Answer:8
Step-by-step explanation:
We have given test consists of 40 multiple choice questions having five options for each question.
Suppose there is one correct answer to each question
therefore probabilty of getting a correct answer on making a guess is [tex]\frac{1}{5}[/tex]
i.e. 1 out of 5 questions is correct
Using binomial distribution
where n=40 p=[tex]\frac{1}{5}[/tex]
mean of binomial distribution is np
therefore mean of no of correct answers=[tex]40\times \frac{1}{5}[/tex]
=8
Solve x'=5t(sqrt(x)) x(0)=1
Answer:
[tex]2\sqrt{x}=\frac{5t^2}{2}+2[/tex]
Step-by-step explanation:
Given: [tex]\frac{\mathrm{d} x}{\mathrm{d} t}=5t\sqrt{x}\,,\, x(0)=1[/tex]
Solution:
A differential equation is said to be separable if it can be written separately as functions of two variables.
Given equation is separable.
We can write this equation as follows:
[tex]\frac{dx}{\sqrt{x}}=5t\,dt[/tex]
On integrating both sides, we get
[tex]\int \frac{dx}{\sqrt{x}}=\int 5t\,dt[/tex]
Formulae Used:
[tex]\int \frac{1}{\sqrt{x}}=2\sqrt{x}\,\,,\,\,\int t\,dt=\frac{t^2}{2}[/tex]
So, we get solution as [tex]2\sqrt{x}=\frac{5t^2}{2}+C[/tex]
Applying condition: x(0) = 1, we get [tex]C=2[/tex]
Therefore, [tex]2\sqrt{x}=\frac{5t^2}{2}+2[/tex]
Analyze the diagram below and complete the instructions that follow.
Find the value of x and the value of y.
A. x=4, y=8
B. x=7 , y=4√2
C. x=4√3 , y= 7√2
D. x=7√3 , y=4√2
Answer:
D. x=7√3 , y=4√2
Step-by-step explanation:
Step 1: Lets assume an imaginary line from A to B to make the triangle ABC( (refer to the attached image).
AB = 4
BD = 3
Step 2: Find the value of y by the sin formula.
opposite = 4
adjacent = BC
hypotenuse = AC
sin (angle) = opposite/hypotenuse
sin (45) = 4/AC
√2/2 = 4/AC
AC = y = 4√2
Step 3: Find the value of x
Tan (45) = opposite/adjacent
1 = 4/CB
CB = 4
x = CB + BD
x = 4 + 3
x = 7
Therefore, the answer is D where x = 7 and y = 4√2
!!
Homewood Middle School has 1200 students, and 730 of these students attend a summer picnic. If two-thirds of the girls in the school and one-half of the boys in the school attend the picnic, how many girls attend the picnic?
Answer:
520 girls attended the picnic
Step-by-step explanation:
Hello
step 1
Let
Homewood middle school total students=1200
A=unknown=total of girls
(2/3)A=total girls attended the picnic
B=unknown= total of boys
(1/2)B=total boys attended the picnic
step2
replace
[tex]A+B=1200\ equation (1) \\\frac{2A}{3}+\frac{B}{2}=730\ equation(2)\\[/tex]
Step 3
find A and B
from equation (1)
[tex]A=1200-B\ equation\ (3)\\\\[/tex]
from equation (2)
[tex]\frac{2A}{3}+\frac{B}{2}=730\\\frac{2A}{3}=730-\frac{B}{2}\\A=(730-\frac{B}{2})*\frac{3}{2} \\1200-B=1095-\frac{3B}{4}\\A=A\\-B+\frac{3B}{4}=1095-1200\\-\frac{B}{4}=-105\\ B=420\\\\hence\\\\A=1200-B\\A=1200-420\\A=780[/tex]
total girls attended the picnic=(2/3)A=(2/3)780=520
520 girls attended the picnic
Answer:
520
Step-by-step explanation:
i also got the problem and got it wrong - i found the answer is 520
A company provided the following data: Selling price per unit: 60 Variable cost per unit 20 Total fixed costs 400,000 HOw many units must be sole to earn a profit of $40,000?
Answer:
The company needs to sell 9000 units in order to turn a profit of $40,000
Step-by-step explanation:
Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.
