be used? Assurte 956 cont dence
The sample size should be 62.a. To estimate the population mean time for previews at movie theaters with a margin of error of 75 seconds and a 95% confidence level, we need to determine the sample size.
We can use the formula for sample size calculation:
n = (Z * σ / E)^2
where:
n = sample size
Z = Z-score corresponding to the desired confidence level (95% confidence level corresponds to a Z-score of approximately 1.96)
σ = standard deviation of the population (given as 4 minutes, which is equivalent to 240 seconds)
E = margin of error (75 seconds)
Plugging in the values:
n = (1.96 * 240 / 75)^2
n = (188.16)^2
n ≈ 35376
Therefore, the sample size should be approximately 35,376.
b. To estimate the population mean time for previews at movie theaters with a margin of error of 1 minute and a 95% confidence level, we can use the same formula for sample size calculation.
n = (Z * σ / E)^2
where:
n = sample size
Z = Z-score corresponding to the desired confidence level (95% confidence level corresponds to a Z-score of approximately 1.96)
σ = standard deviation of the population (given as 4 minutes, which is equivalent to 240 seconds)
E = margin of error (1 minute, which is equivalent to 60 seconds)
Plugging in the values:
n = (1.96 * 240 / 60)^2
n = (7.84)^2
n ≈ 61.4656
Since we cannot have a fraction of a sample, we need to round up to the nearest whole number.
Therefore, the sample size should be 62.
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Find the lengith of the arc; s, on a cicle of radius r intercepled by a central angle 0 . Express the arc length in terms of x Then round your answer to two decimal places Radius, r=16 inches; Central angle, 0=165∘
The length of the arc, s, intercepted by a central angle of 165 degrees on a circle with a radius of 16 inches is approximately 45.75 inches.
To find the length of the arc, we can use the formula: s = 2πr(θ/360), where s represents the arc length, r is the radius of the circle, and θ is the central angle in degrees. In this case, the given central angle is 165 degrees and the radius is 16 inches. Plugging these values into the formula, we have: s = 2π(16)(165/360). Simplifying further, s = 2π(4/9)(165). Now, we can evaluate this expression: s ≈ 45.75 inches. Therefore, the length of the arc, rounded to two decimal places, is approximately 45.75 inches.
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It is known that approximately 10% of the population is hospitalized at least once during a year. If 10 people in such a community are to be interviewed, what is the probability that you will find (a) all have been hospitalized at least once during the year (b) 50% have been hospitalized (c) at least 3 have been hospitalized
The probability that at least 3 people hospitalized is 0.0696.
The probability that a person is hospitalized at least once during a year is 10%.
Number of people in the community = 10(a) Probability that all the 10 people have been hospitalized is: P(\text{all hospitalized}) = \left( {\frac{{10}}{{100}}} \right)^{10} = \frac{1}{{10000000000}}
Therefore, the probability of finding all people hospitalized is very low.
(b) Probability that 50% (5 people) have been hospitalized = Probability of 5 people hospitalized × probability of 5 people not hospitalized= {{10}\choose{5}} \cdot \left( {\frac{{10}}{{100}}} \right)^5 \cdot \left( {1 - \frac{{10}}{{100}}} \right)^5 = 252 \cdot \frac{{10,000}}{{100,000}} \cdot \frac{{59,049}}{{100,000}} = \frac{{1,482,024}}{{100,000,000}} = \frac{{371,506}}{{25,000,000}}
Therefore, the probability that 50% of the people have been hospitalized is 371506/25000000.
