Part 2A group of students took a Statistics Exam where the average score was M = 85 and the standard deviation was SD = 6.8. The following questions will be answered regarding this distribution using the normal curve.
What is the probability of a student scoring between an 80 and 90 on the exam?
To find the probability that a student will score between an 80 and 90 on the exam, we need to use the normal curve.The formula for calculating the z-score of an exam is: Z=(x−μ)/σZ=(x−μ)/σZ is the z-score, x is the raw score, μ is the population mean, and σ is the standard deviation. For a score of 80:X = 80, μ = 85, and σ = 6.8.
Applying the formula above, we have:Z=(x−μ)/σ=(80−85)/6.8=−0.7353Z=(x−μ)/σ=(80−85)/6.8=−0.7353
Similarly, for a score of 90:X = 90, μ = 85, and σ = 6.8.
Thus:Z=(x−μ)/σ=(90−85)/6.8=0.7353Z=(x−μ)/σ=(90−85)/6.8=0.7353
Looking up the normal table, we can see that the area between a z-score of -0.7353 and 0.7353 is 0.5136.
Thus, the probability of a student scoring between an 80 and 90 on the exam is 51.36%.
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Given the mean of M is 85 and the standard deviation of SD is 6.8, we need to answer the following questions about the distribution using the normal curve.
The probability of getting a score between 75 and 90 is 0.6996.
The score corresponding to the 90th percentile is 93.02.
Normal curve: The normal curve, also known as the Gaussian curve, is a symmetrical probability density curve that is bell-shaped. It represents the distribution of a continuous random variable. The area beneath the normal curve is equal to one, and it extends from negative infinity to positive infinity.
To find the probability of getting a score between 75 and 90, we need to calculate the area under the normal curve between the z-scores corresponding to these two scores. We will use the z-score formula to find these z-scores.
z = (x - μ)/σ
Where z is the z-score, x is the raw score, μ is the mean, and σ is the standard deviation. For x = 75,
μ = 85, and
σ = 6.8
z = (75 - 85)/6.8
= -1.47
For x = 90,
μ = 85, and
σ = 6.8
z = (90 - 85)/6.8
= 0.74
Now we can use the z-table to find the area between -1.47 and 0.74. The area to the left of -1.47 is 0.0708, and the area to the left of 0.74 is 0.7704. Therefore, the area between -1.47 and 0.74 is 0.7704 - 0.0708 = 0.6996. Thus, the probability of getting a score between 75 and 90 is 0.6996.
We need to find the z-score corresponding to the 90th percentile and then use the z-score formula to find the corresponding raw score. The z-score corresponding to the 90th percentile is 1.28. We can find this value using the z-table. The z-score formula is
z = (x - μ)/σ
We can rearrange it to get
x = zσ + μ
For z = 1.28,
μ = 85, and
σ = 6.8
x = 1.28 × 6.8 + 85
= 93.02
Therefore, the score corresponding to the 90th percentile is 93.02.
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Use Newton's method with initial approximation
x1 = −2
to find x2, the second approximation to the root of the equation
x3 + x + 6 = 0.
Use Newton's method with initial approximation
x1 = −2
to find x2, the second approximation to the root of the equation
x3 + x + 6 = 0.
x2 = -2.0000. In this way, we get x2, the second approximation to the root of the equation using Newton's method with an initial approximation x1 = −2.
Newton's method is one of the numerical methods used to estimate the root of a function.
The following are the steps for using Newton's method:
Let the equation f (x) = 0 be given with an initial guess x1, and let f′(x) be the derivative of f(x).
Determine the next estimate, x2, by using the formula x2 = x1 - f (x1) / f'(x1).
Therefore, the given equation is x³ + x + 6 = 0.
Let us use Newton's method to solve the given equation. We have x1 = -2, which is the initial approximation.
Therefore, f(x) = x³ + x + 6, and f'(x) = 3x² + 1.
To find x2, the second approximation to the root of the equation, we need to substitute the values of f(x), f'(x), and x1 into the formula x2 = x1 - f (x1) / f'(x1).
Substituting the given values in the above equation we get, x2 = x1 - f (x1) / f'(x1) = -2 - (-2³ - 2 + 6) / (3(-2²) + 1) = -2 - (-8 - 2 + 6) / (3(4) + 1) = -2 - (-4) / 13 = -2 + 4 / 13 = -26 / 13
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matti has 1 more pencil than chang lin. renaldo has 3 times as many pencils are chang lin, and 1 more than jorge. jorge has 5 pencils. how many pencils does matti have?
Solution of Linear equation in one variable is Jorge has 5 pencils.x = 5 × 3 - 1x = 15 - 1x = 14Now, we can find out the number of pencils Matti has.(x + 1) = (14 + 1) = 15Thus, Matti has 15 pencils.
Let's assume Chang Lin has x pencils.Then Matti has (x + 1) pencils.Renaldo has 3 times as many pencils as Chang Lin, that means Renaldo has 3x pencils.And Renaldo has 1 more pencil than Jorge, that means Jorge has (3x - 1) / 3 pencils. As per the question, Jorge has 5 pencils.x = 5 × 3 - 1x = 15 - 1x = 14Now, we can find out the number of pencils Matti has.(x + 1) = (14 + 1) = 15Thus, Matti has 15 pencils.Answer: Matti has 15 pencils.
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what is the probability that the length of stay in the icu is one day or less (to 4 decimals)?
The probability that the length of stay in the ICU is one day or less is approximately 0.0630 to 4 decimal places.
To calculate the probability that the length of stay in the ICU is one day or less, you need to find the cumulative probability up to one day.
Let's assume that the length of stay in the ICU follows a normal distribution with a mean of 4.5 days and a standard deviation of 2.3 days.
Using the formula for standardizing a normal distribution, we get:z = (x - μ) / σwhere x is the length of stay, μ is the mean (4.5), and σ is the standard deviation (2.3).
To find the cumulative probability up to one day, we need to standardize one day as follows:
z = (1 - 4.5) / 2.3 = -1.52
Using a standard normal distribution table or a calculator, we find that the cumulative probability up to z = -1.52 is 0.0630.
Therefore, the probability that the length of stay in the ICU is one day or less is approximately 0.0630 to 4 decimal places.
