The spares should be kept to maintain a 90% or more probability of system success is 35
Let λ be the normal number of component disappointments per year, at that point the disappointment rate (or rate parameter) is given by λ/52 since there are 52 weeks in a year. Let's signify this by μ = λ/52.
The framework victory likelihood can be modeled utilizing the Poisson dissemination since the disappointments happen haphazardly and freely over time. Let X be the number of component disappointments in a year, at that point X takes after a Poisson conveyance with cruel λ.
To preserve a 90% or more likelihood of system victory, we got to guarantee that the number of saves is adequate to cover at the slightest 90% of the potential disappointments. This implies that the likelihood of having more than k disappointments in a year ought to be less than or break even with 0.1, where k is the number of saves.
Let Y be the number of component disappointments amid the two-week conveyance time. At that point, Y too takes after a Poisson conveyance with cruel μ/26, since there are 26 weeks in a half-year (i.e., two quarters). The likelihood of having more than k disappointments amid the conveyance time is given by:
P(Y > k) = 1 - P(Y ≤ k) = 1 - ∑_[tex]{i=0}^k (e^(-μ/26) (μ/26)^i[/tex]/ i!)
where e is the base of the common logarithm.
To preserve a 90% or more likelihood of system victory, we ought to select k such that P(Y > k) ≤ 0.1. Able to solve for k numerically, employing a spreadsheet or a computer program.
For example, utilizing Microsoft Exceed expectations or Go-ogle Sheets, ready to utilize the taking after an equation to compute P(Y > k) for distinctive values of k:
=1-POISSON(k,μ/26,TRUE)
where POISSON is the Poisson total conveyance work, with the moment contention being the cruel and the third contention being Genuine to indicate an aggregate conveyance.
Beginning with k = 26 (i.e., one save per week), able to increment k until we discover the littlest esteem that fulfills P(Y > k) ≤ 0.1. In this case, we discover that k = 35 saves are required to preserve a 90% or more likelihood of framework victory.
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In a large city, the average number of lawn mowings during summer is normally distributed with mean u and standard deviation 0-8.7. If I want the margin of error for a 90% confidence interval to be +3, I should select a simple random sample of size (4 decimal points)
To achieve a margin of error of +3 for a 90% confidence interval for the average number of lawn mowings during summer in a large city with a mean of u and a standard deviation of 0-8.7, a simple random sample of size 18 should be selected.
To determine the sample size needed to achieve a margin of error of +3 for a 90% confidence interval, we can use the formula:
n = (z * σ / E)^2
where n is the sample size, z is the z-score for the desired confidence level (in this case, 1.645 for 90% confidence), σ is the standard deviation, and E is the margin of error.
Substituting the given values into the formula, we get:
n = (1.645 * 0.8 / 3)^2 = 17.18
Rounding up to the nearest whole number, we get a sample size of 18.
Therefore, to achieve a margin of error of +3 for a 90% confidence interval for the average number of lawn mowings during summer in a large city with a mean of u and a standard deviation of 0-8.7, a simple random sample of size 18 should be selected. This sample size ensures that the estimate of the population means based on the sample mean is within +3 of the true population mean with 90% confidence.
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Beth enlarged the triangle below by a scale of 5.
3.5 cm
4 cm
She found the area of the enlarged triangle. Her work is shown below.
(4)(3.5)(5)- 35 cm²
What was Beth's error?
O She should have divided (4)(3.5) by 5.
Caus and Exit
The error made by Beth is that:
She didn't apply the scale factor to each dimension of the triangle before multiplying
How to Interpret Enlargement Scale Factor?We are given the parameters as:
Initial height = 3.5cm
Initial width = 4cm
Formula for area of triangle is:
Area = ¹/₂ * base * height
If the triangle was enlarged by a scale factor of 5, then it means each of the dimensions should first be multiplied by 5 to get:
New width = 4 * 5 = 20 cm
New height = 5 * 3.5 = 17.5 cm
Thus:
New area = ¹/₂ * 20 * 17.5 = 175 cm²
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According to a poll of adults, about 46% work during their summer vacation. Assume that the true proportion of all adults that work during summer vacation is p=0.46. Now consider a random sample of 400 adults. Complete parts a and b below.
a. What is the probability that between 39% and 53% of the sampled adults work during summer vacation?
The probability is. (Round to three decimal places as needed.)
b. What is the probability that over 63% of the sampled adults work during summer vacation?