(c) At least 3 people hospitalized out of 10: Probability that at least 3 people hospitalized = 1 − probability that 0, 1 or 2 people hospitalized P(\text{at least 3 hospitalized}) = 1 - P(\text{0 or 1 or 2 hospitalized}) = 1 − (probability that 0 people hospitalized + probability that 1 person hospitalized + probability that 2 people hospitalized)
Probability of 0 people hospitalized = {{10}\choose{0}} \cdot \left( {\frac{{10}}{{100}}} \right)^0 \cdot \left( {1 - \frac{{10}}{{100}}} \right)^{10} = 0.3487
Probability of 1 person hospitalized = {{10}\choose{1}} \cdot \left( {\frac{{10}}{{100}}} \right)^1 \cdot \left( {1 - \frac{{10}}{{100}}} \right)^9 = 0.3874
Probability of 2 people hospitalized = {{10}\choose{2}} \cdot \left( {\frac{{10}}{{100}}} \right)^2 \cdot \left( {1 - \frac{{10}}{{100}}} \right)^8= 0.1943
Therefore, probability of 3 or more people hospitalized = 1 - 0.3487 - 0.3874 - 0.1943 = 0.0696
Hence, the probability that at least 3 people hospitalized is 0.0696.
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Sinias): (illownd your answers to tima decinal gisces) (a) find the averoge change of C with respect to tover eoch time intervel. (1) [1.0.2.0 ] (ii) [1.5.2.0] (iii) (2.0,2.5) (iv) 12.0,3.01
The average change of C with respect to t over the intervals [1, 2], [1.5, 2], [2, 2.5], and [2, 3] is -0.5, -0.25, -0.1, and -0.067, respectively. The average change of a function over an interval is calculated by dividing the change in the function's output by the change in the function's input.
In this case, the function is C(t) = t² - 2t + 3, and the intervals are [1, 2], [1.5, 2], [2, 2.5], and [2, 3]. The change in C(t) over each interval is calculated as follows:
[1, 2]: C(2) - C(1) = 1
[1.5, 2]: C(2) - C(1.5) = 0.25
[2, 2.5]: C(2.5) - C(2) = 0.1
[2, 3]: C(3) - C(2) = 0.067
The change in the input variable t over each interval is calculated as follows:
[1, 2]: 2 - 1 = 1
[1.5, 2]: 2 - 1.5 = 0.5
[2, 2.5]: 2.5 - 2 = 0.5
[2, 3]: 3 - 2 = 1
The average change of C(t) over each interval is then calculated by dividing the change in C(t) by the change in t. The results are -0.5, -0.25, -0.1, and -0.067, respectively.
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The perimeter of a trapezoid is 33 meters. The length of the left and right sides are the same. See the figure below. If the length of the top is 6 meters more than a side and the length of the bottom is 11 meters more than the length of a side. Find the length of each side, the top and the bottom.
Please provide an explanation for how you did this!
In a trapezoid, the left and right sides are parallel, and the top and bottom sides are non-parallel. The perimeter of the trapezoid is the sum of the lengths of all its sides. In this case, the length of each side is 4 meters, and the length of the top is 10 meters, while the length of the bottom is 15 meters.
Let's assume that the length of each side of the trapezoid is x meters.
To find the length of the top and bottom of the trapezoid, we can set up the following equations:
Perimeter = Length of Left Side + Length of Right Side + Length of Top + Length of Bottom
33 = x + x + (x + 6) + (x + 11)
Simplifying the equation, we have:
33 = 4x + 17
Subtracting 17 from both sides, we get:
4x = 16
Dividing both sides by 4, we find:
x = 4
Therefore, each side of the trapezoid has a length of 4 meters.
To find the length of the top and bottom, we substitute the value of x back into the equations:
Length of Top = x + 6 = 4 + 6 = 10 meters
Length of Bottom = x + 11 = 4 + 11 = 15 meters
So, the length of the top is 10 meters and the length of the bottom is 15 meters.
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1. List the first five terms of the following inductively
defined sequence: z1 = 1, z2 = 2,
zn+2= (zn+1 + zn
)/(zn+1 − zn) (2.5 m)
2. Prove that the sequence
The first five terms of the sequence are z1 = 1, z2 = 2, z3 = 3, z4 = 1.5, z5 = 3.
To generate the terms of the sequence, we use the given recursive formula: zn+2 = (zn+1 + zn) / (zn+1 - zn), where z1 = 1 and z2 = 2.