From the question we can get some important hints before creating our formula. First the Selling Price will be profit so it will be a positive number, but both unit cost and fixed costs are losses so they will be negative values in our formula. Also our formula will depend on the amount sold which we can represent as the variable x. With these hints we can create our formula as the following
[tex](60x-20x)-400,000 = y[/tex]
Where:
x is the amount of units created and soldy is the total profit after selling x-unitsNow that we have our formula the question asks how many units need to be sold in order to earn a profit of $40,000. We can calculate this by replacing the $40,000 with y and solving for x like so,
[tex](60x-20x)-400,000 = 40,000[/tex] .... add 400,000 on both sides
[tex]40x= 360,000[/tex] ... divide both sides by 40
[tex]x= 9000[/tex]
The company needs to sell 9000 units in order to turn a profit of $40,000
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
A youth basketball coach has 12 kids on his team and he selects 5 kids to start each game. To be fair he wants to start a different group of 5 kids each game. How many different ways can he start 5 of the 12 players? If there is 20 games will he be able to start a different group of 5 kids for each game?
[tex]_{12}C_5=\dfrac{12!}{5!7!}=\dfrac{8\cdot9\cdot10\cdot11\cdot12}{120}=792[/tex]
792>20, so yeah, he will be able.
Answer:
792>20, so yeah, he will be able.
Step-by-step explanation:
n a study of speed dating, male subjects were asked to rate the attractiveness of their female dates, and a sample of the results is listed below (1equalsnot attractive; 10equalsextremely attractive). Construct a confidence interval using a 95% confidence level. What do the results tell about the mean attractiveness ratings of the population of all adult females? 7, 7, 3, 8, 5, 6, 6, 9, 9, 8, 6, 9 What is the confidence interval for the population mean mu? 5.8less thanmuless than 8.1 (Round to one decimal place as needed.) What does the confidence interval tell about the population of all adult females? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. We are 95% confident that the interval from nothing to nothing actually contains the true mean attractiveness rating of all adult females. (Round to one decimal place as needed.) B. The results tell nothing about the population of all adult females, because participants in speed dating are not a representative sample of the population of all adult females. C. We are confident that 95% of all adult females have attractiveness ratings between nothing and nothing. (Round to one decimal place as needed.)
The confidence interval is [tex]\fbox{(5.8, 8.1)}[/tex] and [tex]\fbox{\text{Option A}}[/tex] is correct.
Further Explanation:
Given:
The least value is [tex]1[/tex] that is the least attractive.
The highest value is [tex]10[/tex] that is the most attractive.
The observations are,
7, 7, 3, 8, 5, 6, 6, 9, 9, 8, 6, 9.
Calculation:
The sum of all observations is [tex]83[/tex].
The population mean is [tex]\mu[/tex].
The standard deviation [tex]s[/tex] is [tex]1.832[/tex].
The sample mean [tex]\bar^{X}[/tex] is [tex]6.92[/tex].
Level of significance [tex]\alpha[/tex] = [tex]5\%[/tex].
Formula for confidence interval = [tex]\left( \bar{X} \pm t_{n-1, \frac{\alpha}{2}\%} \frac{s}{\sqrt{n}} \right)[/tex]
Confidence interval =[tex]\left( 6.92 \pm t_{12-1, \frac{5}{2}\%} \frac{1.832}{\sqrt{12}} \right)[/tex]
The value of [tex]t_{11, \frac{5}{2}\%[/tex]=[tex]2.201[/tex]
Confidence interval = [tex]( 6.92 \pm 2.201}\times \frac{1.832}{\sqrt{12}}) \right)[/tex]
Confidence interval = [tex]( 6.92 - 2.201}\times 0.5288 ,6.92 + 2.201}\times \0.5288) \right)[/tex]
Confidence interval = [tex](6.92-1.1639,6.92+1.1639)[/tex]
Confidence interval = [tex]\fbox{(5.8, 8.1)}[/tex]
The [tex]95\%[/tex] confidence interval gives us an idea that [tex]95\%[/tex] chances of the true mean or population mean lies in the interval.
A. We are [tex]95\%[/tex] confident that the interval from [tex]5.8[/tex] to [tex]8.1[/tex] actually contains the true mean of attractiveness rating of all adult females.
B. The results tell nothing about the population of all adult females, because participants in speed dating are not a representative sample of the population of all adult females.
C. We are confident that [tex]95\%[/tex] of all adult females have attractiveness ratings between [tex]5.8[/tex] and [tex]8.1[/tex].
[tex]\fbox{\text{Option A}}[/tex] is Correct as we are [tex]95\%[/tex] confident that the interval from [tex]5.8[/tex] to [tex]8.1[/tex] actually contains the true mean of attractiveness rating of all adult females.
Option B is not correct as the confidence interval tells us about the population mean.
Option C is not correct as the individual rating can be more than [tex]5.8[/tex] or less than the [tex]8.1[/tex].