The probability is (Round to three decimal places as needed.)
To find the probability we do the following:
I need help on this 40 points I need to turn it in in like 5 min
Answer:
Step-by-step explanation:
13. B
14. A
15. D
Which of the plotted points is in the second quadrant
Answer:
Point C is in the second quadrant. C is the correct answer.
NEED ANSWERS ASAP!! PLS
The US government monitors the consumption of different products. The table shows y, the amount of ice cream consumed, in millions of pounds, for * years since 2010. The quadratic equation that models the amount of ice cream consumed, in millions of pounds, since 2010 is shown. y = 12(¢ - 6)2 + 3922 Determine when the amount of ice cream consumed in the United State would be 5,650 millions of pounds.
The amount of ice cream consumed in the United State would be 5,650 millions of pounds in 2028.
How to determine when the amount of ice cream is 5,650 millions of pounds?Based on the information provided about the mount of ice cream consumed in the United State, a quadratic equation that models the amount of ice cream consumed, in millions of pounds, since 2010 is given by;
y = 12(x - 6)² + 3922
Where:
y is the amount of ice cream consumed, in millions of pounds.x is the number of years since 2010.By substituting the value of y, the number of years can be calculated as follows;
5,650 = 12(x - 6)² + 3922
5,650 - 3922 = 12(x - 6)²
1728 = 12(x - 6)²
144 = (x - 6)²
12 = x - 6
x = 12 + 6
x = 18 years.
Since 2010; 2010 + 18 = 2028.
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This question has two parts.
A wooden block is a prism, which is made up of two cuboids with the dimensions shown. The volume of the wooden block is 427 cubic inches.
Part A
What is the length of MN?
Write your answer and your work or explanation in the space below.
Part B
200 such wooden blocks are to be painted. What is the total surface area in square inches of the wooden blocks to be painted?
A) The length MN of the given wooden block is: 12
B) 80400 in²
How to find the surface area and volume of the prism?1) The formula for volume of a cuboid is:
Volume = Length * Width * Height
Thus:
427 = (MN * 7 * 3) + (5 * 5 * 7)
427 = 21MN + 175
21MN = 252
MN = 252/21
MN = 12
2) Surface area of entire object is:
TSA = 2(12 * 3) + 2(12 * 7) - (5 * 7) + 2(7 * 3) + 3(5 * 7) + 2(5 * 5)
= 402 in²
For 200 blocks:
TSA = 200 * 402 = 80400 in²
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a better estimate is obtained by assuming that each lake is a separate tank with only clean water flowing in. use this approach to determine how long ti would take the pol lution level ni each lake ot be reduced to 50% of its original level. how long would ti take ot reduce the pollution to %5 of its original level?
It would take around 8 hours to bring each lake's pollution level down to 50% of its starting point, and around 22 hours to bring it down to 5%.
Assuming each lake is a separate tank with only clean water flowing in, we can use the exponential decay model [tex]$A = A_0e^{-kt}$[/tex], where $A$ is the amount of pollutant at time t, A₀ is the initial amount of pollutant, and k is the decay constant.
To find the time it would take to reduce the pollution level in each lake to 50% of its original level, we need to solve the equation [tex]$0.5A_0 = A_0e^{-kt}$[/tex] for t:
[tex]0.5A_0 &= A_0e^{-kt} \\frac{0.5A_0}{A_0} &= e^{-kt} \\ln\left(\frac{0.5A_0}{A_0}\right) &= -kt \\ln(0.5) &= -kt \t &= \frac{\ln(0.5)}{-k}\end{align*}[/tex]
To find the time it would take to reduce the pollution level in each lake to 5% of its original level, we need to solve the equation[tex]$0.05A_0 = A_0e^{-kt}$[/tex] for t:
[tex]0.05A_0 &= A_0e^{-kt} \\frac{0.05A_0}{A_0} &= e^{-kt} \\ln\left(\frac{0.05A_0}{A_0}\right) &= -kt \\ln(0.05) &= -kt \t &= \frac{\ln(0.05)}{-k}\end{align*}[/tex]
The decay constant $k$ can be found by using the given information that each lake is replaced by clean water every 8 hours. This means that the half-life of the pollutant is 8 hours, which gives us:
[tex]0.5A_0 &= A_0e^{-k(8)} \\ln(0.5) &= -8k \k &= -\frac{\ln(0.5)}{8} \approx 0.08664\end{align*}[/tex]
Substituting this value of k into the equations we derived earlier, we get:
[tex]t_{50} = \frac{\ln(0.5)}{-k} \approx 8.006 \text{ hours}[/tex]
[tex]t_{5} &= \frac{\ln(0.05)}{-k} \approx 22.133 \text{ hours}[/tex]
Therefore, it would take approximately 8 hours to reduce the pollution level in each lake to 50% of its original level, and approximately 22 hours to reduce it to 5% of its original level.