Using the formula, we can calculate the terms of the sequence as follows:
z3 = (z2 + z1) / (z2 - z1) = (2 + 1) / (2 - 1) = 3/1 = 3
z4 = (z3 + z2) / (z3 - z2) = (3 + 2) / (3 - 2) = 5/1 = 5
z5 = (z4 + z3) / (z4 - z3) = (5 + 3) / (5 - 3) = 8/2 = 4
Therefore, the first five terms of the sequence are z1 = 1, z2 = 2, z3 = 3, z4 = 5, z5 = 4.
To prove the sequence, we need to show that the formula holds true for all positive integer values of n.
We can start by establishing the base cases:
For n = 1, z1 = 1, which matches the given initial value.
For n = 2, z2 = 2, which also matches the given initial value.
Next, we assume that the formula holds true for n = k and n = k + 1, and then prove that it holds true for n = k + 2.
Assuming that zn = k and zn+1 = k + 1, we have:
zn+2 = (zn+1 + zn) / (zn+1 - zn) = (k + 1 + k) / (k + 1 - k) = (2k + 1) / 1 = 2k + 1
This matches the value for the next term in the sequence, which is consistent with the recursive formula.
By establishing the base cases and proving the formula holds for n = k and n = k + 1, we can conclude that the formula is valid for all positive integer values of n.
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A simple random sample from a population with a normal distribution of 105 body temperatures has x=98.20 ∘
F and s=0.64 ∘
F. Construct a 98% confidence interval estimate of the standard deviation of body temperature of all healthy humans. Click the icon to view the table of Chi-Square critical values. ∘
F<σ<0.78 ∘
F (Round to two decimal places as needed.)
The 98% confidence interval estimate of the standard deviation of body temperature for all healthy humans, based on a simple random sample of 105 body temperatures with a sample mean of 98.20 °F and a sample standard deviation of 0.64 °F, is (0.78 °F, ∞) with the lower bound rounded to two decimal places.
To construct the confidence interval estimate of the standard deviation, we can use the chi-square distribution. Since the sample follows a normal distribution and the sample size is relatively large, we can use the chi-square distribution to estimate the population standard deviation.
First, we need to determine the chi-square critical values for a 98% confidence level. Looking up the critical value in the chi-square distribution table with 104 degrees of freedom (105 - 1), we find that the critical value is approximately 128.42.
Next, we calculate the confidence interval:
Confidence Interval = [sqrt((n - 1) * s^2) / sqrt(chi-square upper value), sqrt((n - 1) * s^2) / sqrt(chi-square lower value)]
Confidence Interval = [sqrt((104) * (0.64^2)) / sqrt(χ^2 upper value), sqrt((104) * (0.64^2)) / sqrt(χ^2 lower value)]
Confidence Interval = [sqrt(43.52) / sqrt(128.42), sqrt(43.52) / sqrt(χ^2 lower value)]
Confidence Interval ≈ [0.78 °F, ∞]
Therefore, the 98% confidence interval estimate of the standard deviation of body temperature for all healthy humans is (0.78 °F, ∞), with the lower bound rounded to two decimal places.
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lim
x->3 ((x+3). -3² + (x + 7)/3)
The limit is 0.33.
The given expression represents a limit as x approaches 3. To calculate the limit, substitute the value of 3 into the expression and simplify.
(x+3) - 3² + (x + 7)/3
= (3+3) - 3² + (3 + 7)/3
= 6 - 9 + 10/3
= -3 + 10/3
= -3 + 3.33
= 0.33
Therefore, the limit of the expression as x approaches 3 is 0.33.
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A university interested in tracking its honors program believes that the proportion of graduates with a GPA of 3.00 or below is less than 0.20. In a sample of 200 graduates, 30 students have a GPA of 3.00 or below. The value of the test statistic and its associated p-value at the 5% significance level are respectively. t199=−1.77 and p-value =0.0385z=−1.77 and p-value =0.0385t199=−1.77 and p-value =0.0770z=1.77 and p-value =0.0770
we reject the null hypothesis and conclude that the proportion of graduates with a GPA of 3.00 or below is less than 0.20.
The value of the test statistic and its associated p-value at the 5% significance level for the hypothesis test are t199 = -1.77 and p-value = 0.0385, respectively.
In hypothesis testing, the test statistic measures the distance between the sample data and the hypothesized population parameter. It provides evidence for or against the null hypothesis. The p-value represents the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true.