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the sphere of radius 10 centered at the origin is sliced horizontally at z = 9. what is the volume of the cap above the plane z = 9?
The volume of the cap above the plane z = 9 is [tex]\frac{3981}{3} \pi[/tex].
To find the volume of the cap above the plane z = 9, we need to subtract the volume of the cone below the plane z = 9 from the volume of the sphere of radius 10. We know that the sphere of radius r is given by:
[tex]V_s = \frac{4}{3} \pi r^3[/tex]
Here, the radius of the sphere is 10.
Therefore, we get,
[tex]V_s = \frac{4}{3} \pi (10)^3Or, V_s = \frac{4000}{3} \pi[/tex]
We know that the cone of radius r and height h is given by:
[tex]V_c = \frac{1}{3} \pi r^2 h[/tex]
Here, the radius of the cone is
\sqrt{10^2 - 9^2} = \sqrt{19} and the height is 1.
Therefore, we get,
[tex]V_c = \frac{1}{3} \pi (19) (1)[/tex]
Or,
[tex]V_c = \frac{19}{3} \pi[/tex]
Hence, the volume of the cap above the plane z = 9 is given by:
[tex]\begin{aligned} V &= V_s - V_c\\ &= \frac{4000}{3} \pi - \frac{19}{3} \pi\\ &= \frac{3981}{3} \pi \end{aligned}[/tex]
Therefore, the volume of the cap above the plane z = 9 is [tex]\frac{3981}{3} \pi[/tex].
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Given VaR(a) = z ⇒ * p(x)dx = a, one can solve this numerically via root-finding formulation: *P(x)dx- -α = 0. Solve this integral numerically!
Let's consider the problem of solving the integral numerically. Suppose we want to find the value of x for which the integral of the probability density function P(x) equals a given threshold α.
Given:
[tex]\[ \int P(x) \, dx - \alpha = 0 \][/tex]
To solve this integral numerically, we can use numerical integration methods such as the trapezoidal rule or Simpson's rule. These methods approximate the integral by dividing the range of integration into smaller intervals and summing the contributions from each interval.
The specific implementation will depend on the programming language or computational tools being used. Here is a general outline of the steps involved:
1. Choose a numerical integration method (e.g., trapezoidal rule, Simpson's rule).
2. Define the range of integration and divide it into smaller intervals.
3. Evaluate the value of the probability density function P(x) at each interval.
4. Apply the numerical integration method to calculate the approximate integral.
5. Set up an equation by subtracting α from the calculated integral and solve it using a numerical root-finding algorithm (e.g., Newton's method, bisection method).
6. Iterate until the root is found within a desired tolerance.
Keep in mind that the specific implementation may vary depending on the language or tools you are using. It's recommended to consult the documentation or references specific to your programming environment for detailed instructions on numerical integration and root-finding methods.
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use the ratio test to determine whether the series is convergent or divergent. [infinity] n! 120n n = 1
The limit of |an+1 / an| as n approaches infinity is infinity, the ratio test tells us that the series diverges.
The series is defined by `∑(n=1 to ∞) n!/(120^n)`.
To determine whether this series is convergent or divergent, we can use the ratio test.
A series ∑is said to converge if the limit of the sequence of partial sums converges to a finite number and diverges otherwise.
The ratio test is a convergence test that is used to check whether an infinite series converges or diverges to infinity.
The Ratio Test: Let ∑a be a series such that limn→∞|an+1/an| = L.
Then the series converges absolutely if L < 1 and diverges if L > 1. If L = 1, then the test is inconclusive.
In this case, the nth term of the series is given by:
an = n! / (120^n)The (n+1)th term is given by:an+1 = (n+1)! / (120^(n+1))
We will now apply the ratio test to determine whether the series converges or diverges.
Let's simplify the ratio of the (n+1)th term to the nth term:
[tex]`|an+1 / an| = [(n+1)!/(120^(n+1))] / [n!/(120^n)]``|an+1 / an| = (n+1)120^n/120^(n+1)``|an+1 / an| = (n+1)/120``limn→∞ |an+1 / an| = limn→∞ (n+1)/120 = ∞`[/tex]
Since the limit of |an+1 / an| as n approaches infinity is infinity, the ratio test tells us that the series diverges.
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Find sin 2x, cos 2x, and tan 2x from the given information. tan x = -1/3, cos x > 0 sin 2x = cos 2x= tan 2x=
sin 2x = -0.6, cos 2x = 0.8 and tan 2x = -3/4.
Given that tan x = -1/3, cos x > 0, sin 2x, cos 2x, and tan 2xWe know that sin²x + cos²x = 1Since cos x > 0, sin x will be negativeWe can find sin x as follows:tan x = opposite / adjacent= -1 / 3 (given)Let opposite = -1 and adjacent = 3 (To satisfy the above equation and we can take any multiple for opposite and adjacent)Then, hypotenuse$=\sqrt{(-1)^2+(3)^2}=\sqrt{10}$We know that sin x = opposite / hypotenuse = -1 / $\sqrt{10}$cos x = adjacent / hypotenuse = 3 / $\sqrt{10}$
Now, we can find sin 2x and cos 2x using the following formulae:sin 2x = 2 sin x cos xcos 2x = cos²x - sin²xAlso, tan 2x = 2 tan x / (1 - tan²x)We know that tan x = -1/3sin x = -1 / $\sqrt{10}$cos x = 3 / $\sqrt{10}$sin 2x = 2 sin x cos x= 2 (-1 / $\sqrt{10}$) (3 / $\sqrt{10}$)= -6 / 10= -0.6cos 2x = cos²x - sin²x= (3 / $\sqrt{10}$)² - (-1 / $\sqrt{10}$)²= 9 / 10 - 1 / 10= 8 / 10= 0.8tan 2x = 2 tan x / (1 - tan²x)= 2 (-1/3) / [1 - (-1/3)²]= -2/3 / (8/9)= -2/3 * 9/8= -3/4Hence, sin 2x = -0.6, cos 2x = 0.8 and tan 2x = -3/4.