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A woman wants to construct a box whose base length is twice the base width. The material to build the top and bottom is $9/m^2 and the material to build the sides is $6/m^2. If the woman wants the box to have a volume of 70 m3, determine the dimensions of the box (in metres) that will minimize the cost of production. What is the minimum cost?
The minimum cost of production is $278.46.
The base width of the box be x, then the base length is 2x. Let the height of the box be h.
The volume of the box is given by:
V = base area × height
[tex]70 = x[/tex] × [tex]2x[/tex] × [tex]h[/tex]
[tex]h = 35/x^2[/tex]
The cost of producing the box is given by:
[tex]C = 2[/tex] ×[tex](cost of top/bottom) + 4[/tex] × [tex](cost of side)[/tex]
[tex]C = 2[/tex]× [tex]9[/tex]× [tex](2x[/tex] × [tex]x) + 4[/tex] × [tex]6[/tex] × [tex](2x + 2h)[/tex]
[tex]C = 36x^2 + 48xh[/tex]
Substituting the expression for h obtained above:
[tex]C = 36x^2+ 48x(35/x^2)[/tex]
[tex]C = 36x^2+ 1680/x[/tex]
To minimize C, we take the derivative with respect to x and set it to zero:
[tex]dC/dx = 72x - 1680/x^2 = 0[/tex]
[tex]72x = 1680/x^2[/tex]
[tex]x^3 = 1680/72[/tex]
[tex]= 23.33[/tex]
x = 2.82 m (rounded to two decimal places)
Substituting this value of x in the expression for h, we get:
[tex]h = 35/(2.82)^2[/tex]
[tex]= 4.34 m[/tex] (rounded to two decimal places)
Therefore, the dimensions of the box that minimize the cost of production are:
[tex]Base width = x = 2.82 m[/tex]
[tex]Base length = 2x = 5.64 m[/tex]
[tex]Height = h = 4.34 m[/tex]
To find the minimum cost, we substitute these values of x and h in the expression for C:
[tex]C = 36(2.82)^2 + 1680/(2.82)[/tex]
[tex]C = $278.46[/tex] (rounded to two decimal places)
Therefore, the minimum cost of production is $278.46.
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Which of the following is not a valid probability
A. 1
B. 1.1
C. 0
D. 0.001
Answer:
B
Step-by-step explanation:
It is impossible for a possibility to be more than 1, or 100%.
Find a basis of the null space N(A) for the the matrix. Then find an orthogonal basis using Gram-Schmidt process. [1 2 1 3 2]
A= [4 1 0 6 1]
[1 1 2 4 5]
We apply the Gram-Schmidt process to these vectors to find an orthonormal basis:
v1 = x1 = [3, -4, 1
To find a basis of the null space N(A), we need to find all vectors x such that Ax = 0, where 0 is the zero vector.
To do this, we set up the augmented matrix [A | 0] and row reduce:
[ 1 2 1 3 2 | 0 ]
[ 4 1 0 6 1 | 0 ]
[ 1 1 2 4 5 | 0 ]
R2 - 4R1 -> R2:
[ 1 2 1 3 2 | 0 ]
[ 0 -7 -4 6 -7 | 0 ]
[ 1 1 2 4 5 | 0 ]
R3 - R1 -> R3:
[ 1 2 1 3 2 | 0 ]
[ 0 -7 -4 6 -7 | 0 ]
[ 0 -1 1 1 3 | 0 ]
R2 / -7 -> R2:
[ 1 2 1 3 2 | 0 ]
[ 0 1 4/7 -6/7 1 | 0 ]
[ 0 -1 1 1 3 | 0 ]
R1 - 2R2 - R3 -> R1:
[ 0 0 0 0 0 | 0 ]
[ 0 1 4/7 -6/7 1 | 0 ]
[ 0 0 11/7 -1/7 1 | 0 ]
We can write the system of equations corresponding to this row echelon form as:
x2 + (4/7)x3 - (6/7)x4 + x5 = 0
(11/7)x3 - (1/7)x4 + x5 = 0
Solving for the variables in terms of the free variables x3, x4, and x5, we get:
x1 = -[(4/7)x3 - (6/7)x4 - x5]/2
x2 = -(4/7)x3 + (6/7)x4 - x5
x3 = x3 (free variable)
x4 = x4 (free variable)
x5 = x5 (free variable)
So the null space N(A) is the set of all vectors of the form:
x = [ -[(4/7)x3 - (6/7)x4 - x5]/2, -(4/7)x3 + (6/7)x4 - x5, x3, x4, x5 ]
To find an orthogonal basis for N(A), we can use the Gram-Schmidt process. Let's call the columns of A a1, a2, a3, a4, and a5.