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Answer Details:
Grade: College Statistics
Subject: Mathematics
Chapter: Confidence Interval
Keywords:
Probability, Statistics, Speed dating, Females rating, Confidence interval, t-test, Level of significance , Normal distribution, Central Limit Theorem, t-table, Population mean, Sample mean, Standard deviation, Symmetric, Variance.
The confidence interval for the population mean attractiveness rating of all adult females is 5.8 to 8.1, and we are 95% confident that this interval contains the true mean attractiveness rating.
In this study of speed dating, male subjects were asked to rate the attractiveness of their female dates on a scale from 1 (not attractive) to 10 (extremely attractive). A sample of attractiveness ratings was provided: 7, 7, 3, 8, 5, 6, 6, 9, 9, 8, 6, 9. The objective is to construct a confidence interval with a 95% confidence level to estimate the true mean attractiveness rating of all adult females.
The calculated confidence interval is 5.8 to 8.1. This means that we are 95% confident that the interval from 5.8 to 8.1 contains the true mean attractiveness rating of all adult females. The margin of error, represented by the range of the interval, provides a level of precision in our estimation.
The confidence interval suggests that the true mean attractiveness rating for all adult females falls within the specified range. It does not provide an exact value for the mean, but it indicates a plausible interval within which the true mean is likely to be found. Additionally, the confidence interval does not imply a causal relationship, but rather a statistical estimate based on the observed sample.
Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent wear a necklace. If one of the students is chosen randomly, what is the probability that this student is wearing (a) a ring or a necklace?
Answer:
0.4
Step-by-step explanation:
Given
60 % wear neither ring nor a necklace
20 % wear a ring
30 % wear necklace
This question can be Solved by using Venn diagram
If one person is choosen randomly among the given student the probability that this student is wearing a ring or necklace is
[tex]P\left ( wear \ ring\ or \ necklace )+P\left ( neither\ ring\ or\ necklace)=1[/tex]
[tex]P\left ( wear \ ring\ or\ necklace )=1-0.6=0.4[/tex]
The sum of probabilty is equal to 1 because it completes the set
Therefore the required probabilty is 0.4
Final answer:
The probability that a randomly chosen student is wearing either a ring or a necklace is 40%. This conclusion is based on the complement of the given percentage of students who wear neither, assuming that there is no overlap in the 20% and 30% who wear rings and necklaces respectively.
Explanation:
To find the probability that a randomly chosen student is wearing a ring or a necklace, we need to understand that 'or' in probability means either one or the other, or both. According to the question, 60% of the students wear neither, which means 40% of the students wear either a ring, a necklace, or both. Since 20% wear a ring and 30% wear a necklace, we might be tempted to add these percentages to get 50%. However, doing so could potentially double-count students who wear both a ring and a necklace.
Without additional information, we can simply state that the probability that a student is wearing a ring or a necklace is the complement of the probability of a student wearing neither, which is 40%. Here we're assuming that students either wear a ring or a necklace or both, as there is no mention of wearing neither in the probabilities given.
Elyria Warehousing desired to locate a central warehouse to serve five Ohio markets. Placed on a grid system, its five markets had coordinates and demand as shown below.
Location Coordinates (miles)
Demand (units)
(45, 65)
2,200
(120, 55)
900
(160, 95)
1,300
(100, 200)
1,750
(195, 175)
3,100
Calculate the Y-coordinate of the center-of-gravity location using the center-of-gravity technique.
69 miles
131 miles
53 miles
197 miles
171 miles
Answer:
The correct option is 2.
Step-by-step explanation:
According to the the center-of-gravity technique, the coordinates of the center-of-gravity location are
[tex](\frac{\sum x_iL_i}{\sum L_i},\frac{\sum y_iL_i}{\sum L_i})[/tex]
Where ([tex](x_i,y_i)[/tex] represent the coordinates and [tex]L_i[/tex] is demand.
We have to find the Y-coordinate of the center-of-gravity location.
The sum of product of demand and corresponding y coordinates is
[tex]\sum y_iL_i=65\times 2200+55\times 900+95\times 1300+200\times 1750+175\times 3100=1208500[/tex]
The sum of demanded units is
[tex]\sum L_i=2200+900+1300+1750+3100=9250[/tex]
The Y-coordinate of the center-of-gravity location is
[tex]y_0=\frac{\sum y_iL_i}{\sum L_i}[/tex]
[tex]y_0=\frac{1208500}{9250}[/tex]
[tex]y_0=130.6486[/tex]
[tex]y_0\approx 131[/tex]
The Y-coordinate of the center-of-gravity location is 131. Therefore the correct option is 2.