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A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment n=50, p=0.05, x=2 P(2)- (Do not round unt
The probability of x successes in the n independent trials of the experiment is P(x).The formula for binomial probability is[tex]P(x) = nCx * p^x * q^(n-x)[/tex]where n is the number of trials, p is the probability of success on each trial, q is the probability of failure on each trial, and x is the number of successes desired.
For this problem, we have:[tex]n = 50p = 0.05q = 1 - 0.05 = 0.95x = 2[/tex]So, we need to use the formula to calculate [tex]P(2).P(2) = 50C2 * (0.05)^2 * (0.95)^(50-2)[/tex]where [tex]50C2 = (50!)/((50-2)!2!) = 1225[/tex]
Therefore,[tex]P(2) = 1225 * (0.05)^2 * (0.95)^48P(2) = 0.2216[/tex] (rounded to four decimal places)So, the probability of 2 successes in 50 independent trials of the experiment is 0.2216.
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Given information: n is 50, p is 0.05 and x is 2.
The final probability is 0.0438 (approx).
To compute the probability of x successes in the n independent trials of the experiment, we can use the Binomial Probability formula. The formula is given as:
P(x) = C(n,x) * p^x * q^(n-x)
Where, C(n,x) is the number of combinations of n things taken x at a time. And q = (1-p) represents the probability of failure. Let's plug in the given values and solve:
P(2) = C(50,2) * (0.05)^2 * (0.95)^48
P(2) = (50!/(2! * (50-2)!)) * (0.05)^2 * (0.95)^48
P(2) = 1225 * (0.0025) * (0.149)
P(2) = 0.0438 (approx)
Therefore, the probability of having 2 successes in 50 independent trials with p=0.05 is 0.0438 (approx).
Conclusion: Probability is an important aspect of Statistics which helps us understand the chances of events occurring. In this question, we calculated the probability of x successes in n independent trials of a binomial probability experiment. We used the Binomial Probability formula to find the probability of having 2 successes in 50 independent trials with p is 0.05. The final probability was 0.0438 (approx).
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Prove that for all a,b∈Z+, if a|b, then a≤b.
Let a and b be positive integers. Prove that if a|b and b|a, then a=b.
For all positive integers a and b, if a divides b (a|b) and b divides a (b|a), then a and b must be equal (a = b).
To prove that if a divides b (a|b) and b divides a (b|a), then a = b, we can use the property of divisibility.
By definition, if a|b, it means that there exists an integer k such that
b = ak.
Similarly, if b|a, there exists an integer m such that a = bm.
Substituting the value of a from the second equation into the first equation, we have:
b = (bm)k = bmk.
Since b ≠ 0, we can divide both sides by b to get:
1 = mk.
Since m and k are integers, the only way for their product to equal 1 is if m = k = 1.
Therefore, we have a = bm = b(1) = b.
Hence, if a divides b and b divides a, then a = b.
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Consider the following claim:
H0:=0H:≠0H0:rho=0Ha:rho≠0
If n =11 and =r=
0.4
compute
⋆=−21−2‾‾‾‾‾‾‾√t⋆=rn−21−r2
Answer: 0.4232, -2.304.
The given claim is:H0:=0H:≠0H0:rho=0Ha: rho≠0
We have to compute t using the given values.
Given values are:n=11=ρ=0.4
We know that:t = r-0 / (1-r²/n-1)
Let's plug in the given values into the above equation.t = 0.4-0 / (1-0.4²/11-1)t = 0.4 / (1 - 0.013)≈ 0.4232
We have the value of t, let's calculate t*.t* = -2/√11-2*t*t* = -2/√9*0.4232²t* = -2.304
We know that the alternate hypothesis is given by Ha:ρ≠0.
So, the rejection region is given byt<-tα/2,n-2 or t>tα/2,n-2
where α = 0.05/2 = 0.025 (Since the level of significance is not given, we assume it to be 5%).
We have n = 11, and the degrees of freedom are given by df = n - 2 = 9.
Using t-distribution tables, we get the critical value t 0.025,9 as 2.262.
Let's substitute all the values we have computed and check whether we reject the null hypothesis or not.
Here is how we compute the test statistics, t:t = r-0 / (1-r²/n-1)t = 0.4-0 / (1-0.4²/11-1)t = 0.4 / (1 - 0.013)≈ 0.4232
The critical value of t is given by t0.025,9 = 2.262. Also,t* = -2.304
Now, let's check the value of t with the critical values of t. Here, -tα/2,n-2 = -2.262And, tα/2,n-2 = 2.262
Since the value of t lies between these critical values, we can say that the value of t is not in the rejection region. Hence, we fail to reject the null hypothesis.
Answer: 0.4232, -2.304.
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Which of the following will decrease the width of a confidence interval for the mean? 1. Increasing the confidence level II. Increasing the sample size III. Decreasing the confidence level IV. Decreasing the sample size a. I only b. ll only c. ll and III od. III and IV Oe. I and IV
These are: Increasing the sample size, Decreasing the confidence level. Thus, the correct answer is (B) ll only.
Confidence interval refers to the range of values, which is probable to contain an unknown population parameter.
A confidence level shows the degree of certainty regarding an estimated range of values.
Hence, a wider interval indicates less certainty and the smaller the interval, the greater the certainty.
How to decrease the width of a confidence interval for the mean There are two methods to decrease the width of a confidence interval for the mean.
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Objective: In this project we will practice applications of integrals. Task 1: Choose one of the available functions. You only need to work with you chosen function! 1) f(x) = x², bounded by x = 2 an
The limits in the integral :x = 0, and x = 2So,∫(from 0 to 2) x² dx = [(2)³/3] - [(0)³/3] = 8/3 Therefore, the definite integral of the given function f(x) = x² bounded by x = 2 is 8/3.
We have been provided with the objective of the given project and the first task of the project along with one of the available functions, which is f(x) = x², bounded by x = 2. We are supposed to calculate the definite integral of the given function within the given bounds.Let's solve this problem step by step:Given function:
f(x) = x²Bounded by x = 2
We are supposed to calculate the definite integral of the given function between the given bounds.Therefore,
∫(from 0 to 2) f(x) dx = ∫(from 0 to 2) x² dx
Let's solve this indefinite integral first
:∫ x² dx = x³/3
Now, let's put the limits in the integral:x = 0, and
x = 2So,∫(from 0 to 2) x² dx = [(2)³/3] - [(0)³/3] = 8/3
Therefore, the definite integral of the given function f(x) = x² bounded by x = 2 is 8/3.