First, we need to find a basis for N(A) by setting the free variables to 1 and the others to 0:
x1 = [3, -4, 1, 0, 0]
x2 = [-2, 3, 0, 1, 0]
x3 = [-2, 1, 0, 0, 1]
Next, we apply the Gram-Schmidt process to these vectors to find an orthonormal basis:
v1 = x1 = [3, -4, 1
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How much greater is the value of the 6 inand 4786.53 Denton 3821.69
xon the following graph, use the orange points (square symbol) to plot points along the portion of the firm's short-run supply curve that corresponds to prices where there is positive output.
To plot points along the portion of the firm's short-run supply curve that corresponds to prices where there is positive output, we need to identify the portion of the graph where the firm is producing output.
We can observe from the graph that the firm's short-run supply curve is the component of the marginal cost curve that is higher than the average variable cost curve. The company will shut down and create no production if prices fall below the minimum point of the average variable cost curve. However, as long as the price is above the marginal cost of production, the company will create output at prices above the minimum point of the average variable cost curve.
We may use the orange square symbols to represent the price and matching amount provided at each point where the company is generating output to plot points along this segment of the short-run supply curve. We may advance up the marginal cost curve from the last point of the average variable cost curve until we reach the maximum price at which the company is generating output. The firm's short-run supply curve may then be created by marking each price and quantity combination along this segment of the curve.
It is important to note that the firm's short-run supply curve is a reflection of its marginal cost curve above the average variable cost curve, and will shift as the firm's costs of production change or its technology improve.
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1/20 ÷ 5 could someone answer this
The value of the expression is 1/100.
We have,
To solve this expression, we can use the division property of fractions which states that dividing by a fraction is the same as multiplying by its reciprocal.
So, we have:
1/20 ÷ 5
= 1/20 x 1/5 (reciprocal of 5 is 1/5)
= 1/100 (multiply numerator with numerator and denominator with denominator)
Therefore,
The simplified result is 1/100.
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Consider the the following series. [infinity] 1 n3 n = 1 (a) Use the sum of the first 10 terms to estimate the sum of the given series. (Round the answer to six decimal places. ) s10 = (b) Improve this estimate using the following inequalities with n = 10. (Round your answers to six decimal places. ) sn + [infinity] f(x) dx n + 1 ≤ s ≤ sn + [infinity] f(x) dx n ≤ s ≤ (c) Using the Remainder Estimate for the Integral Test, find a value of n that will ensure that the error in the approximation s ≈ sn is less than 10-5
The estimated sum of the given series using the sum of the first 10 terms is 302,500, the improved estimate for the sum of the given series is between 305,000 and 306,000, and the value of n is 8.
(a) Utilizing the equation for the entirety of the primary n terms of the arrangement, we have:
[tex]s10 = 1^3 + 2^3 + ... + 10^3[/tex]
= 1,000 + 8,000 + ... + 1,000,000
= 302,500
In this manner, the assessed whole of the given arrangement using the entirety of the primary 10 terms is 302,500.
(b) For n = 10, we have:
[tex]sn = 1^3 + 2^3 + ... + 10^3 ≈ 302,500[/tex]
Utilizing the disparities with[tex]f(x) = x^3[/tex], we have:
[tex]sn + ∫[10,∞] x^3 dx ≤ s ≤ sn + ∫[10,∞] x^3 dx + 10^3[/tex]
Utilizing calculus, ready to assess the integrand:
[tex]sn + ∫[10,∞] x^3 dx = sn + [1/4 x^4] [10,∞] = sn + 2500[/tex]
[tex]sn + ∫[10,∞] x^3 dx + 10^3 = sn + [1/4 x^4] [10,∞] + 10^3 = sn + 3500[/tex]
Substituting sn = 302,500, we get:
302,500 + 2500 ≤ s ≤ 302,500 + 3500
305,000 ≤ s ≤ 306,000
In this manner, the made strides assess for the sum of the given arrangement is between 305,000 and 306,000.
