Answer:
a) 4.2%
b) 14.1%
Step-by-step explanation:
a) 0.9³⁰ = 0.0423911583
b) 30C1 × 0.1 × 0.9²⁹ = 0.1413038609
Answer:
a.) 4.2%
b.) 14.1%
Step-by-step explanation:
We solve using the probability distribution formula for selection and this formula uses the combination formula for estimation.
When choosing a random selection of "r" items from a sample of "n" items, The formula is generally denoted by:
P(X=r) = nCr × p^r × q^n-r.
Where p = probability of success
q= probability of failure.
From the given question,
n = number of samples =30,
p = Probability that a student is late = 10% = 0.1,
q=0.9
a.) when no student is late, that is when r = 0, then
P(X=0) = 30C0 × 0.1^0 × 0.9^30
P(X=0) = 0.0424 = 4.24 ≈ 4.2%
b.) when exactly one student is late, that is when r=1, then
P(X=1) = 30C1 × 0.1¹ × 0.9^29
P(X=1) = 0.1413 = 14.13 ≈ 14.1%
Scarborough Faire Herb Farm is a small company specializing in selling organic fresh herbs, teas and herbal crafts. Currently, basil is their top selling herb, with $45,000 in sales last year. Parsley is their second biggest seller with $27,000 in sales. Total sales last year were $170,000 and Scarborough Faire forecasts sales to increase by 10% this year. If basil sales remain the same as last year but total sales grow as percentage will basil sales be this year?
Answer:
24.06% of total sales
Step-by-step explanation:
Total sales, which were $170,000 originally, are expected to grow by 10%. The expected value of total sales this year is:
[tex]S=\$170,000*1.10 = \$187,000[/tex]
If Basil sales remain at the same value of $45,000, the percentage of sales corresponding to basil is:
[tex]B=\frac{\$45,000}{\$187,000}*100\%\\B=24.06\%[/tex]
Therefore, basil will correspond to 24.06% of total sales.
Basil sales will represent approximately 24.06% of Scarborough Faire Herb Farm's projected total sales this year, given that total sales are forecasted to increase by 10% and basil sales remain unchanged.
Explanation:The student's question concerns the percentage of total sales that basil sales will represent for Scarborough Faire Herb Farm in the current year, assuming a 10% increase in total sales from the previous year and unchanged basil sales. Firstly, we calculate the projected total sales for this year by increasing last year's total sales by 10%. The calculation is as follows:
Total sales last year: $170,000Forecasted increase: 10%Projected total sales this year: $170,000 + ($170,000 × 0.10) = $170,000 + $17,000 = $187,000Since the basil sales are to remain the same as last year ($45,000), we now calculate what percentage this represents of the projected total sales for this year:
Basil sales this year: $45,000 (unchanged)Percentage of basil sales out of total sales: ($45,000 / $187,000) × 100%Final percentage: 24.06% (rounded to two decimal places)Thus, basil sales will account for approximately 24.06% of the Scarborough Faire Herb Farm's total sales this year.
Children inherit many traits from their parents. Some inherited traits are not desirable. For example, if both parents are carriers, there is a 25% chance any child will have sickle-cell disease. If we assume no identical twins in a family, and both individuals in a couple are carriers, what is the probability none of their four children has the disease
The probability that none of their four children has the disease = 0.316
Step-by-step explanation:
Step 1 :
The probability that the child will have the specified disease if both the parents are carriers = 25% = 0.25
So the probability that the child does not have the specified disease = 1 - 0.25 = 0.75
Step 2 :
There are four children. The probability that each child has the disease is independent of the condition of the other children
The probability of more than one independent events happening together can be obtained by multiplying probability of individual events
So, the probability that none of their 4 children has the sickle cell disease = [tex](0.75)^{4}[/tex] = 0.316
Step 3 :
Answer :
The probability no children has the specified cell disease = 0.316
The following information relates to Franklin Freightways for its first year of operations (data in millions of dollars): Pretax accounting income:$200 Pretax accounting income included: Overweight fines (not deductible for tax purposes) 5 Depreciation expense 70 Depreciation in the tax return using MACRS: 110 The applicable tax rate is 40%. There are no other temporary or permanent differences. Franklin Freightways experienced ($ in millions) a: Multiple Choice Tax liability of $66. Tax liability of $36. Tax liability of $70.6. Tax benefit of $10 due to the NOL.
Answer:
Franklin Freightways experienced $66 million.
Step-by-step explanation:
= ( Pretax accounting income + Overweight fines - Temporary difference: Depreciation) * tax rate
($200 + 5 - $40) × 40%
=Tax liability of $66.
A box contains five slips of paper, marked $1, $1, $10, $25, and $25. The winner of a contest selects two slips of paper at random and then gets the larger of the dollar amounts on the two slips. Define a random variable w by w = amount awarded. Determine the probability distribution of w. (Hint: Think of the slips as numbered 1, 2, 3, 4, and 5, so that an outcome of the experiment consists of two of these numbers.)
Answer:
w = $1 or $10 or $25
Probability distribution of w = ($1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $10,$1 $10,$1 $10,$10 $10,$25 $10,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25)
Step-by-step explanation:
Since the winner gets the larger amount of the two slips picked, random variable 'w' which is the amount awarded could be $1 or $10 or $25.
The probability distribution of 'w' is the values that the statistic takes on. Which could be: $1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $1,$1 $1,$1 $1,$10 $1,$25 $1,$25 $10,$1 $10,$1 $10,$10 $10,$25 $10,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25 $25,$1 $25,$1 $25,$10 $25,$25 $25,$25
Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that all are able to solve the problem
The probability that all three students solve the problem is calculated by multiplying their individual success probabilities together. The total probability in this case is 64.6%.
Explanation:Your question pertains to probability, a topic in Mathematics. When three students independently attempt to solve a problem, and you have the probabilities of their success, the probability that all three will successfully solve the problem is determined by the product of their respective probabilities.