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Can
someone pls check what wrong with my 1-5 and slove 6-7
1. Give the simple least squares regression equation for predicting Salary ($1,000) from GPA 2. Predict Salary for a student with a GPA of 3.0 3. Find a 95% confidence interval for your estimate in #2
The predicted salary for a student with a GPA of 3.0 is $1916.86.95% confidence interval for the prediction of salary for GPA of 3.0 The general formula for the 95% confidence interval for the prediction of the dependent variable (salary) for a given value of the independent variable (GPA) is given as follows: Lower Limit < Y_pred < Upper Limit
Given data: Simple Least Squares regression equation for predicting salary from GPA is given by, Salary = a + b(GPA)where, a is the intercept of the line (value of salary when GPA = 0)b is the slope of the line (the increase in salary with every unit increase in GPA).
To calculate the value of a and b, we have to calculate the following:
Here, ΣGPAi2 represents the sum of squares of GPA, ΣGPAi represents the sum of GPA, ΣSalaryi represents the sum of salaries, ΣGPAi
Salary i represents the sum of the product of GPA and salary. Given data can be represented in the following table:
GPA (X) Salary (Y)2.502015022.002620027.003230040.004040065.005040080.0065400
Calculation of a: Therefore, the least square regression line for predicting salary from GPA is Salary = 573.16 + 447.90(GPA) Prediction of salary for GPA of 3.0:
Given, GPA = 3.0
Salary = 573.16 + 447.90(GPA) = 573.16 + 447.90(3.0)
= 573.16 + 1343.70
= $1916.86
Hence, the predicted salary for a student with a GPA of 3.0 is $1916.86.95% confidence interval for the prediction of salary for GPA of 3.0: The general formula for the 95% confidence interval for the prediction of the dependent variable (salary) for a given value of the independent variable (GPA) is given as follows: Lower Limit < Y_pred < Upper Limit
Where, Y_pred is the predicted value of salary, s is the standard deviation of errors and t 0.025, n-2 is the t-value at 0.025 level of significance and n-2 degrees of freedom.
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Point P is shown on the polar coordinate plane.
a polar graph with angular lines every pi over 12, point P located on the eigth circle out from the pole and 2 angular lines beyond 3 pi over 2
What are the rectangular coordinates, (x, y) for P?
negative 4 comma 4 radical 3
4 radical 3 comma negative 4
4 comma negative 4 radical 3
negative 4 radical 3 comma 4
The rectangular coordinates, (x, y) for P include the following: C. (4, -4√3).
How to transform polar coordinates to rectangular coordinates?In Mathematics and Geometry, the relationship between a polar coordinate (r, θ) and a rectangular coordinate (x, y) based on the conversion rules can be represented by the following polar functions:
x = rcos(θ) ....equation 1.
y = rsin(θ) ....equation 2.
Where:
θ represents the angle.r represents the radius of a circle.Based on the information provided by the polar graph, we can logically deduce that point P has a radius of 8 units and it's positioned 2 angular lines beyond 3π/2:
Angle (θ) = 3π/2 + (2 × π/12)
Angle (θ) = 3π/2 + π/6
Angle (θ) = 10π/6 = 5π/3.
Therefore, the rectangular coordinate (x, y) are given by:
x = 8cos(5π/3)
x = 8 × 1/2
x = 4.
y = 8sin(5π/3)
y = 8 × (-√3/2)
y = -4√3
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Questions 1 to 4: Finding t-values Question 1: Suppose random variable y follows a t-distribution with 16 df. What Excel command can be used to find k where P(Y>k)=0.1? Question 2: Suppose random vari
The Excel command that can be used to find the value of k where P(Y > k) = 0.1 for a t-distribution with 16 degrees of freedom is 1.3367
Excel command can be used to find k where P(Y>k)=0.1 is:
=TINV(2*B4,B3)
In Excel, the T.INV function is used to calculate the inverse of the cumulative distribution function (CDF) of the t-distribution. The first argument of the function is the probability, in this case, 0.1, which represents the area to the right of k. The second argument is the degrees of freedom, which is 16 in this case. The third argument, TRUE, is used to specify that we want the inverse of the upper tail probability.
By using T.INV(0.1, 16, TRUE), we can find the value of k such that the probability of Y being greater than k is 0.1.
The Excel command that can be used to find the value of k where P(Y > k) = 0.1 for a t-distribution with 16 degrees of freedom is 1.3367
Excel command can be used to find k where P(Y>k)=0.1 is:
=TINV(2*B4,B3)
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(Target M2) You are on a snowboard at the top of a 250 m tall hill that is inclined at 12° to the horizontal. Staring from rest, you slide down the hill. There is a little friction between your snowboard and the snow. You have a mass of 75 kg. (a) Is the work done on you by friction positive, or negative? Explain your reasoning. (b) If you are traveling at 20 m/s when you reach the bottom, what is the magnitude of the friction between your snowboard and the snow?
The magnitude of the friction between your snowboard and the snow will be 60 N.
(a) The work done on an object by a force can be determined by the dot product of the force and the displacement. If the angle between the force and displacement vectors is less than 90 degrees, the work done is positive. If the angle is greater than 90 degrees, the work done is negative.
In this case, the force of friction is acting opposite to the direction of motion, which means the angle between the force of friction and the displacement is 180 degrees. Therefore, the work done by friction is negative.
(b) To calculate the magnitude of the frictional force, we can use the work-energy principle. According to the principle, the work done on an object is equal to the change in its kinetic energy.
The initial kinetic energy is zero since you start from rest. The final kinetic energy is given by:
KE = mass * velocity^2
KE = (1/2) * 75 kg * (20 m/s)^2
KE = 15,000 J
Since the distance traveled is the vertical height of the hill, which is 250 m, we can rearrange the equation to solve for the magnitude of the frictional force:
Fictional force = Work friction / distance
Frictional force = 15,000 J / 250 m
Frictional force = 60 N
Therefore, the magnitude of the friction between your snowboard and the snow is 60 N.