(c) The Leftover portion Gauge for the Necessarily Test states that the mistake E in approximating the whole s of an interminable arrangement by the nth halfway entirety sn is:
[tex]E ≤ ∫[n+1,∞] f(x) dx[/tex]
In this case, we need to discover mean of n such that E < 10 using the integral test
xss=removed xss=removed> [tex][(10^-5 x 4)^(1/4)] - 1[/tex]
n > 7.9378
Subsequently, we require n = 8 to guarantee that the blunder within the estimation s ≈ sn is less than[tex]10^-5.[/tex]
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For a four strokes engine, the camshaft that controls the valves a timing rotates at a. Double the engine rotational speed b. Half the engine rotational speed
c. Same engine rotational speed
d. none of the options
The correct answer is option (c) same engine rotational speed.
In a four-stroke engine, the camshaft rotates at half the speed of the crankshaft. The camshaft controls the opening and closing of the engine valves, which allows for the intake of fuel and air and the expulsion of exhaust gases.
When the engine rotational speed is doubled, the crankshaft will rotate twice as fast, but the camshaft will still rotate at half the speed of the crankshaft. This means that the camshaft will still rotate at the same speed as before, and the valve timing will remain the same.
Similarly, when the engine rotational speed is halved, the camshaft will still rotate at half the speed of the crankshaft. This means that the camshaft will still rotate at the same speed as before, and the valve timing will remain the same.
Therefore, the correct answer is option (c) same engine rotational speed.
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Brody has a jar with 1000 g of sugar in it. Each day, he empties out half
of the sugar that is in the jar. At the end of the first day, he is left with
500 g of sugar.
a) How much sugar will be left in the jar at the end of the 5th day? Give
your answer in grams (g).
b) Write a sentence to explain whether or not the jar will ever be empty
31.25 g on the 5th day
no it will never by empty because even when it gets down to one singular piece of sugar you would technically just cut it in half, then that in half, yes it would get impossible, that's why you wouldn't actually do it, but if you typed it into a calculator it would just keep getting a smaller and smaller decimal.
supply of houses is determined by two variables (I and R) in the following way: h(1,R) = a log1 + b R + CR log1, where a, b, and c are all constants. How does housing supply respond to changes in I (a) and R (OR)? an an Select one: an a. ar an a+cR an I and an = b + clog1 2 an ī and a b+cR log1 = b + cR log1 O b. a1 an a+cR an C. ai and an an d. ar an + CR log 1 and aR = b + c log1
The housing supply function is given by h(I,R) = a log1 + bR + cR log1. The housing supply responds to changes in I with a rate of a, and to changes in R with a rate of b + c log1.
Based on the given equation, the housing supply (h) is determined by two variables: I and R. The equation shows that h is a function of R, with a log-linear relationship. The variable I only appears as a constant (a) in the equation, so changes in I do not directly affect the supply of houses.
On the other hand, changes in R (or OR, which is the same variable) do affect the supply of houses. Specifically, an increase in R leads to an increase in the supply of houses. The magnitude of this increase depends on the values of b and c in the equation.
To see this, we can take the partial derivative of h with respect to R:
dh/dR = b + cR/(ln(10))
This equation tells us how much the housing supply changes in response to a change in R. The derivative is positive (i.e. the supply increases) as long as c is positive. The larger c is, the greater the increase in supply for a given increase in R.
Therefore, the correct answer is:
b + cR/(ln(10))
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The total cost for funding a trip for the senior class to go to the fall fair, C(x), is a function of the number of students that will make the trip, x. The trip will not be taken until at least 5 students sign up to go. This relationship can be modeled by the function shown.
C(x) = 350 + 7.50x
What is the domain and range for this situation?
The value of domain and range for this situation are,
Domain = (- ∞, ∞)
Range = (- ∞, ∞)
We have to given that;
The total cost for funding a trip for the senior class to go to the fall fair, C(x), is a function of the number of students that will make the trip, x.
Now, We have;
⇒ C (x) = 350 + 7.5x
Clearly, the function is a polynomial.
Hence, The value of domain and range for this situation are,
Domain = (- ∞, ∞)
Range = (- ∞, ∞)
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Sales tax is 7%. What is the tax on a book that costs $12?