The center-of-gravity Y-coordinate is calculated by summing the product of the Y-coordinates and their respective demands, then dividing by the sum of all demands. The Y-coordinate of the center-of-gravity location is approximately 131 miles.
Explanation:The center-of-gravity location is calculated by using a specific formula: (Sum of (Demand * Y-coordinate) /Sum of Demand). Let's use the Y-coordinates given and their corresponding demands.
It would look like this:
((2200*65) + (900*55) + (1300*95) + (1750*200) + (3100*175)) / (2200+900+1300+1750+3100)
Do the math to find out the Y-coordinate:
143,000 + 49,500 + 123,500 + 350,000 + 542,500 = 1,208,500
2200+900+1300+1750+3100 = 9250
1,208,500/9250 = 130.59
So the Y-coordinate of the center-of-gravity location is approximately 131 miles.
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Convert the binary expansion of each of the following integers to a decimal expansion The decimal expansion of (1 1001)_2 is
[tex]1\cdot2^4+1\cdot2^3+1\cdot 2^0=16+8+1=25[/tex]
[tex]11001_2=25_{10}[/tex]
Final answer:
To convert (11001)_2 to decimal, multiply each binary digit by the corresponding power of 2 based on its position, starting from the right. Add the products to get the decimal value, which in this case is 25.
Explanation:
To convert the binary number (11001)_2 to its decimal equivalent, you must understand that each digit represents a power of 2, starting from the rightmost digit which is the least significant bit (LSB). The leftmost digit is the most significant bit (MSB). Now, let's convert (11001)_2 to decimal:
The rightmost digit (1) is in the 20 place, so it is worth 1*20 = 1.
The next digit to the left (0) is in the 21 place, so it is worth 0*21 = 0.
Continuing to the left, the next digit (0) is in the 22 place, so it's worth 0*22 = 0.
The next digit (1) is in the 23 place, so it's worth 1*23 = 8.
Finally, the leftmost digit (1) is in the 24 place, so it's worth 1*24 = 16.
Add up all the values: 16+0+0+8+1 = 25. So, the decimal expansion of (11001)_2 is 25.
Find the remainder when a is divided by m
a = 207^321 + 689! ; m=7
Answer with explanation:
⇒689!=689×688×687×686×........×7×6×5×4×3×2×1
So, 689! is divisible by 7.
We have to find the remainder when,
[tex]\frac{a}{m}=\frac{207^{321}+689!}{7}\\\\=\frac{207^{321}}{7}+\frac{689!}{7}\\\\=\frac{(210-3)^{321}}{7}+0\\\\=\frac{_{0}^{321}\textrm{C}\times (210)^{321}\times (3)^{0} -_{1}^{321}\textrm{C}\times (210)^{320}\times (3)^{1}+_{2}^{321}\textrm{C}\times (210)^{319}\times (3)^{2}-----(-1)\times _{321}^{321}\textrm{C}\times (321)^{0}\times (3)^{321}}{7}\\\\ \text{As ,210 is divisible by 7}\\\\=\frac{ (3)^{321}}{7}[/tex]
⇒[tex]3^7[/tex] ,when divided by 7, gives remainder 3.
[tex]\frac{ (3)^{321}}{7}=\frac{ (3)^{45\times 7+6}}{7}\\\\=3+\frac{3^6}{7}\\\\=3+1\\\\=4[/tex]
So,Remainder when
[tex]\frac{207^{321}+689!}{7}[/tex] is 4.
Write parametric equations for a circle of radius 2, centered at the origin that is traced out once in the clockwise direction for 0 ≤ t ≤ 4π. Use the module to verify your result. (Enter your answer as a comma-separated list of equations. Let x and y be in terms of t.)
Answer:
We know that equation of a circle with origin as it's center is given by
[tex]x^{2}+y^{2}=r^{2}\\\\\therefore x^{2}+y^{2}=2^{2}\\\\(\frac{x}{2})^{2}+(\frac{y}{2})^{2}=1\\\\[/tex]
Comparing with [tex]sin^{2}(\theta )+cos^{2}(\theta )=1[/tex] we get
[tex]\frac{x}{2}=sin(\theta )\\\\\therefore x=2sin(\theta )\\\\\frac{y}{2}=cos(\theta )\\\\\therefore y=cos(\theta )[/tex]
Since [tex]sin(\theta ),cos(\theta )[/tex] have a period of 2π but in the given question we need to increase the period to 4π thus we reduce the argument by 2
[tex]\therefore x=2sin(\frac{\theta }{2})\\\\y=2cos(\frac{\theta }{2})[/tex]
The parametric equations for a clockwise circle of radius 2 centered at origin traced out once for a full revolution (0 ≤ t ≤ 4π) are x = 2 cos(-t/2), y = -2 sin(t/2). This can be confirmed by substituting -t/2 for t in Pythagorean Identity sin²(t) + cos²(t) = 1 which results in 1, proving the correctness of the equations.