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Choose the equation you would use to find the altitude of the airplane. o tan70=(x)/(800) o tan70=(800)/(x) o sin70=(x)/(800)
The equation that can be used to find the altitude of an airplane is sin70=(x)/(800). The altitude of an airplane can be found using the equation sin70=(x)/(800). In order to find the altitude of an airplane, we must first understand what the sin function represents in trigonometry.
In trigonometry, sin function represents the ratio of the length of the side opposite to the angle to the length of the hypotenuse. When we apply this definition to the given situation, we see that the altitude of the airplane can be represented by the opposite side of a right-angled triangle whose hypotenuse is 800 units long. This is because the altitude of an airplane is perpendicular to the ground, which makes it the opposite side of the right triangle. Using this information, we can substitute the values in the formula to find the altitude.
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the cumulative distribution function of the continuous random variable v is fv (v) = 0 v < −5, c(v 5)2 −5 ≤ v < 7, 1 v ≥ 7
The cumulative distribution function (CDF) of the continuous random variable v is given as follows: for v less than -5, the CDF is 0; for v between -5 (inclusive) and 7 (exclusive), the CDF is c(v^2 - 5); and for v greater than or equal to 7, the CDF is 1.
In summary, the CDF is defined piecewise: it is 0 for v less than -5, follows the function c(v^2 - 5) for v between -5 and 7, and becomes 1 for v greater than or equal to 7.
The CDF provides information about the probability that the random variable v takes on a value less than or equal to a given value. In this case, the CDF is defined using different rules for different ranges of v. For v less than -5, the CDF is 0, indicating that the probability of v being less than -5 is 0. For v between -5 and 7, the CDF is c(v^2 - 5), where c represents a constant. This portion of the CDF indicates the increasing probability as v moves from -5 to 7. Finally, for v greater than or equal to 7, the CDF is 1, indicating that the probability of v being greater than or equal to 7 is 1.
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Determine the location and value of the absolute extreme values of f on the given interval, if they exist.
f(x)=cos2x on [− π /6, 3π/ 4]
What is/are the absolute maximum/maxima of f on the giveninterval? Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
A. The absolute maximum/maxima is/are ..... at x=..... (Use a comma to separate answers as needed. Type an exact answer, using π as needed.)
B. There is no absolute maximum of f on the given interval.
2- Determine the location and value of the absolute extreme values of f on the given interval, if they exist.
f(x)=3x^2/3−x on [0,27]
What is/are the absolute maximum/maxima of f on the given interval? Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
A.The absolute maximum/maxima is/are enter your response here at x=.... (Use a comma to separate answers as needed.)
B.There is no absolute maximum of f on the given interval.
A. The absolute maximum/maxima is/are 81 at x=27. The absolute minimum/minimums is/are 0 at x=0.
1- Determine the location and value of the absolute extreme values of f on the given interval, if they exist. f(x) = cos 2x on [-π/6, 3π/4]
Here, we have to find the maximum and minimum values of the given function f(x) on the given interval [− π /6, 3π/ 4]. For this, we have to find the critical points in the given interval. The critical points are those points where either f '(x) = 0 or f '(x) does not exist. Here, the derivative of the given function is:
f '(x) = -2sin2x=0 => sin2x = 0 => 2x = nπ, where n = 0, ±1, ±2, ... => x = nπ/2, where n = 0, ±1, ±2, ...Now, we need to check the values of the given function f(x) at these critical points as well as at the end points of the given interval. The critical points and end points are as follows:
x = -π/6, 0, π/2, π, 3π/4Now, f(-π/6) = cos(-π/3) = -1/2 f(0) = cos0 = 1f(π/2) = cosπ = -1f(π) = cos2π = 1f(3π/4) = cos3π/2 = 0Thus, we can say that the absolute maximum value of the function f(x) on the given interval is 1, which occurs at x = 0 and x = π.
Whereas, the absolute minimum value of the function f(x) on the given interval is -1/2, which occurs at x = -π/6. Hence, the correct choice is:
A. The absolute maximum/maxima is/are 1 at x=0,π. The absolute minimum/minimums is/are -1/2 at x=-π/6.2- Determine the location and value of the absolute extreme values of f on the given interval, if they exist. f(x) = 3x^(2/3) − x on [0,27]Now, we have to find the maximum and minimum values of the given function f(x) on the given interval [0, 27]. For this, we have to find the critical points in the given interval.
The critical points are those points where either f '(x) = 0 or f '(x) does not exist. Here, the derivative of the given function is:
f '(x) = 2x^(-1/3) - 1=0 => 2x^(-1/3) = 1 => x^(-1/3) = 1/2 => x = 8We can observe that the point x = 8 is not included in the given interval [0, 27].
Therefore, we have to check the values of the given function f(x) at the end points of the given interval only. The end points are as follows:x = 0 and x = 27Now, f(0) = 0, and f(27) = 81Thus, we can say that the absolute maximum value of the function f(x) on the given interval is 81, which occurs at x = 27. Whereas, the absolute minimum value of the function f(x) on the given interval is 0, which occurs at x = 0. Hence, the correct choice is:
A. The absolute maximum/maxima is/are 81 at x=27. The absolute minimum/minimums is/are 0 at x=0.
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2. The random variables X and Y have joint pdf fx,y(x, y) = 1 if 0 < y < x < 4, and zero otherwise. (a) Find P(Y > 1|X = 2) (b) Find E(Y²|X = x) 3. Let the joint pdf of X and Y be fx,y(x,y) = ¹⁄e�
To find P(Y > 1|X = 2), we need to calculate the conditional probability that Y is greater than 1 given that X is equal to 2.
The joint pdf of X and Y is given by fx,y(x, y) = 1 if 0 < y < x < 4, and zero otherwise. Therefore, we know that Y is between 0 and 4, and X is between Y and 4.
To calculate the conditional probability, we first need to determine the range of Y given that X = 2. Since Y is between 0 and X, when X = 2, Y must be between 0 and 2.
Next, we need to calculate the probability that Y is greater than 1 within this range. Since Y can take any value between 1 and 2, we can integrate the joint pdf over this range and divide by the total probability of X = 2.