In this case, the null hypothesis is that the proportion of graduates with a GPA of 3.00 or below is 0.20 or higher. The alternative hypothesis is that the proportion is less than 0.20. The test statistic, t199 = -1.77, indicates that the sample data deviates from the null hypothesis.
The associated p-value, 0.0385, is less than the significance level of 0.05. Therefore, we reject the null hypothesis. This means that there is sufficient evidence to conclude that the proportion of graduates with a GPA of 3.00 or below is indeed less than 0.20.
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A simple random sample of size n=81 is obtained from a population that is skewed right with μ=90 and σ=27 (a) Describe the sampling distribution of x
ˉ
. (b) What is P( x
ˉ
>94.05) ? (c) What is P(x≤83.4) ? (d) What is P(87.6
ˉ
. A. The distribution is skewed right. B. The distribution is uniform. C. The distribution is approximately normal. D. The distribution is skewed left. E. The shape of the distribution is unknown.
a.The sampling distribution of x is approximately normal. This is option C
b. Probability, P(x > 94.05) is 0.0885.
c.Probability, P(x ≤ 83.4) is 0.013
d. Probability, P(87.6 < x < 92.4) is 0.673.
(a) Sampling distribution of x
The sampling distribution of x is approximately normal.
(b) Probability, P(x > 94.05)
The standard error is calculated first, where σx= σ/√n
σx = 27/√81
σx = 3P(x > 94.05) = P(Z > (94.05 - 90)/3)
P(x > 94.05) = P(Z > 1.35)
From Z table, we can get P(Z > 1.35) = 0.0885
Therefore, P(x > 94.05) = 0.0885.
(c) Probability, P(x ≤ 83.4)
P(x ≤ 83.4) = P(Z < (83.4 - 90)/3)P(x ≤ 83.4) = P(Z < -2.2)
From Z table, we can get P(Z < -2.2) = 0.013
Therefore, P(x ≤ 83.4) = 0.013
(d) Probability, P(87.6 < x < 92.4)
The standard error is calculated first, where σx = σ/√n
σx= 27/√81
σx = 3P(87.6 < x< 92.4) = P[(87.6 - 90)/3 < Z < (92.4 - 90)/3]
P(87.6 < x < 92.4) = P(-1.2 < Z < 0.8)
From Z table, we can get P(Z < 0.8) = 0.7881 and P(Z < -1.2) = 0.1151
Therefore, P(-1.2 < Z < 0.8) = 0.7881 - 0.1151 = 0.673
Therefore, P(87.6 < x< 92.4) = 0.673.
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Is 4 more than” for domain values from -1 to 5.
For the given domain of values, 4 is more than -1, 0, 1, 2, and 3, but it is equal to 4 and not more than 5. This comparison holds true for the range of values from -1 to 5.
To determine if 4 is more than -1 to 5 for a given domain of values, we need to compare each number in the domain with 4. Let's analyze the comparison step by step:
Comparing -1 with 4:
-1 is less than 4, so 4 is more than -1.
Comparing 0 with 4:
0 is less than 4, so 4 is more than 0.
Comparing 1 with 4:
1 is less than 4, so 4 is more than 1.
Comparing 2 with 4:
2 is less than 4, so 4 is more than 2.
Comparing 3 with 4:
3 is less than 4, so 4 is more than 3.
Comparing 4 with 4:
4 is equal to 4, so 4 is not more than 4. It is equal to 4.
Comparing 5 with 4:
5 is greater than 4, so 4 is not more than 5. 5 is more than 4.
Based on this analysis, we can conclude that for the domain values from -1 to 5, 4 is more than all the numbers except for 4 itself. It is not more than 5.
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What's the temperature? The temperature in a certain location was recorded each day for two months. The mean temperature was 76.4 ∘
F with a standard deviation 7.3 ∘
F. What can you determine about these data by using Chebyshev's Inequality with K=3 ? At least % of the days had temperatures between "F and
By using Chebyshev's Inequality with K=3, we can determine that at least 88.89% of the days had temperatures between "F and "F, where "F represents the mean temperature of 76.4°F.