Therefore, the probability that all three students - the first with a probability of 0.95, the second with 0.85, and the third with 0.80 - will successfully solve the problem is calculated as follows:
0.95 * 0.85 * 0.80 = 0.646
Hence, there is a 64.6% probability that all three students will successfully solve the problem.
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Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y′+p(t)y=g(t) have jump discontinuities. If t0 is such a point of discontinuity, then it is necessary to solve the equation separately for t < t0 and t > t0. Afterward, the two solutions are matched so that y is continuous at t0; this is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y′ continuous at t0.
Solve the initial value problem.
y' + 6y = g(t), y(0) = 0
where
g(t) = 1, 0 ≤ t ≤ 1,
= 0, t > 0.
For [tex]0\le t\le1[/tex], we have
[tex]y'+6y=1\implies e^{6t}y'+6e^{6t}=(e^{6t}y)'=e^{6t}\implies y=\dfrac16+Ce^{-6t}[/tex]
Given that [tex]y(0)=0[/tex], we have
[tex]0=\dfrac16+C\implies C=-\dfrac16[/tex]
so that
[tex]y=\dfrac16(1-e^{-6t})[/tex]
For [tex]t>1[/tex] (I think you mistakenly wrote [tex]t>0[/tex], which overlaps with the first definition of [tex]g(t)[/tex]), we have
[tex]y'+6y=0\implies e^{6t}y'+6e^{6t}y=(e^{6t}y)'=0\implies y=Ke^{-6t}[/tex]
We want this to be a continuation of the previously found solution [tex]y[/tex] at [tex]t=1[/tex], which means we need to pick [tex]K[/tex] such that
[tex]\dfrac16(1-e^{-6})=Ke^{-6}\implies K=\dfrac16(e^6-1)[/tex]
Then the solution to the IVP is
[tex]y(t)=\begin{cases}\frac16(1-e^{-6t})&\text{for }0\le t\le1\\\frac{e^6-1}6e^{-6t}&\text{for }t>1\end{cases}[/tex]
Alternatively, we can get around treating [tex]g(t)[/tex] piecemeal and resorting to the Laplace transform. Write
[tex]g(t)=\begin{cases}1&\text{for }0\le t\le1\\0&\text{for }t>1\end{cases}=u(t)-u(t-1)[/tex]
where
[tex]u(t-c)=\begin{cases}0&\text{for }t<c\\1&\text{for }t\ge c\end{cases}[/tex]
is the unit step function.
Take the Laplace transform of both sides of the ODE:
[tex]y'+6y=g(t)\overset{\text{L.T.}}{\implies}(sY-y(0))+6Y=\mathcal L_s\{g(t)\}[/tex]
where [tex]Y=Y(s)[/tex] is the Laplace transform of [tex]y(t)[/tex].
We have
[tex]\mathcal L_s\{g(t)\}=\displaystyle\int_0^\infty g(t)e^{-st}\,\mathrm dt=\int_0^1e^{-st}\,\mathrm dt=\dfrac{1-e^{-s}}s[/tex]
so that
[tex](s+6)Y=\frac{1-e^{-s}}s\implies Y=\dfrac{1-e^{-s}}{s(s+6)}=\dfrac{1-e^{-s}}6\left(\dfrac1s-\dfrac1{s+6}\right)[/tex]
Taking the inverse transform yields
[tex]y=\dfrac{1-u(t-1)}6-\dfrac{e^{-6t}}6(e^tu(t-1)-1)[/tex]
[tex]y=\dfrac{1-e^{-6t}}6+\dfrac{e^{6-6t}-1}6u(t-1)[/tex]
which is equivalent to the same solution found earlier; for [tex]0\le t\le1[/tex], [tex]u(t-1)=0[/tex], so that [tex]y=\frac{1-e^{-6t}}6[/tex]; for [tex]t>1[/tex], [tex]u(t-1)=1[/tex], and [tex]y=\frac{1-e^{-6t}}6+\frac{e^{6-6t}-1}6=\frac{(e^6-1)e^{-6t}}6[/tex].
The given differential equation needs to be solved separately for two time ranges because of the piecewise-defined function g(t). Solution for the corresponding equations are founded using the techniques of homogeneous equation solutions and the integrating factor method. These solutions are then matched at the point of continuity.
Explanation:The given differential equation is a first-order linear differential equation of the form y′+p(t)y=g(t). We need to solve this equation considering two cases due to the piecewise definition of g(t).
Case 1: For 0 ≤ t ≤ 1, g(t) = 1. The corresponding homogeneous equation is y' + 6y = 0, with the solution being y(t) = Ce-6t. We find the particular solution using the integrating factor method, yielding y(t) = t/6 - 1/36 + Ce-6t. Substituting the initial condition y(0) = 0 helps us solve for C, giving the final solution for this range as y(t) = t/6 - 1/36.
Case 2: For t > 1, g(t) = 0. The homogeneous solution is the same as in Case 1, but in this case, no particular solution needs to be added, so the solution is y(t) = Ce-6t. The constant is determined by making the function continuous at t=1. We ultimately get y(t) = (1-e-6(t-1))/36.
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Terri and Donna both sell crafts at two different craft shows each weekend. Terri is charged a 5% commission on the amount of money she earns and pays $35 for her booth. Donna is charged a 3% commission on the amount of money she earns and pays $55 for her booth. On the last weekend in November, Terri and Donna both earned the same amount of money at their craft shows. They both paid their respective craft shows the same total amount of money for their booths and commission.
Set up a system of equations to model the amount of money Terri and Donna pay each weekend at the craft shows. Let x represent the money earned from sales, let T represent the total amount Terri pays in one weekend, and let D represent the total amount Donna pays in one weekend.
What is the solution to the system of equations found in Part A? Give your answer as an ordered pair.
What does the solution of the system of equations found in Part B represent in the context of this situation? Be sure to explain the meaning of the values in the solution.
Answer:
(x, T) = (x, D) = (1000, 85)each booth pays $85 in fees on rental and sales of $1000Step-by-step explanation:
A. Given
T = 0.05x +35 . . . . Terri's cost of operating a craft booth
D = 0.03x +55 . . . . Donna's cost of operating a craft boot
T = D
where x is the dollar amount of sales.