Explanation:The parametric equations for a circle of radius 2, centered at the origin, traced out once in the clockwise direction for 0 ≤ t ≤ 2π are x = 2 cos(t), y = 2 sin(t). However, as your question indicates the path traced out in the clockwise direction, the equations would be changed to x = 2 cos(-t) and y = 2 sin(-t). This is achieved by replacing t with -t in the original equations.
In the context of the question, parametric equations which are traced out once for a full revolution (0 ≤ t ≤ 4π in the negative or clockwise direction) are x = 2 cos(-t/2), y = -2 sin(t/2). This is because time is needed twice as much to make a full revolution, so 2t is replaced with t/2.
To verify these equations, you can use the Pythagorean Identity sin²(t) + cos²(t) = 1, substituting -t/2 for t in this identity equation, you will find that the result indeed equals 1, confirming these are the correct parametric equations.
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A and B and n x n matrices such that AB = 0. Prove that if A is invertible then B is not invertible.
Answer and Step-by-step explanation:
Since we have given that
AB = 0
where A and B are n x n matrices.
Consider determinant on both sides,
[tex]\mid AB\mid=\mid 0\mid\\\\\mid A\mid \mid B\mid =0\\\\either\ \mid A\mid =0\ or\ \mid B\mid =0[/tex]
since A is invertible, then |A| ≠ 0
so, it means |B| = 0.
Hence, B is not invertible.
Hence proved.
The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability. The probability of exactly 44 green marbles A. The area between 43.5 and 44.5 B. The area between 43.5 and 45.5 C. The area between 44 and 44.5 D. The area between 43.5 and 44
Answer:
The correct option is A.
Step-by-step explanation:
It is given that the probability of exactly 44 green marbles.
We need to use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability.
A continuity correction is an adjustment that is made when a discrete distribution is approximated by a continuous distribution.
For example: In discrete x=7, then in continuous 6.5<x<7.5. It means we need to subtract 0.5 in the number to find the lower limit and we need to add 0.5 in the given number to find the upper limit.
Similarly,
In discrete x=44, then in continuous
44-0.5<x<44+0.5
43.5<x<44.5
The area between 43.5 and 44.5. Therefore the correct option is A.
Convert 185 to base three
Answer:
20212
Step-by-step explanation:
Divide 185 by [tex]3^4[/tex] as [tex]3^4[/tex] is closest to 185
After dividing we get remainder 23 and quotient 2
Divide the remainder 23 by [tex]3^3[/tex] we get remainder 23 and quotient 0
Divide the remainder 23 by [tex]3^2[/tex] we get remainder 5 and quotient 2
Divide the remainder 5 by [tex]3^1[/tex] we get remainder 2 and quotient 1
Divide the remainder 2 by [tex]3^0[/tex] we get remainder 0 and quotient 2
Taking all the quotient values we get 20212
Hence, 185₁₀=20212₃
Problem Page
A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t) = 120t - 16t^2 . After how long will it reach its maximum height?
Do not round your answer.
Answer: After 3.75 seconds
Step-by-step explanation:
For a Quadratic function in the form [tex]f(x)=ax^2+bx+c[/tex], if [tex]a<0[/tex] then the parabola opens downward.
Rewriting the given function as:
[tex]h(t) = - 16t^2+120t[/tex]
You can identify that [tex]a=-16[/tex]
Since [tex]a<0[/tex] the the parabola opens downward.
Therefore, we can conclude that the x-coordinate of the vertex is the time in seconds in which the ball will reach the maximum height. You can find it with this formula:
[tex]t=\frac{-b}{2a}[/tex]
Substitute values, we get
[tex]t=\frac{-120}{2(-16)}=3.75\ seconds[/tex]
Final answer:
The time at which the ball reaches its maximum height is calculated using the vertex formula for a quadratic equation. In this case, the ball achieves maximum height after 3.75 seconds.
Explanation:
The function given is a quadratic equation in the form h(t) = 120t - 16t2, representing the height of a ball thrown vertically upward after t seconds. To find the time when the ball reaches its maximum height, we need to determine the vertex of the parabola represented by this equation.