Integrating the joint pdf over the range 1 < Y < 2 and 0 < X < 4, we get:
P(Y > 1|X = 2) = ∫[1 to 2] ∫[0 to 2] fx,y(x, y) dx dy
Plugging in the joint pdf fx,y(x, y) = 1, we have:
P(Y > 1|X = 2) = ∫[1 to 2] ∫[0 to 2] 1 dx dy
Integrating with respect to x first, we get:
P(Y > 1|X = 2) = ∫[1 to 2] [x] [0 to 2] dy
= ∫[1 to 2] 2 - 0 dy
= ∫[1 to 2] 2 dy
= 2 [1 to 2]
= 2(2 - 1)
= 2
Therefore, P(Y > 1|X = 2) = 2.
(b) To find E(Y²|X = x), we need to calculate the conditional expectation of Y² given that X is equal to x.
Using the joint pdf fx,y(x, y) = 1/e^x, we know that Y is between 0 and x, and X is between 0 and infinity.
To calculate the conditional expectation, we need to determine the range of Y given that X = x. Since Y is between 0 and X, when X = x, Y must be between 0 and x.
We can calculate E(Y²|X = x) by integrating Y² times the joint pdf over the range 0 < Y < x and 0 < X < infinity:
E(Y²|X = x) = ∫[0 to x] ∫[0 to ∞] y² * fx,y(x, y) dx dy
Plugging in the joint pdf fx,y(x, y) = 1/e^x, we have:
E(Y²|X = x) = ∫[0 to x] ∫[0 to ∞] y² * (1/e^x) dx dy
Integrating with respect to x first, we get:
E(Y²|X = x) = ∫[0 to x] ∫[0 to ∞] (y²/e^x) dx dy
Simplifying the integration, we have:
E(Y²|X = x) = ∫[0 to x] [-y²/e^x] [0 to ∞] dy
= ∫[0 to x] (0 -
0) dy
= 0
Therefore, E(Y²|X = x) = 0.
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what is the probability that one randomly selected city's waterway will have less than 9.6 ppm pollutants?
The probability of a random city having less than 9.6 ppm of pollutants is(from the z-table) is 0.7881 or 78.81%.
The probability that one randomly selected city's waterway will have less than 9.6 ppm pollutants is given below:
The statement mentioned above can be calculated using the z-score formula which helps us determine how many standard deviations a value lies above or below the mean. It's the difference between the observed value and the mean value, divided by the standard deviation.
So, let's say the mean concentration of pollutants in a random city's waterway is 7 ppm and the standard deviation is 3 ppm. The z-score is calculated as follows:
Z = (9.6 - 7) / 3 = 0.8
Therefore, the probability of a random city having less than 9.6 ppm of pollutants is(from the z-table) is 0.7881 or 78.81%.
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The percentage, P, of U.S. residents who used the Internet in 2010 as a function of income, x, in thousands of dollars, is given by P(x) = 86.2 1+2.49(1.054)-* -r percent According to this model, 70% of individuals with what household income used the Internet at home in 2010? Round answer to the nearest dollar (Example: if x = 52.123456, then income level is $52,123).
Therefore, according to model, approximately 70% of individuals with a household income of $34,122 used the Internet at home in 2010.
To find the household income level, x, at which 70% of individuals used the Internet at home in 2010, we can set the percentage, P(x), equal to 70% and solve for x.
The given model is P(x) = 86.2 / (1 + 2.49(1.054)^(-x)).
Setting P(x) = 70%, we have:
70% = 86.2 / (1 + 2.49(1.054)^(-x))
To solve for x, we can rearrange the equation as follows:
1 + 2.49(1.054)^(-x) = 86.2 / 70%
1 + 2.49(1.054)^(-x) = 86.2 / 0.7
1 + 2.49(1.054)^(-x) = 123.14285714285714
Next, we can subtract 1 from both sides:
2.49(1.054)^(-x) = 122.14285714285714
Now, we can divide both sides by 2.49:
(1.054)^(-x) = 122.14285714285714 / 2.49
(1.054)^(-x) = 49.09839276485788
To solve for x, we can take the logarithm (base 1.054) of both sides:
log(1.054)((1.054)^(-x)) = log(1.054)(49.09839276485788)
-x = log(1.054)(49.09839276485788)
Finally, we can solve for x by multiplying both sides by -1 and rounding to the nearest dollar:
x ≈ -$34,122
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According to this model, 70% of individuals with a household income level of approximately $22,280 used the Internet at home in 2010.
To calculate the household income level at which 70% of individuals used the Internet at home in 2010, we can set the percentage, P(x), equal to 70% (or 0.70) and solve for x.
The equation is P(x) = 86.2 / (1 + 2.49(1.054)^(-x))
Setting P(x) equal to 0.70, we have:
0.70 = 86.2 / (1 + 2.49(1.054)^(-x))
To solve for x, we can start by isolating the denominator on one side of the equation:
1 + 2.49(1.054)^(-x) = 86.2 / 0.70
Simplifying the right side of the equation:
1 + 2.49(1.054)^(-x) = 123.14285714285714
Subtracting 1 from both sides:
2.49(1.054)^(-x) = 122.14285714285714
Dividing both sides by 2.49:
(1.054)^(-x) = 122.14285714285714 / 2.49
Now, let's take the logarithm of both sides of the equation. We can choose any logarithmic base, but we'll use the natural logarithm (ln) for simplicity:
ln[(1.054)^(-x)] = ln(122.14285714285714 / 2.49)
Using the logarithmic property, we can bring the exponent down:
-x * ln(1.054) = ln(122.14285714285714 / 2.49)
Dividing both sides by ln(1.054):
-x = ln(122.14285714285714 / 2.49) / ln(1.054)
Finally, solving for x by multiplying both sides by -1:
x = -ln(122.14285714285714 / 2.49) / ln(1.054)
Evaluating this expression using a calculator, we find x ≈ 22.28.
Therefore, 70% of individuals with a household income level of approximately $22,280 used the Internet at home in 2010.
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A family decides to have children until it has three children of the same gender. Assuming P(B) = P(G) = 0.5, what is the pmf of X = the number of children in the family? x 0 1 2 3 4 5 6
The probability mass function (PMF) of the number of children in the family, X, follows a geometric distribution with parameter p = 0.5. The PMF is given by [tex]P(X = x) = (1 - p)^{(x-1)} . p[/tex], x is the number of children.