__
B. Solution
Subtracting the equation for D from that of T, we get ...
T - D = 0
(0.05x +35) -(0.03x +55) = 0 = 0.02x -20
0 = x -1000 . . . . . divide by 0.02
x = 1000
T = D = 0.05(1000) +35 = 85
(x, T) = (x, D) = (1000, 85)
__
C. Meaning
According to the given definitions of the variables, each booth pays a total of $85 in fees for sales of $1000.
The system of equations is T = 0.05x + 35 and D = 0.03x + 55. The solution is (1000, 85), meaning Terri and Donna each earned $1000 from sales and paid $85 total in booth and commission fees.
Explanation:The system of equations to model the amount of money Terri and Donna pay each weekend at the craft shows can be written as follows: T = 0.05x + 35 and D = 0.03x + 55.
Since we know that Terri and Donna both paid the same total amount of money for their booths and commission, it means T = D. Or, we can equate the two equations: 0.05x + 35 = 0.03x + 55. Solving for x gives us x = 1000. Substituting x = 1000 into T = 0.05x + 35 equation, we get T (and D) = 85. So, the solution to the system of equations is (1000, 85).
In the context of this situation, the solution means that Terri and Donna both earned $1000 from sales, and each paid $85 total for their booth and commission fees.
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A light bulb has a lifetime that is exponential with a mean of 200 days. When it burns out a janitor replaces it immediately. In addition there is a handyman who comes at times of a Poisson process at rate .01 and replaces the bulb as "preventive maintenance." (a) How often is the bulb replaced? (b) In the long run what fraction of the replacements are due to failure?
Answer:
(a) The number of bulbs often replaces is 66.67.
(b) The fraction of the replacements that are due to failure, in the long run, is [tex]\frac{1}{3}[/tex].
Step-by-step explanation:
Let X = lifetime of a bulb and Y = time after which the bulb is replaced.
It is provided that X follows Exponential distribution with mean lifetime of a bulb is, 200 days.
And the rate at which the bulb is replaced is, 0.01 also following an Exponential distribution.
(a)
A bulb is replaced only after it burns out or a handyman comes at times of a Poisson process and replaces it.
Then min (X, Y) follows an Exponential distribution with parameter [tex](\frac{1}{200}+0.01)[/tex].
The mean of an Exponential distribution with parameter θ is:
[tex]Mean=\frac{1}{\theta}[/tex]
Compute the mean of min (X, Y) as follows:
[tex]Mean =\frac{1}{(\frac{1}{200}+0.01)} =\frac{1}{0.015}= 66.67[/tex]
Thus, the number of bulbs often replaces is 66.67.
(b)
Compute the probability of the event (X < Y) as follows:
[tex]P(X<Y)=\frac{0.005}{0.015} =\frac{1}{3}[/tex]
Thus, the fraction of the replacements that are due to failure, in the long run, is [tex]\frac{1}{3}[/tex].
Answer:
(a) The number of bulbs often replaces is 66.67.
(b) The fraction of the replacements that are due to failure, in the long run, is .
Step-by-step explanation:
marco and paolo are working with expressions with rational exponents. marco claims that 642/3=512.which of them is correct? use the rational exponent to justify your answer
Answer:
poalo is correct
Step-by-step explanation
Answer:
Neither Marco nor Paolo is correct, because neither one used the rational exponent property correctly.
Step-by-step explanation:
I just did the test and it told me that I got it right.
Hope that helps! Please mark me brainliest.
Consider the following data on x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127 y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105 Use the general software output to decide whether there is a useful linear relationship between rainfall and runoff.
There is a useful linear relationship between rainfall and runoff and the relationship is y = 0.85x - 2.65
Deciding whether there is a useful linear relationship between rainfall and runoff
From the question, we have the following parameters that can be used in our computation:
x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127
y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105
Using a graphing tool, we have the following summary
Sum of X = 799Sum of Y = 649Mean X = 53.2667Mean Y = 43.2667Sum of squares (SSX) = 20086.9333Sum of products (SP) = 17313.9333The regression equation can be represented as
y = bx + a
Where
b = SP/SSX = 17313.93/20086.93 = 0.86
a = MY - bMX = 43.27 - (0.86*53.27) = -2.65
So, we have
y = 0.85x - 2.65
Hence, there is a useful linear relationship between rainfall and runoff
What point is between 4,16 and 16,16
The point between (4,16) and (16,16) is (10,16) as calculated using the midpoint formula. This point is exactly halfway between the given points.
To determine a point between the two points (4,16) and (16,16), we need to calculate the midpoint.
The formula for finding the midpoint M between two points (x1, y1) and (x2, y2) is:[tex]M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)[/tex]
Putting the coordinates (4,16) and (16,16) into the midpoint formula:[tex]M = \left(\frac{4 + 16}{2}, \frac{16 + 16}{2}\right) = (10, 16)[/tex]
Therefore, the point that lies between (4,16) and (16,16) is (10,16).
Solve using normalcdf
Answer:
0.209
Step-by-step explanation:
Find the sample mean and standard deviation.
μ = 1050
s = 218 / √50 = 30.8
Find the z-score.
z = (1075 − 1050) / 30.8
z = 0.81
Find the probability.
P(Z > 0.81) = 1 − 0.7910 = 0.2090
Suppose the probability that a company will be awarded a certain contract is .25, the probability that it will be awarded a second contract is .21 and the probability that it will get both contracts is .13. What is the probability that the company will win at least one of the two contracts?
Answer:
33% probability that the company will win at least one of the two contracts
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a company is awarded the first contract.
B is the probability that a company is awarded the second contract.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a company is awarded the first contract but not the second and [tex]A \cap B[/tex] is the probability that a company is awarded both contract.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
The probability that it will get both contracts is .13.