The standard form of a quadratic equation is ax2 + bx + c. In this form, the time to reach the maximum height occurs at t = -b/(2a). Applying this to our equation, where a = -16 and b = 120, we calculate t = -120/(2*(-16)) = 120/32 = 3.75 seconds. Therefore, the ball reaches its maximum height after 3.75 seconds.
1.) Given P(A) = 0.28, P(B) = 0.01, P(B|A) = 0.47, what is P(A and B)?
Answer:
p(A and B) =0.1316
Step-by-step explanation:
We know that [tex]p(B|A)=\frac{p(A\cap B)}{p(A)}\\\\p(B\cap A )=p(B|A)\times p(A)[/tex]
Applying values we get
[tex]p(A and B)=0.47\times 0.28\\\\p(AandB)=0.1316[/tex]
The time between unplanned shutdowns of a power plant has an exponential distribution with a mean of 30 days. Find the probability that the time between two unplanned shutdowns is between 18 and 24 days.
The probability that the time between two unplanned shutdowns of a power plant is between 18 and 24 days, following an exponential distribution with a mean of 30 days, is approximately 0.0033.
For an exponential distribution, the probability density function (PDF) is given by:
[tex]\[ f(x) = \lambda e^{-\lambda x} \][/tex]
where [tex]\( \lambda \)[/tex] is the rate parameter, and for an exponential distribution, [tex]\( \lambda = \frac{1}{\text{mean}} \)[/tex].
Given a mean of 30 days, [tex]\( \lambda = \frac{1}{30} \)[/tex].
Now, to find the probability that the time between two unplanned shutdowns is between 18 and 24 days, integrate the PDF over this interval:
[tex]\[ P(18 < X < 24) = \int_{18}^{24} \lambda e^{-\lambda x} \, dx \]\[ P(18 < X < 24) = \int_{18}^{24} \frac{1}{30} e^{-\frac{x}{30}} \, dx \][/tex]
Let's complete the calculations step by step.
[tex]\[ P(18 < X < 24) = \int_{18}^{24} \frac{1}{30} e^{-\frac{x}{30}} \, dx \]\[ P(18 < X < 24) = -\frac{1}{30}e^{-\frac{x}{30}} \Big|_{18}^{24} \][/tex]
Now, evaluate the integral at the upper and lower limits:
[tex]\[ P(18 < X < 24) = -\frac{1}{30} \left(e^{-\frac{24}{30}} - e^{-\frac{18}{30}}\right) \]\[ P(18 < X < 24) = -\frac{1}{30} \left(e^{-0.8} - e^{-0.6}\right) \][/tex]
Using a calculator:
[tex]\[ P(18 < X < 24) \approx -\frac{1}{30} \left(0.44933 - 0.54881\right) \]\[ P(18 < X < 24) \approx -\frac{1}{30} \times (-0.09948) \]\[ P(18 < X < 24) \approx 0.003316 \][/tex]
Therefore, the probability that the time between two unplanned shutdowns is between 18 and 24 days is approximately 0.0033 (rounded to four decimal places).
5. The Fibonacci numbers fn are defined as follows: fi -1, f2-1, and f- fn-1+ f-2 for n 2 3, that is, from the third number on each number is the sum of the previous two numbers. The first ten Fibonacci numbers are: Find f11, f12, f13, and fia.
Answer with explanation:
The Fibonacci series is as follows:and it's first ten entries are
0,1,1,2,3,5,8,13,21,34,.....
You can see that after first two terms, each term that is from third term,sum of previous two consecutive terms.
1→1
2→0
3→1=1+0
4→2=1+1
5→3=2+1
6→5=3+2
7→8=5+3
8→13=5+8
9→21=13+8
10→34=21+13
[tex]\rightarrow f_{11}=f_{10}+f_{9}\\\\=34+21\\\\f_{11}=55\\\\\rightarrow f_{12}=f_{11}+f_{10}\\\\=55+34\\\\f_{12}=89\\\\\rightarrow f_{13}=f_{12}+f_{11}\\\\=89+55\\\\f_{13}=144\\\\\rightarrow f_{ia}=f_{i(a-1)}+f_{i(a-2)}[/tex]
(1 pt) In a study of red/green color blindness, 950 men and 2050 women are randomly selected and tested. Among the men, 89 have red/green color blindness. Among the women, 6 have red/green color blindness. Construct the 99% confidence interval for the difference between the color blindness rates of men and women.