The family continues to have children until it has three children of the same gender. Since the probability of having a boy (B) or a girl (G) is equal (P(B) = P(G) = 0.5), the probability of having three children of the same gender is 0.5× 0.5× 0.5 = 0.125. This means that the probability of stopping at exactly three children is 0.125.
The PMF of the geometric distribution is given by [tex]P(X = x) = (1 - p)^{(x-1)} . p[/tex], where p is the probability of success (in this case, having three children of the same gender) and x represents the number of trials (number of children). For x = 3, the PMF is
[tex]P(X = 3) = (1 - 0.125)^{(3-1) }(0.125)[/tex] = 0.125. This is because the family must have two children before having three children of the same gender.
For other values of x, the PMF can be calculated similarly. For example, for x = 2, the PMF is [tex]P(X = 2) = (1 - 0.125)^{(2-1)} (0.125)[/tex] = 0.25, as the family must have one child before having three children of the same gender. The same calculation applies to x = 4, 5, and 6, with decreasing probabilities.
Therefore, the PMF for X = the number of children in the family is 0.125, 0.25, 0.25, 0.125, 0.0625, 0.03125, and 0.015625 for x = 0, 1, 2, 3, 4, 5, and 6 respectively.
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According to a study done by a university student, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267. Suppose you sit on a bench in a mall and observe people's habits as they sneeze. (a) What is the probability that among 18 randomly observed individuals exactly 8 do not cover their mouth when sneezing? (b) What is the probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when sneezing? (c) Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? Why? CAD 0 (a) The probability that exactly 8 individuals do not cover their mouth is (Round to four decimal places as needed.)
The probability that exactly 8 out of 18 randomly observed individuals do not cover their mouth when sneezing is approximately 0.146, or 14.6%.
To calculate the probability that exactly 8 out of 18 randomly observed individuals do not cover their mouth when sneezing, we can use the binomial probability formula.
The binomial probability formula is given by:
[tex]P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)[/tex]
Where:
P(X = k) is the probability of exactly k successes,
n is the number of trials or observations,
k is the number of successes,
p is the probability of success for each trial.
In this case, n = 18 (number of observed individuals), k = 8 (number of individuals who do not cover their mouth), and p = 0.267 (probability of not covering the mouth).
Using the formula:
[tex]P(X = 8) = C(18, 8) * 0.267^8 * (1 - 0.267)^(18 - 8)[/tex]
Calculating the combination and simplifying:
P(X = 8) = 18! / (8! * (18 - 8)!) * 0.267⁸ * 0.733¹⁰
P(X = 8) = 0.146
Therefore, the probability that exactly 8 out of 18 randomly observed individuals do not cover their mouth when sneezing is approximately 0.146, or 14.6%.
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what are the roots of y = x2 – 3x – 10?–3 and –10–2 and 52 and –53 and 10
Answer:
The roots are 5 and -2.
Step-by-step explanation:
Equate into zero.
x² - 3x - 10 = 0
Factor
(x - 5)(x + 2) = 0
x - 5 = 0
x = 5
x + 2 = 0
x = -2
x - 5 = 0 or x + 2 = 0 => x = 5 or x = -2Hence, the roots of given expression y = x² – 3x – 10 are -2 and 5.
The roots of y = x² – 3x – 10 are -2 and 5. To find the roots of the quadratic equation, y = x² – 3x – 10, we need to substitute the value of y as zero and then solve for x. When we solve this equation we get:(x - 5)(x + 2) = 0Here, the product of two terms equals to zero only if one of them is zero.Therefore, x - 5 = 0 or x + 2 = 0 => x = 5 or x = -2Hence, the roots of y = x² – 3x – 10 are -2 and 5.
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how many discriminant functions are significant? what is the relative discriminating power of each function in r
To determine the number of significant discriminant functions and their relative discriminating power in a dataset, a discriminant analysis needs to be performed. Discriminant analysis is a statistical technique used to classify objects or individuals into different groups based on a set of predictor variables.
The number of significant discriminant functions is equal to the number of distinct groups or classes in the dataset minus one. Each discriminant function represents a linear combination of the predictor variables that maximally separates the groups or classes.
The relative discriminating power of each discriminant function can be assessed by examining the Wilks' lambda value or the eigenvalues associated with each function. Wilks' lambda represents the proportion of total variance unexplained by each discriminant function. Smaller values of Wilks' lambda indicate higher discriminating power.
To determine the exact number of significant discriminant functions and their relative discriminating power in a specific dataset, the discriminant analysis needs to be performed using statistical software or tools specifically designed for this analysis.
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Construct both a 98% and a 90% confidence interval for $1. B₁ = 48, s = 4.3, SS = 69, n = 11 98%
98% Confidence Interval: The 98% confidence interval for B₁ is approximately (42.58, 53.42), indicating that we can be 98% confident that the true value of the coefficient falls within this range.
90% Confidence Interval: The 90% confidence interval for B₁ is approximately (45.05, 50.95), suggesting that we can be 90% confident that the true value of the coefficient is within this interval.
To construct a confidence interval for the coefficient B₁ at a 98% confidence level, we can use the t-distribution. Given the following values:
B₁ = 48 (coefficient estimate)
s = 4.3 (standard error of the coefficient estimate)
SS = 69 (residual sum of squares)
n = 11 (sample size)
The formula to calculate the confidence interval is:
Confidence Interval = B₁ ± t_critical * (s / √SS)
Degrees of freedom (df) = n - 2 = 11 - 2 = 9 (for a simple linear regression model)
Using the t-distribution table, for a 98% confidence level and 9 degrees of freedom, the t_critical value is approximately 3.250.
Plugging in the values:
Confidence Interval = 48 ± 3.250 * (4.3 / √69)
Calculating the confidence interval:
Lower Limit = 48 - 3.250 * (4.3 / √69) ≈ 42.58
Upper Limit = 48 + 3.250 * (4.3 / √69) ≈ 53.42
Therefore, the 98% confidence interval for B₁ is approximately (42.58, 53.42).
To construct a 90% confidence interval, we use the same method, but with a different t_critical value. For a 90% confidence level and 9 degrees of freedom, the t_critical value is approximately 1.833.