This means that [tex]A \cap B = 0.13[/tex]
The probability that it will be awarded a second contract is .21
This means that [tex]B = 0.21[/tex]
[tex]B = b + (A \cap B)[/tex]
[tex]0.21 = b + 0.13[/tex]
[tex]b = 0.08[/tex]
The probability that a company will be awarded a certain contract is .25
This means that [tex]A = 0.25[/tex]
[tex]A = a + (A \cap B)[/tex]
[tex]0.25 = a + 0.13[/tex]
[tex]a = 0.12[/tex]
What is the probability that the company will win at least one of the two contracts?
[tex]A \cup B = a + b + A \cap B = 0.12 + 0.08 + 0.13 = 0.33[/tex]
33% probability that the company will win at least one of the two contracts
The probability that the company will win at least one of the two contracts is [tex]\(\frac{17}{50}\) or 0.34.[/tex]
To find the probability that the company will win at least one of the two contracts, we can use the principle of inclusion-exclusion. The principle states that the probability of the union of two events (in this case, winning either contract) is the sum of the probabilities of each event occurring individually, minus the probability of both events occurring together.
Let [tex]\(P(A)\)[/tex] be the probability that the company will win the first contract, [tex]\(P(B)\)[/tex] be the probability that the company will win the second contract, and [tex]\(P(A \cap B)\)[/tex] be the probability that the company will win both contracts. We are given:
[tex]\(P(A) = 0.25\), \(P(B) = 0.21\), \(P(A \cap B) = 0.13\).[/tex]
The probability that the company will win at least one contract, [tex]\(P(A \cup B)\)[/tex], is given by:
[tex]\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\][/tex]
Substituting the given probabilities:
[tex]\[P(A \cup B) = 0.25 + 0.21 - 0.13\] \[P(A \cup B) = 0.46 - 0.13\] \[P(A \cup B) = 0.33\][/tex]
To express this probability as a fraction, we can write 0.33 as [tex]\(\frac{33}{100}\)[/tex], which simplifies to[tex]\(\frac{17}{50}\)[/tex] when reduced to its simplest form.
Therefore, the probability that the company will win at least one of the two contracts is [tex]\(\frac{17}{50}\)[/tex] or 0.34
solution to 2/3x = -5/2 + 2
Answer:
x=-0.75
Step-by-step explanation:
Combine Like terms
2/3x= -5/2+2
2/3x= -2.5+2
2/3x= -0.5
Multiply both sides by 3
2x= -1.5
Divide both sides by 2
x= -0.75
Answer:
x = -4/3
Step-by-step explanation:
The LCM for the fractions is 6x.
2/3x = -5/2 + 2
Multiply through by 6x
(6x) 2/3x = (6x) -5/2 + (6x)2
4 = -15x + 12x
4 = -3x
x = -4/3
Enjoy Maths!
If mABC= 184°, what is m∠ABC?
88°
90°
84°
92°
mADC=360-mABC=360-184=176⁰
So, measure of angle ABC= mADC/2=176/2=88⁰
The required measure of angle ABC is 88° for the given figure. The correct answer is option A.
What is an arc?The arc is a portion of the circumference of a circle. The circumference of a circle is the distance or perimeter around a circle
The measure of the arc is given as follows:
mABC = 184°
According to the given figure, mADC is the part of the full circle which is complete arc that measure of angle is 360°.
mADC = 360° - mABC
mADC = 360° - 184°
mADC = 176°
As we know that the measure of angle ABC is equal to half of mADC.
The measure of angle ABC = mADC/2
The measure of angle ABC = 176/2
The measure of angle ABC = 88°
Therefore, the required measure of angle ABC is 88°.
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A study was made of seat belt use among children who were involved in car crashes that caused them to be hospitalized. It was found that children not wearing any restraints had hospital stays with a mean of 7.37 days and a standard deviation of 1.25 days with an approximately normal distribution. (a) Find the probability that their hospital stay is from 5 to 6 days, rounded to five decimal places. .10756 (b) Find the probability that their hospital stay is greater than 6 days, rounded to five decimal places.
Answer:
a) [tex]P(5<X<6)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{5-7.37}{1.25}<Z<\frac{6-7.37}{1.26})=P(-1.90<z<-1.10)[/tex]
And we can find this probability with this difference:
[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)[/tex]
And using the norma standard distribution or excel we got:
[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)=0.136-0.029=0.107[/tex]
b) [tex]P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)[/tex]
And using the complement rule we got:
[tex]P(Z>-1.096) =1-P(Z<-1.096) = 1-0.137= 0.863[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(7.37,1.25)[/tex]
Where [tex]\mu=7.37[/tex] and [tex]\sigma=1.25[/tex]
We are interested on this probability
[tex]P(5<X<6)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(5<X<6)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{5-7.37}{1.25}<Z<\frac{6-7.37}{1.26})=P(-1.90<z<-1.10)[/tex]
And we can find this probability with this difference:
[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)[/tex]
And using the norma standard distribution or excel we got:
[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)=0.136-0.029=0.107[/tex]
Part b
For this case we want this probability:
[tex] P(X>6)[/tex]
And we can use the z score and we got:
[tex]P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)[/tex]
And using the complement rule we got:
[tex]P(Z>-1.096) =1-P(Z<-1.096) = 1-0.137= 0.863[/tex]
To find the probability that the hospital stay is from 5 to 6 days, we need to standardize the values using the z-score formula. The probability that their hospital stay is from 5 to 6 days is approximately 0.10756. The probability that their hospital stay is greater than 6 days is approximately 0.86301.
Explanation:To find the probability that the hospital stay is from 5 to 6 days, we first need to standardize the values using the z-score formula.
z = (x - µ) / σ
Let's calculate the z-scores for x = 5 and x = 6.
For x = 5:
z = (5 - 7.37) / 1.25 = -1.896
For x = 6:
z = (6 - 7.37) / 1.25 = -1.096
Next, we can use the standard normal distribution table or a calculator to find the probabilities associated with these z-scores:
P(z < -1.896) = 0.02999
P(z < -1.096) = 0.13699
To find the probability that the hospital stay is from 5 to 6 days, we subtract P(z < -1.096) from P(z < -1.896):
P(5 < x < 6) = P(z < -1.896) - P(z < -1.096) = 0.02999 - 0.13699 = 0.10756
Therefore, the probability that their hospital stay is from 5 to 6 days is approximately 0.10756, rounded to five decimal places.