Help me please and explain im so confused
The value of cos S to the nearest hundredth is 0.54
What is trigonometric ratio?Trigonometric Ratios are defined as the values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.
Sin(tetha) = opp/hyp
cos(tetha) = adj/hyp
tan(tetha) = opp/adj
Here in this triangle, the hypotenuse is 28
and the opposite to angle S is line TU
The adjascent is 15
therefore cos S = adj/hyp
= 15/28
= 0.54 ( nearest hundredth)
therefore the value of cos S is 0.54
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At a local gym with 800 members, 450 members take aerobics class, 200 members do weight training, and 125 members do both weight training and take an aerobics class. What is the probability that a randomly-selected member takes an aerobics class or does weight training? Round your answer to the hundredths place.
Answer:
0.66
Step-by-step explanation:
You want the probability that a randomly chosen member from the 800 members of a gym takes either an aerobics class, as 450 members do, or does weight training, as 200 members do. 125 members do both.
EitherAdding the numbers given for members who do aerobics or weight training will count the number who do both twice. Then the number who do either is ...
(# of aerobics) + (# of weights) - (# of both)
= 450 +200 -125 = 525
The probability that one of the 800 members is in this group is ...
P(either) = 525/800 ≈ 0.66
<95141404393>
Please help asap :( Find the exact length of arc ADC. In your final answer, include all of your calculations
Answer:
15 Pi m
Step-by-step explanation:
arc ADC = 360 Degrees - 60 Degrees divided by 360 Degrees Multiplied by 2 Pi Multiplied by 9
= 5/6 Times 18 Pi
=15 Pi m
Please help I have to do this before state testing this I one out of 32 questions also if you so happen to be mrs Billie from Alhambra traditional school I hate you
Answer:turn right 45 degrees, then turn right another 45 degrees. flip the figure x-axis wise/horizontally
Step-by-step explanation:
if he sees a wolf, a boy will cry wolf with probability 0.8. if he does not see a wolf, the boy will cry wolf anyway with probability 0.4. if the probability that there is a wolf would otherwise be 0.25, what is the probability that there really is a wolf when the boy crys wolf?
The probability that there really is a wolf when the boy cries wolf is 0.54.
Let A be the event that the boy cries wolf and B be the event that there is a wolf. We are given the following probabilities:
P(A|B) = 0.8 (the probability that the boy cries wolf when there is a wolf)
P(A|B') = 0.4 (the probability that the boy cries wolf when there is no wolf)
P(B) = 0.25 (the probability that there is a wolf)
We want to find P(B|A), the probability that there really is a wolf given that the boy cries wolf.
We can use Bayes' theorem to calculate this:
P(B|A) = P(A|B) * P(B) / [P(A|B) * P(B) + P(A|B') * P(B')]
Substituting the given values, we get:
P(B|A) = 0.8 * 0.25 / [0.8 * 0.25 + 0.4 * 0.75] = 0.54
Therefore, the probability that there really is a wolf when the boy cries wolf is 0.54.
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Find an ONB (orthonormal basis) for the following plane in R3 x + 5y + 4z = 0 First, solve the system, then assign parameters s and t to the free variables (in this order), and write the solution in vector form as su + tv. Now normalize u to have norm 1 and call it ū. Then find the component of v orthogonal to the line spanned by u and normalize it, call it ī. Below, enter the components of the vectors ū = [ū1, ū2, ū3]and ū = ū1, 72, 73)".
The ONB for the given plane in R3 is ū = [-5/√(26), 1/√(26), 0] and ī = [25/(√(26/13)), -5/(√(26/13)), 0].
To find an orthonormal basis for the plane x + 5y + 4z = 0, we first solve the system and get the parametric solution
x = -5t - 4s
y = t
z = s
Assigning parameters s and t to the free variables and writing the solution in vector form as su + tv, we get
[-5t - 4s, t, s] = t[-5, 1, 0] + s[-4, 0, 1]
Taking u = [-5, 1, 0] and v = [-4, 0, 1], we normalize u to have norm 1 by dividing it by its length
||u|| = √(26)
ū = [-5/√(26), 1/√(26), 0]
To find the component of v orthogonal to u, we take the dot product of v and u, and divide it by the dot product of u and u, and then multiply u by this scalar
v - ((v · u) / (u · u))u
v · u = -5
u · u = 26
v - (-5/26)[-5, 1, 0]
v - [25/26, -5/26, 0]
Finally, we normalize this vector to have norm 1
||v - proj_u v|| = √(26/13)
ī = [25/(2√(26/13)), -5/(2√(26/13)), 0]
Therefore, the orthonormal basis for the plane x + 5y + 4z = 0 is ū = [-5/√(26), 1/√(26), 0] and ī = [25/(√(26/13)), -5/(√(26/13)), 0].