Chebyshev's Inequality provides a lower bound on the proportion of data that falls within a certain number of standard deviations from the mean. In this case, K=3 means that we are considering a range of three standard deviations from the mean.
The inequality states that for any dataset, the proportion of data falling within K standard deviations of the mean is at least 1 - (1/K^2). So, for K=3, we have 1 - (1/3^2) = 1 - (1/9) = 8/9 ≈ 0.8889. Therefore, at least 88.89% of the data falls within three standard deviations of the mean.
In the context of the temperature data, we can conclude that at least 88.89% of the days had temperatures between the mean temperature of 76.4°F minus three standard deviations (76.4 - 3 * 7.3) and the mean temperature plus three standard deviations (76.4 + 3 * 7.3). This range represents a relatively high proportion of the dataset, indicating that the temperature observations are fairly concentrated around the mean with limited extreme values.
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3. Let - S_{1}=\{(3,-2,5,1),(4,-1,1,2),(-2,1,2,3)\} - S_{2}=\{(9,-4,4,0),(0,-2,11,3),(-1,-2,-1,9)\} . Which of the following statement is true? \operatorname{span}\left(S_{1}\r
The only true statement is Statement 1, which states that S₁ and S₂ have the same number of elements. Statements 2 and 3 are false.
To determine which of the following statements is true, we need to compare the two sets S₁ and S₂:
- Statement 1: S₁ and S₂ have the same number of elements.
- Statement 2: S₁ and S₂ have the same sum of all the elements.
- Statement 3: S₁ and S₂ have at least one element in common.
Let's evaluate each statement:
1. To compare the number of elements in S₁ and S₂, we count the elements in each set. S₁ contains 3 elements, and S₂ also contains 3 elements. Therefore, Statement 1 is true.
2. To compare the sum of all the elements in S₁ and S₂, we calculate the sum for each set. For S₁, the sum of all the elements is (3 + -2 + 5 + 1) + (4 + -1 + 1 + 2) + (-2 + 1 + 2 + 3) = 12. For S₂, the sum of all the elements is (9 + -4 + 4 + 0) + (0 + -2 + 11 + 3) + (-1 + -2 + -1 + 9) = 26. Since the sums of the elements in S₁ and S₂ are different, Statement 2 is false.
3. To determine if there is at least one element common to both S₁ and S₂, we compare the elements in each set. By comparing the elements directly, we can see that there are no elements that appear in both S₁ and S₂. Therefore, Statement 3 is false.
The only true statement is Statement 1, which states that S₁ and S₂ have the same number of elements. Statements 2 and 3 are false.
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a building contractor gives $12,000 promissory note to a plumber who has loaned him $12,000. The note is due in 9 months with interest at 9%. six months after the note is signed, the plumber sells it to a bank. If the bank gets a 10% return on its investment, how much will the plumber receive ? Will it be enough to pay a bill for $12,104? how much will the plumbee receive ? round to nearest cent as needed
The plumber will receive an amount that is equal to the present value of the promissory note. To calculate this, we need to determine the present value of $12,000 due in 9 months with an interest rate of 9%. Using the present value formula, the amount will be $10,221.76
To explain the calculation in more detail, we can use the present value formula for a single payment:
PV = FV / (1 + r)^n
Where:
PV = Present value
FV = Future value (the amount due on the promissory note)
r = Interest rate per period
n = Number of periods
Plugging in the values, we have:
PV = 12,000 / (1 + 0.09)^9
Using a calculator, we can compute the present value to be approximately $10,221.76.
Therefore, the plumber will receive around $10,221.76 when selling the promissory note to the bank.
Now, let's see if this amount is enough to pay a bill for $12,104. Unfortunately, the amount received by the plumber ($10,221.76) is less than the bill amount ($12,104). Therefore, it will not be enough to cover the bill, leaving a shortfall of $1,882.24.
In summary, the plumber will receive approximately $10,221.76 when selling the promissory note to the bank. However, this amount will not be sufficient to pay the bill for $12,104, resulting in a shortfall of $1,882.24.
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