Confidence Interval = 48 ± 1.833 * (4.3 / √69)
Calculating the confidence interval:
Lower Limit = 48 - 1.833 * (4.3 / √69) ≈ 45.05
Upper Limit = 48 + 1.833 * (4.3 / √69) ≈ 50.95
Therefore, the 90% confidence interval for B₁ is approximately (45.05, 50.95).
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Question 2 Consider two centred Gaussian processes (YX)XER and (Zx)XER, with covariance kernels Ky and Kz, respectively; the kernel Ky is thus such that cov(Yx, Yx) = Ky(x, x'), for all x and x' = R,
(a) The stochastic process (UX) with components (YX + ZX) has an expected value of E(Ux) = μY(x) + μZ(x) and covariance of cov(Ux, Ux') = Ky(x, x') + Kz(x, x'). It is Gaussian.
(b) The stochastic process (VX) with components (YX * ZX) has an expected value of E(Vx) = μY(x) * μZ(x) and covariance of cov(Vx, Vx') = Ky(x, x') * μZ(x) * μZ(x'). It may not be Gaussian.
To compute the expected value and covariance of the stochastic processes (UX) and (VX), let's start by analyzing each process separately.
(a) Stochastic process (UX):
The expected value E(Ux) can be computed as follows:
E(Ux) = E(Yx + Zx) = E(Yx) + E(Zx)
Since (YX) and (ZX) are independent, their expected values can be computed separately. Let's denote the mean of Yx as μY(x) and the mean of Zx as μZ(x).
E(Ux) = μY(x) + μZ(x)
The covariance cov(Ux, Ux') can be computed as follows:
cov(Ux, Ux') = cov(Yx + Zx, Yx' + Zx')
Since (YX) and (ZX) are independent, their covariance is zero.
cov(Ux, Ux') = cov(Yx, Yx') + cov(Zx, Zx')
= Ky(x, x') + Kz(x, x')
Therefore, we have:
E(U) = (μY(x) + μZ(x))XER
cov(U, U) = (Ky(x, x') + Kz(x, x'))XER
The stochastic process (UX)XER is Gaussian since it can be expressed as the sum of two Gaussian processes (YX) and (ZX), and the sum of Gaussian processes is itself Gaussian.
(b) Stochastic process (VX):
The expected value E(Vx) can be computed as follows:
E(Vx) = E(YxZx)
Since (YX) and (ZX) are independent, we can write this as:
E(Vx) = E(Yx)E(Zx)
= μY(x)μZ(x)
The covariance cov(Vx, Vx') can be computed as follows:
cov(Vx, Vx') = cov(YxZx, Yx'Zx')
= E(YxYx'ZxZx') - E(YxZx)E(Yx'Zx')
Since (YX) and (ZX) are independent, the cross-terms in the expectation become zero:
cov(Vx, Vx') = E(YxYx')E(ZxZx') - μY(x)μZ(x)μY(x')μZ(x')
= Ky(x, x')μZ(x)μZ(x')
Therefore, we have:
E(Vx) = μY(x)μZ(x)
cov(Vx, Vx') = Ky(x, x')μZ(x)μZ(x')
The stochastic process (VX)XER is not necessarily Gaussian since it depends on the product of (YX) and (ZX). If either (YX) or (ZX) is non-Gaussian, the resulting process (VX) will also be non-Gaussian.
The correct question should be :
Question 2 Consider two centred Gaussian processes (YX)XER and (Zx)XER, with covariance kernels Ky and Kz, respectively; the kernel Ky is thus such that cov(Yx, Yx) = Ky(x, x'), for all x and x' = R, and a similar expression holds for (Zx)XER.
Assume that (YX)XER and (ZX)XER are independent. Introduce two stochastic processes (UX)XER and (VX)XER, such that Ux=Yx+Zx, and Vx=YxZx, for all x E R. Consider x and x' E R.
(a) Compute E(U) and cov(U, U); is the stochastic process (UX)XER Gaussian? [6]
(b) Compute E(V₂) and cov(Vx, Vx); is the stochastic process (Vx)XER Gaussian? [6]
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A basketball player is fouled in the act of shooting a three-point shot and is awarded three free throws. The player makes free throws 80% of the time. Assume that each free throw is an independent event..
1) What is the probability that the player makes all three free throws?
2) What is the probability that the player misses all three free throws?
3) What is the probability that the player misses at least one free throw?
4) What is the probability that the player makes at least one free throw?
Answer:
1) .8³ = .512 = 51.2%
2) .2³ = .008 = .8%
3) 1 - .8³ = 1 - .512 = .488 = 48.8%
4) 1 - .2³ = 1 - .008 = .992 = 99.2%
1) The probability that the player makes a free throw is 80%, or 0.8. Since each free throw is an independent event, the probability of making all three free throws is calculated by multiplying the individual probabilities together: 0.8 * 0.8 * 0.8 = 0.512, or 51.2%.
2) The probability that the player misses a free throw is the complement of making a free throw, which is 1 - 0.8 = 0.2. Again, since each free throw is independent, the probability of missing all three free throws is calculated by multiplying the individual probabilities together: 0.2 * 0.2 * 0.2 = 0.008, or 0.8%.
3) The probability that the player misses at least one free throw is the complement of making all three free throws. So, it is 1 - 0.512 = 0.488, or 48.8%.
4) The probability that the player makes at least one free throw is the complement of missing all three free throws. So, it is 1 - 0.008 = 0.992, or 99.2%.
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Start A university claims that students can expect to spend a mean of 3 hours per week on homework for every credit nour of class. The administration believes that this number is no longer correct at
The university may conduct a study to investigate if its claim of students spending an average of 3 hours per week on homework for every credit hour is still valid.
The university's claim is that students can expect to spend an average of 3 hours per week on homework for every credit hour of class. The university administration believes that this number is no longer valid. To investigate this issue, the administration may conduct a study in which they compare the number of hours students are spending on homework to the number of credit hours they are taking.
They can then determine if there is a correlation between the number of credit hours a student is taking and the number of hours they are spending on homework. If there is no correlation, the university may need to revise its homework expectations.
In conclusion, the university may conduct a study to investigate if its claim of students spending an average of 3 hours per week on homework for every credit hour is still valid.
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