To find the probability that their hospital stay is greater than 6 days, we can use the standard normal distribution table or a calculator to find the probability associated with the z-score for x = 6:
P(z > -1.096) = 1 - P(z < -1.096) = 1 - 0.13699 = 0.86301
Therefore, the probability that their hospital stay is greater than 6 days is approximately 0.86301, rounded to five decimal places.
Which product is positive?
Suppose the null hypothesis, H0, is: a sporting goods store claims that at least 70% of its customers do not shop at any other sporting goods stores. What is the Type I error in this scenario? a. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, less than 70% of its customers do not shop at any other sporting goods stores. b. The sporting goods store thinks that at least 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores. c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores. d. The sporting goods store thinks that at least 70% of its customers do not shop at any other sporting goods stores when, in fact, less than 70% of its customers do not shop at any other sporting goods stores.
Answer:
Null hypothesis: [tex]p \geq 0.7[/tex]
Alternative hypothesis: [tex]p<0.7[/tex]
A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.
So the correct option for this case would be:
c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.
Step-by-step explanation:
Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.
Type II error, also known as a "false negative" is the error of not rejecting a null hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.
Solution to the problem
On this case we want to test if the sporting goods store claims that at least 70^ of its customers, so the system of hypothesis would be:
Null hypothesis: [tex]p \geq 0.7[/tex]
Alternative hypothesis: [tex]p<0.7[/tex]
A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.
So the correct option for this case would be:
c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.
When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v squared with the force acting in opposition to the motion of the object. A shell of mass 2 kg is shot upward from the ground with an initial velocity of 600 m/sec. If the magnitude of the force due to air resistance is (0.1)v squared, when will the shell reach its maximum height above the ground? What is the maximum height? Assume the acceleration due to gravity to be 9.81 m divided by s squared.
The maximum height reached by the shell is approximately 18255.79 meters.
To find when the shell reaches its maximum height and the value of the maximum height, we need to consider the forces acting on the shell and analyze its motion.
1. Force due to gravity:
The force due to gravity is given by the formula:
Force_gravity = mass × acceleration_due_to_gravity
Here, the mass of the shell is 2 kg, and the acceleration due to gravity is 9.81 m/s².
2. Force due to air resistance:
The force due to air resistance is given by the formula:
Force_air_resistance = (0.1) × velocity²
Here, the velocity of the shell is given as 600 m/s.
Using Newton's second law, we can calculate the net force acting on the shell:
Net force = Force_gravity - Force_air_resistance
When the shell reaches its maximum height, the net force is equal to zero because there is no acceleration at that point. Therefore, we can set the net force equation to zero and solve for the time:
0 = Force_gravity - Force_air_resistance
mass × acceleration_due_to_gravity = (0.1) × velocity²
2 kg × 9.81 m/s² = (0.1) × (600 m/s)²
Simplifying further, we find:
19.62 = 0.1 × 360,000
Time = 600 m/s / 9.81 m/s²
Time ≈ 61.15 seconds
Therefore, the shell will reach its maximum height approximately 61.15 seconds after being shot upward.
To find the maximum height, we can use the kinematic equation:
h = v₀t - (1/2)gt²
Substituting the given values into the equation, we find:
h = (600 m/s) × (61.15 s) - (1/2) × (9.81 m/s²) × (61.15 s)²
h ≈ 18255.79 meters
Therefore, the maximum height reached by the shell is approximately 18255.79 meters.
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The shell will reach its maximum height after approximately 4.51 seconds and this maximum height is around 100 meters.
Explanation:This problem involves the physics of motion under the influence of gravity and air resistance. An important concept here is the 'terminal velocity', which is the velocity at which the upward force (due to the initial impulse provided to the shell) balances out the downward force (due to gravity and air resistance).
First, we must establish that the terminal velocity 'v' of the shell upwards will be achieved when the sum of the forces acting on it are zero. This gives us:
0 = -0.1v^2 + 2 * 9.81
. Solving this quadratic equation reveals v = sqrt(2*9.81/0.1) m/s ≈ 44.3 m/s.
After achieving this terminal velocity, the shell will start decelerating at a rate of 9.81 m/s^2 (the gravity acceleration). The time 't' that takes for the shell to stop moving upwards (so to reach its maximum height) can be calculated using the formula:
t = v / gravity = 44.3/9.81 ≈ 4.51 seconds
.
As for the maximum height 'h', it can be calculated using this formula:
h = v * t + 0.5 * (-gravity) * t^2
By inserting the values of v, gravity, and t in this equation, we find h ≈ 100 m.
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A certain college graduate borrows 7864 dollars to buy a car. The lender charges interest at an annual rate of 13%. Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k dollars per year, determine the payment rate that is required to pay off the loan in 3 years. Also determine how much interest is paid during the 3-year period.
Answer:
Therefore rate of payment = $ 3145.72
Therefore the rate of interest = =$1573.17
Step-by-step explanation:
Consider A represent the balance at time t.
A(0)=$ 7864.
r=13 % =0.13
Rate payment = $k
The balance rate increases by interest (product of interest rate and current balance) and payment rate.
[tex]\frac{dB}{dt} = rB-k[/tex]
[tex]\Rightarrow \frac{dB}{dt} - rB=-k[/tex].......(1)
To solve the equation ,we have to find out the integrating factor.
Here p(t)= the coefficient of B =-r
The integrating factor [tex]=e^{\int p(t) dt[/tex]
[tex]=e^{\int (-r)dt[/tex]
[tex]=e^{-rt}[/tex]
Multiplying the integrating factor the both sides of equation (1)
[tex]e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}[/tex]
[tex]\Rightarrow e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt[/tex]
Integrating both sides
[tex]\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt[/tex]
[tex]\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C[/tex] [ where C arbitrary constant]
[tex]\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}[/tex]
Initial condition B=7864 when t =0
[tex]\therefore 7864= \frac{k}{r} - Ce^0[/tex]
[tex]\Rightarrow C= \frac{k}{r} -7864[/tex]
Then the general solution is
[tex]B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}[/tex]
To determine the payment rate, we have to put the value of B(3), r and t in the general solution.