Answer: (0.066,0.116)
Step-by-step explanation:
The confidence interval for proportion is given by :-
[tex]p_1-p_2\pm z_{\alpha/2}\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}[/tex]
Given : The proportion of men have red/green color blindness = [tex]p_1=\dfrac{89}{950}\approx0.094[/tex]
The proportion of women have red/green color blindness = [tex]p_2=\dfrac{6}{2050}\approx0.003[/tex]
Significance level : [tex]\alpha=1-0.99=0.01[/tex]
Critical value : [tex]z_{\alpha/2}=z_{0.005}=\pm2.576[/tex]
Now, the 99% confidence interval for the difference between the color blindness rates of men and women will be:-
[tex](0.094-0.003)\pm (2.576)\sqrt{\dfrac{0.094(1-0.094)}{950}+\dfrac{0.003(1-0.003)}{2050}}\approx0.091\pm 0.025\\\\=(0.09-0.025,0.09+0.025)=(0.066,\ 0.116)[/tex]
Hence, the 99% confidence interval for the difference between the color blindness rates of men and women= (0.066,0.116)
A round trip takes 3.5 hours going one way and 2 hours to return, if the trip back is at a speed 15 mph faster than the speed of the first trip. Find the speeds each way and the distance between the places.
Answer:
first trip 20 mph, trip back 35 mph, distance = 70 miles.
Step-by-step explanation:
A round trip takes 3.5 hours going one way and 2 hour to return.
Let the speed of first trip = v mph
Given that the trip back is at a speed 15 mph faster than the speed of first trip.
Therefore, the speed of trip back = ( v+15 ) mph
Let the Distance between places = d miles
Now we use the formula
[tex]Time=\frac{Distance}{Velocity}[/tex]
[tex]\frac{Distance}{Velocity}[/tex] = 3.5
and [tex]\frac{D}{V+15}[/tex] = 2
dividing these two
[tex]\frac{v+15}{v}[/tex] = [tex]\frac{3.5}{2}[/tex]
2v + 30 = 3.5v
v = 20 mph
d = 3.5 × 20 = 70 miles
The speed of the first trip is 20 mph and the speed of trip back (20+15) 35 mph. and the distance is 70 miles.
A sample of 100 wood and 100 graphite tennis rackets are taken from the warehouse. If 1212 wood and 2020 graphite are defective and one racket is randomly selected from the sample, find the probability that the racket is wood or defective.
Answer: The probability that the racket is wood or defective is 0.6.
Step-by-step explanation:
Since we have given that
Number of wood tennis rackets = 100
Number of graphite tennis rackets = 100
Total number of rackets = 200
Number of wood are defective = 12
Number of graphite are defective = 20
Total number of defectives = 32
We need to find the probability that the racket is wood or defective.
Let A be the event of wood tennis rackets.
Let B be the event of defective.
So, it becomes,
[tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\P(A\cup B)=\dfrac{100}{200}+\dfrac{32}{200}-\dfrac{12}{200}\\\\P(A\cup B)=\dfrac{100+32-12}{200}=\dfrac{120}{200}=0.6[/tex]
Hence, the probability that the racket is wood or defective is 0.6.
there are 6 sets of coins on the table. There is only one coin in the first set, 2 coins in the second set, 3 coins in the third set, 4 coins in the fourth set, 5 coins in the fifth set, and 6 coins in the sixth set. In each step, it is possible to add a coin to any two sets. Is it possible that after some steps the number of coins in each of the sets will be equal
if you answer correctly in first 10 minutes ill give brainliest
Answer:
no
Step-by-step explanation:
The total number of coins in all sets is 21, an odd number. In order for there to be equal numbers in all sets, the total number of coins must be even. Each step adds an even number of coins, so there is no number of steps that will add an odd number of coins to make the total be even.
Assuming that it is known from previous studies that σ = 4.5 grams, how many mice should be included in our sample if we wish to be 95% confident that the mean weight of the sam- ple will be within 3 grams of the population mean for all mice subjected to this protein diet?
Answer:
The required amount of mice is 11.82 gram.
Step-by-step explanation:
Given : Assuming that it is known from previous studies that σ = 4.5 grams. If we wish to be 95% confident that the mean weight of the sample will be within 3 grams of the population mean for all mice subjected to this protein diet.
To find : How many mice should be included in our sample?
Solution :
The formula used in the situation is
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Where, z value at 95% confidence interval is z=1.96
[tex]\mu=3[/tex] gram is the mean of the population
[tex]\sigma=4.5[/tex] gram is the standard deviation of the sample
Substituting the value, to find x sample mean
[tex]1.96=\frac{x-3}{4.5}[/tex]
[tex]1.96\times 4.5=x-3[/tex]
[tex]8.82=x-3[/tex]
[tex]x=8.82+3[/tex]
[tex]x=11.82[/tex]
Therefore, The required amount of mice is 11.82 gram.