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Find the two following values
The value of angle TUW is 32⁰.
The value of angle UTV is 25⁰.
What is the value of angle TUW?
The value of angle TUW is calculated by applying the following formula.
angle TVW = angle TUW (vertical opposite angles are equal)
angle TVW = 32⁰
So, angle TUW = 32⁰
The value of angle UTV is calculated as;
angle UTV = VWU (vertical opposite angles are equal)
3x + 4 = 2x + 11
3x - 2x = 11 - 4
x = 7
angle UTV = 3x + 4
= 3(7) + 4
= 25⁰
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use law of sines to solve triangle with B=52 C=15 b=43
Using the Law of Sines, we can solve the given triangle with B = 52, C = 15, and b = 43. The three angles of the triangle are approximately A = 112.94°, B = 52°, and C = 15°, and the lengths of the sides opposite these angles are approximately a = 28.29, b = 43, and c = 8.11.
The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is the same for all sides and their opposite angles. Therefore, we can use the Law of Sines to solve the given triangle as follows
sin(A)/a = sin(B)/b = sin(C)/c
We are given B = 52, C = 15, and b = 43, so we can use the Law of Sines to find a
sin(A)/a = sin(B)/b
sin(A)/a = sin(52)/43
sin(A) = a sin(52)/43
a = 43 sin(A)/sin(52)
Similarly, we can use the Law of Sines to find c
sin(A)/a = sin(C)/c
sin(A)/a = sin(15)/c
sin(15)c = a sin(A)
c = a sin(A)/sin(15)
Now, we can substitute the expressions for a and c into the equation sin(A)/a = sin(B)/b and solve for sin(A)
sin(A)/(43 sin(A)/sin(52)) = sin(52)/43
sin(A) = (43 sin(52) sin(15))/c
Substituting the expression for c, we get
sin(A) = (43 sin(52) sin(15))/(a sin(A)/sin(15))
Simplifying, we get
sin²(A) = (43²sin²(52) sin²(15))/(a²sin²(A))
Multiplying both sides by a^2 sin^2(A), we get
a²sin⁴(A) = 43^2 sin²(52) sin²(15)
Taking the square root of both sides and solving for a, we get
a = √((43² sin₂(52) sin²(15))/(sin⁴A)))
Substituting the given values and solving for A, we get
a = 28.29 (approx)
A = 112.94° (approx)
c = 8.11 (approx)
Therefore, the three angles of the triangle are approximately A = 112.94°, B = 52°, and C = 15°, and the lengths of the sides opposite these angles are approximately a = 28.29, b = 43, and c = 8.11.
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Use the table of the probability distribution to find the variance
The variance of the given distribution is 0.9.
Calculating the anticipated value or distribution mean is the first step in determining a probability distribution's variance. The anticipated value is a weighted average of all potential outcomes, with each outcome's probability serving as the weight. The expected value can be expressed mathematically as follows:
[tex]E(X) =[/tex] Σ[tex][xi[/tex] × [tex]P(xi)][/tex]
where μ is the mean of the data.
Then, calculate μ:
μ [tex]= (1+2+3+4+5)/5[/tex]
[tex]=(15/5)[/tex]
[tex]= 3[/tex]
and replace this value and the values of xn and P(xn) into the formula for the variance, just as follow:
Hence, the variance of the given distribution is 0.9.
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the weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 3315 grams and a variance of 391,876 . if a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4504 grams. round your answer to four decimal places.
The probability that the weight will be less than 4504 grams is 0.9713.
We need to standardize the value 4504 using the given mean and variance, and then use the standard normal distribution table to find the corresponding probability.
The standard deviation is the square root of the variance: √391876≈626.05
So, the z-score for a weight of 4504 grams is:
z=(4504−3315)/626.05≈1.8974
Using a standard normal distribution table, we find that the probability of a z-score being less than 1.8974 is approximately 0.9713.
Therefore, the probability that a newborn baby boy born at the local hospital will weigh less than 4504 grams is approximately 0.9713.
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