Here B(3)=0, r=0.13 and t=3
[tex]B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}[/tex]
[tex]\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0[/tex]
⇒k≈3145.72
Therefore rate of payment = $ 3145.72
Therefore the rate of interest = ${(3145.72×3)-7864}
=$1573.17
The payment rate that is required to pay off the loan in 3 years is approximately 2949.51 dollars/year. The interest paid during the 3-year period is about 984.53 dollars.
Explanation:To solve this problem, the formula for continuously compounded interest should be used which is A = P * e^(rt), where A is the value of the investment at a future time t, P is the principal amount, r is the annual interest rate, and t is the time the money is invested or borrowed for.
First, set up the equation: 7864 = k * (1/(0.13)) * (e^(0.13 * 3) - 1), then solve for k: k = 7864 * (0.13) / (e^(0.13 * 3) - 1) ≈ 2949.51 dollars/year.
Using the formula A = P * e^(rt), the total amount paid is A = 2949.51 * 3 = 8848.53 dollars. The interest paid during the 3-year period is calculated by subtracting the loan amount from the total amount paid, that is 8848.53 - 7864 = 984.53 dollars.
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Identify the correct statements for the given functions of set {a, b, c, d} to itself.
1.
f(a) = b, f(b) = a, f(c) = c, f(d) = d is a one-to-one function as each element is an image of exactly one element.
2.
f(a) = b, f(b) = a, f(c) = c, f(d) = d is not a one-to-one function as a is mapped to b and b is mapped to a.
3.
f(a) = b, f(b) = b, f(c) = d, f(d) = c is a one-to-one function as each element is mapped to only one element.
4.
f(a) = b, f(b) = b, f(c) = d, f(d) = c is not a one-to-one function as b is an image of more than one elements.
5.
f(a) = d, f(b) = b, f(c) = c, f(d) = d is a one-to-one function as each element is mapped to only one element.
6.
f(a) = d, f(b) = b, f(c) = c, f(d) = d is not a one-to-one function as d is an image of more than one element.
Answer:
The correct answers are
option(1), option (4),option (5)
Step-by-step explanation:
One to one: Every image has exactly one unique pre-image in domain.
(1)
f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is one to one.
(2)
f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is one to one.
(3)
f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Therefore it is not a one to one mapping.
(4)
f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Therefore it is not a one to one mapping.
(5)
f(a) = d, f(b)=b, f(c)=d, f(d)=c
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is a one to one mapping.
(6)
f(a)=d, f(b)=b,f(c)=c,f(d)=d
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is a one to one mapping.
f(a) = b, f(b) = a, f(c) = c, f(d) = d and f(a) = d, f(b) = b, f(c) = c, f(d) = d are one-to-one function and f(a) = b, f(b) = b, f(c) = d, f(d) = c is not one-to-one function. Then the correct statements are 1, 4, and 5.
What is a function?The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.
One-to-one - every image has exactly one unique pre-image in the domain.
1. f(a) = b, f(b) = a, f(c) = c, f(d) = d
b has a pre image in a domian that is a
a has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
2. f(a) = b, f(b) = a, f(c) = c, f(d) = d
b has a pre image in a domian that is a
a has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
3. f(a) = b, f(b) = b, f(c) = d, f(d) = c
b has a pre image in a domian that is a
b has a pre image in a domian that is b
d has a pre image in a domian that is c
c has a pre image in a domian that is d
Here all elements do have not a unique pre-image in a domain.
Therefore it is not one-to-one.
4. f(a) = b, f(b) = b, f(c) = d, f(d) = c
b has a pre image in a domian that is a
b has a pre image in a domian that is b
d has a pre image in a domian that is c
c has a pre image in a domian that is d
Here all elements do have not a unique pre-image in a domain.
Therefore it is not one-to-one.
5. f(a) = d, f(b) = b, f(c) = c, f(d) = d
d has a pre image in a domian that is a
b has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
6. f(a) = d, f(b) = b, f(c) = c, f(d) = d
d has a pre image in a domian that is a
b has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
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A cupcake stand has 40 chocolate, 30 coconut and 20 banana cupcakes. Alice chooses 20 cupcakes at random to create a box as a present for her friend.
What is the probability that she chose:
(a) Eight banana and 6 coconut cupcakes?
(b) At least 2 chocolate cupcakes?
(c) All cupcakes of the same kind?
Answer:
a) 0.00563
b) 1
c) 0
Step-by-step explanation:
Total = 40+30+20 =90
a) (20C8×30C6×40C6)/90C20
= 0.00563
b) 1 - (no chocolate + 1 chocolate)
1 - [(50C20) + (40C1×50C19)]/90C20
1 - 0.00002478
= 0.9999752187
c) [40C20+20C20+30C20]/90C20
= 0.0000000027045
This is about permutations and combinations.
a) Probability = 0.00563
b) Probability = 0.99997522
c) Probability = 0.0000000027045
We are told cupcakes at the stand are;Chocolate = 40
Coconut = 30
Banana = 20
Total number of chocolates = 40 + 30 + 20
Total number of chocolates = 90
a) Probability that she will choose 8 banana and 6 chocolate cakes if she chooses 20 cupcakes at random will be;
(20C₈ × 30C₆ × 40C₆)/90C₂₀
(125970 × 593775 × 3838380)/50980740277700939310
This gives us 0.00563
b) Probability of at least 2 chocolate cupcakes is;
1 - [P(no chocolate) + P(1 chocolate)]
P(no chocolate) = (50C₂₀)/90C₂₀
P(1 chocolate) = (40C₁ × 50C₁₉)/90C₂₀
Thus;
1 - [P(no chocolate) + P(1 chocolate)] = 1 - [(40C₁ × 50C₁₉) + 50C₂₀]/90C₂₀
This gives us; 0.99997522
c) Probability of getting all cupcakes of same kind is;
(40C₂₀ + 20C₂₀ + 30C₂₀)/90C₂₀
⇒ 0.0000000027045
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An average light bulb manufactured at The Lightbulb Company lasts and average of 300 days, with a standard deviation of 50 days. Suppose the lifespan of a light bulb from this company is normally distributed. (a) What is the probability that a light bulb from this company lasts less than 210 days? More than 330 days?(b) What is the probability that a light bulb from this company lasts between 280 and 380 days?(c) How would you characterize the lifespan of the light bulbs whose lifespans are among the shortest 2% of all bulbs made by this company?(d) If a pack of 6 light bulbs from this company are purchased, what is the probability that exactly 4 of them last more than 330 days?
Answer:
a). 0.03593, 0.27425
(bl 0.0703
(c) the lifespan of such bulb is less than 210 days
(d) 0.0298.
Step-by-step explanation:
Given that U = 300, sd= 50
Z = X-U/sd
(a) We find the probability
Pr(X<210) = Pr(Z<(210-300)/50)
= Pr(Z<-1.8)
= 0.03593
Pr(X >330) = Pr(Z >(330-300)/50))
= Pr(Z> 0.6)
= 1- PR(Z<0.6) = 1 - 0.72575
Pr(X>330)= 0.27425
(b) Pr(280< X< 330) = Pr( 280 < Z< 380)
= Pr([280-300/50] < Z < [380-300/50])
= Pr(0.4 < Z < 1.6)
= Pr(Z< 1.6) - Pr(Z < 0.4)
=0.72575 - 0.65542
= 0.07033
= 0.0703
(c) the lifespan of such bulbs is less than 210 days
(d) given n = 6, x = 4 ,
Pr(X>330) = 0.27425 = p
q = 1-p = 0.72575
Pr(X=4) = 6C4 × (0.27425)⁴ ×(0.72575)²
Pr(X=4) = 10 × 0.005657 × 0.52671
Pr(X= 4) = 0.029795
Pr(X= 4) = 0.0298
Assume lim f(x)-6 and lim g(x)-9. Compute the following limit and state the limit laws used to justify the computation.
x→3 x →3
Lim 3√f(x).g(x)+10
x →3
Using the limit laws of addition and multiplication, the limit of the given equation is solved by combining the calculated limits of the function and the constant, giving an answer of approximately 13.78.
Explanation:To solve this problem, we can use the limit laws of addition and multiplication. The limit laws state that the limit of the sum of two functions is equal to the sum of their individual limits, and the limit of the product of two functions is equal to the product of their individual limits.
So, based on these laws, we first separate the function into two parts: the limit of 3√(f(x).g(x)) as x approaches 3 and the limit of 10 as x approaches 3.
Given lim f(x)=>6 and lim g(x)=>9 when x approaches 3, we multiply these values together to get a product of 54. The cubed root of 54 is approximately 3.78.
The constant 10 has a limit of 10, as constants maintain their value irrespective of the limit.
Adding these values together, 3.78 + 10, gives us a limit of approximately 13.78.
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A builder of houses needs to order some supplies that have a waiting time Y for delivery, with a continuous uniform distribution over the interval from 1 to 4 days. Because she can get by without them for 2 days, the cost of the delay is fixed at $400 for any waiting time up to 2 days. After 2 days, however, the cost of the delay is $400 plus $50 per day (prorated) for each additional day. That is, if the waiting time is 3.5 days, the cost of the delay is $400 $50(1.5)
Find the expected value of the builder’s cost due to waiting for supplies.
Answer:
The Expected cosy of the builder is $433.3
Step-by-step explanation:
$400 is the fixed cost due to delay.
Given Y ~ U(1,4).
Calculating the Variable Cost, V
V = $0 if Y≤ 2
V = 50(Y-2) if Y > 2
This can be summarised to
V = 50 max(0,Y)
Cost = 400 + 50 max(0, Y-2)
Expected Value is then calculated by;
Waiting day =2
Additional day = at least 1
Total = 3
E(max,{0, Y - 2}) = integral of Max {0, y - 2} * ⅓ Lower bound = 1; Upper bound = 4, (4,1)
Reducing the integration to lowest term
E(max,{0, Y - 2}) = integral of (y - 2) * ⅓ dy Lower bound = 2; Upper bound = 4 (4,2)
E(max,{0, Y - 2}) = integral of (y) * ⅓ dy Lower bound = 0; Upper bound = 2 (2,0)
Integrating, we have
y²/6 (2,0)
= (2²-0²)/6
= 4/6 = ⅔
Cost = 400 + 50 max(0, Y-2)
Cost = 400 + 50 * ⅔
Cost = 400 + 33.3
Cost = 433.3
Answer:
the expected value of the builder’s cost due to waiting for supplies is $433.3
Step-by-step explanation:
Due to the integration symbol and also for ease of understanding, i have attached the explanation as an attachment.
onsider a random number generator designed for equally likely outcomes. If numbers between 0 and 99 are chosen, determine which of the following is not correct. a. If 100 numbers are generated comma each integer between 0 and 99 must occur exactly once. b. For each random number generated comma each integer between 0 and 99 has probability 0.01 of being selected. c. If a very large number of random numbers are generated comma then each integer between 0 and 99 would occur close to 1 % of the time. d. The cumulative proportion of times that a 0 is generated tends to get closer to 0.01 as the number of random numbers generated gets larger and larger.
Answer:
The following option is not correct:
(a) If 100 numbers are generated comma each integer between 0 and 99 must occur exactly once.
Step-by-step explanation:
This is not correct, as there is a possibility of an integer being generated twice when 100 numbers are generated.
This can be explained with an example such as stated below:
3 numbers are to be generated.
The number generated can either be 1, 2, or 3.
The probability for all three numbers to be generated once when the generator is run 3 times is:
(1/3)*(1/3)*(1/3) * Number of ways to arrange the three numbers
Thus this probability will be:
Probability = (1/3)^3 * 3!
Probability = 0.222
Since the probability here is not equal to 1, the probability for the same thing happening at a larger scale will also not be 1.
Final answer:
In a random number generator for numbers between 0 and 99, each integer should occur exactly once when 100 numbers are generated.
Explanation:
The correct statement among the options is:
a. If 100 numbers are generated, each integer between 0 and 99 must occur exactly once. This statement holds true in a random number generator designed for equally likely outcomes.Let's break it down:
Option a: To ensure each number between 0 and 99 appears once in 100 numbers, the generator should evenly distribute the outcomes.Option b: The probability of selecting each integer should indeed be 0.01 in an equally likely random number generator.Option c: With a large number of random numbers, each integer between 0 and 99 would indeed occur close to 1% of the time due to the even distribution.Option d: The cumulative proportion of selecting 0 getting closer to 0.01 is a characteristic of equally likely outcomes over a large number of trials.According to Ohm’s Law, the voltage V , current I , and resistance R in a circuit are related by the equation V = I R , where the units are volts, amperes, and ohms. Assume that the voltage is constant with V = 20 V. Calculate the average rate of change of I with respect to R for the interval from R = 6 to R = 6.1 . (Use decimal notation. Give your answer to three decimal places.)
Answer:
The average rate of change as follows of I is -0.185.
Step-by-step explanation:
The relation between voltage V, current I, and resistance R in a circuit, according to the Ohm's law is:
[tex]V = IR[/tex]
It is provided that:
V = 20 V
The interval between which R varies is, R = 8 to R = 8.1.
Compute the value of I as follows:
[tex]V=IR\Rightarrow I=\frac{I}{R}\Rightarrow I=\frac{20}{R}[/tex]
Compute the average rate of change as follows of I as follows:
[tex]\frac{\delta I}{\delta R}=\frac{(I\times8.1)-(I\times8)}{8.1-8}[/tex]
[tex]=\frac{1}{0.1}[\frac{12}{8.1}-\frac{12}{8}][/tex]
[tex]=\frac{12}{0.1}[\frac{8-8.1}{64.8}][/tex]
[tex]=-0.1852[/tex]
Thus, the average rate of change as follows of I is -0.185.
The accounting records of Blossom Company show the following data. Beginning inventory 3,090 units at $5 Purchases 7,930 units at $8 Sales 9,390 units at $10 Determine cost of goods sold during the period under a periodic inventory system using the FIFO method. (Round answer to 0 decimal places, e.g. 1,250.)
Answer:
The cost of goods sold is 65,850
Step-by-step explanation:
FIFO Perpetual chart is attached.
FIFO Perpetual chart shows purchases, sales and balance of the period.
The cost of goods sold is:
3,090 units x $5=$15,450
6,300 units x $8=$50,400
Total=65,850
Seventy independent messages are sent from an electronic transmission center.Messages are processed sequentially, one after another. Transmission time of each message is Exponential with parameter λ = 5 min−1. Find the probability that all 70 messages are transmitted in less than 12 minutes. Use the Central Limit Theorem.
Answer:
The probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Step-by-step explanation:
Let X = the transmission time of each message.
The random variable X follows an Exponential distribution with parameter λ = 5 minutes.
The expected value of X is:
[tex]E(X)=\frac{1}{\lambda}=\frac{1}{5}=0.20[/tex]
The variance of X is:
[tex]V(X)=\frac{1}{\lambda^{2}}=\frac{1}{5^{2}}=0.04[/tex]
Now define a random variable T as:
T = X₁ + X₂ + ... + X₇₀
According to a Central limit theorem if a large sample (n > 30) is selected from a population with mean μ and variance σ² then the sum of random variables X follows a Normal distribution with mean, [tex]\mu_{s} = n\mu[/tex] and variance, [tex]\sigma^{2}_{s}=n\sigma^{2}[/tex].
Compute the probability of T < 12 as follows:
[tex]P(T<12)=P(\frac{T-\mu_{T}}{\sqrt{\sigma_{T}^{2}}}<\frac{12-(70\times0.20)}{\sqrt{70\times 0.04}})\\=P(Z<-1.19)\\\=1-P(Z<1.19)\\=1-0.883\\=0.117[/tex]
*Use a z-table for the probability.
Thus, the probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Following are the solution to the given question:
Let X signify the letter's transmission time [tex]i^{th}[/tex] , and [tex]i =1,2,3,...,70.[/tex] this is assumed that the [tex]X_i \sim Exp (\lambda =5)[/tex].
[tex]\to Mean =\frac{1}{\lambda} =\frac{1}{5}=0.2\\\\ \to Variance =\frac{1}{\lambda^2} =\frac{1}{5^2} =\frac{1}{25}=0.04\\\\[/tex]
Let [tex]T = X_1 + X_2 +...+ X_{70}[/tex], be the total transmission time. According to the Central Limit Theorem, T has a Normal distribution of mean, Variance.mean [tex]= E(T) = 70 \times 0.2=14[/tex]
Variance [tex]=V(T) =70 \times 0.04 = 2.8[/tex]
Therefore
[tex]\to P(T<12) = P (Z < \frac{12-14}{\sqrt{2.8}})[/tex]
[tex]= P(Z<-1.195) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (by sy memetry) \\\\=0.5-P(0<Z<1.19)\\\\=0.5-0.3830 \\\\=0.1170\\\\[/tex]
Therefore, the final answer is "0.1170".
Learn more about the Central Limit Theorem:
brainly.com/question/898534
The daily sales S (in thousands of dollars) that are attributed to an advertising campaign are given by S = 11 + 3 t + 3 − 18 (t + 3)2 where t is the number of weeks the campaign runs. (a) Find the rate of change of sales at any time t.
Answer:
The parking rate at the hospital a parking garage are 50 cents for the 1st hour and 25 cents for each additional hour if guan part in the hospital parking garage for 8 hours how much will the total of charge for parking be