A market research firm conducts telephone surveys with a 41% historical response rate. a. What is the probability that in a new sample of 400 telephone numbers, at least 135 individuals will cooperate and respond to the questions? In other words, what is the probability that the sample proportion will be at least 135/400 = 0.3375?
Answer:
probability is 0.778
Step-by-step explanation:
Given data in question
success of probability (p) = 41% = 0.41
no of sample (n) = 400
individual = 150
to find out
probability that the sample proportion will be at least 135/400
solution
first we calculate the mean i.e
mean = p × n
mean = 400 × 0.41
mean = 164
now we calculate standard deviation = [tex]\sqrt{400(0.41) 0.59}[/tex]
standard deviation = 9.83666
we know P(X ≥ 135) = P(Z ≥ (150 - 164)/9.83666)
P(X ≥ 135) = P(Z ≥ (-1.42324)
from table 1 find corresponding probability
P(X ≥ 135) = 1 - P(Z ≥ (-1.42324)
= 1 - 0.9222 = 0.778
probability is 0.778
In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2455 subjects randomly selected from an online group involved with ears. 931 surveys were returned. Construct a 99% confidence interval for the proportion of returned surveys.
The 99% confidence interval for the proportion of returned surveys is between 35.378% and 40.4213%, based on 931 responses from 2455 surveyed subjects.
Explanation:To construct a 99% confidence interval for the proportion of returned surveys, we will use the formula for a proportion confidence interval which includes the sample proportion (π), the z-value that corresponds to the desired level of confidence, and the standard error of the proportion.
The sample proportion (π) can be calculated as the number of returned surveys divided by the total number of surveys sent:
π = 931 / 2455 = 0.379
The z-value for a 99% confidence interval is approximately 2.576. The standard error (SE) of π is calculated using the formula:
SE = √(π(1 - π) / n)
SE = √(0.379(1-0.379)/2455)
SE = √(0.379 * 0.621 / 2455)
SE = √(0.2353 / 2455)
SE = √(0.0000958)
SE = 0.009788
Now, we can calculate the margin of error (ME):
ME = z * SE
ME = 2.576 * 0.009788
ME = 0.025213
Finally, the 99% confidence interval (CI) is calculated as:
CI = π ± ME
CI = 0.379 ± 0.025213
CI = [0.353787, 0.404213]
We can be 99% confident that the true proportion of returned surveys falls between 35.378% and 40.4213%.
The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 6 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.
Answer: 0.375
Step-by-step explanation:
The given interval : (0,6) [in minutes]
Let X represents the waiting time of a passenger.
We know that the cumulative uniform distribution function for interval (a,b) is given by :_
[tex]F(x)=\begin{cases}0,&\text{ for } x<a\\\frac{x-a}{b-a},& \text{for } a\leq x\leq 1\\1,& \text{for }x>b\end{cases}[/tex]
Then , the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes. is given by :_
[tex]P(2.25<x)=\dfrac{2.25-0}{6-0}=0.375[/tex]
Hence, the required probability : 0.375
Answer:
Step-by-step explanation:
the answer is 0.625 reason is the probability is 62.5% and when you take
(6-2.25)/(6-0)
= (3.75)/(6)
=0.625
A refrigerator contains 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, and 2 mangos. Imagine you stick your hand into the refrigerator and pull out a piece of fruit at random. What is the chance you don't get an apple? 10/4410/44 6/446/44 38/44
The chance you don't get an apple is:
[tex]\dfrac{38}{44}[/tex]
Step-by-step explanation:We know that the probability of an outcome is the chance of getting an outcome and it is calculated by:
Taking the ratio of number of favorable outcomes to total number of outcomes.
A refrigerator contains 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, and 2 mangoes.
Total number of fruits in the refrigerator i.e. total number of outcomes are: 6+5+10+3+7+11+2
= 44
Also the number of favorable outcomes i.e. number of fruits which are not apples in the refrigerator are: 44-6=38
This means that the probability of not getting an apple is:
[tex]\text{Probability(not\ getting\ an\ apple)}=\dfrac{38}{44}[/tex]
Final answer:
The chance you do not get an apple is38/44, or roughly86.36.
Explanation:
The chance you do not get an apple is38/44.
To calculate this, add up the total number of fruits in the refrigerator banning apples, which is 5 oranges 10 bananas 3 pears 7 peaches 11 catches 2 mangos = 38 fruits. The total number of fruits in the refrigerator is 6 apples 5 oranges 10 bananas 3 pears 7 peaches 11 catches 2 mangos = 44 fruits.
simplifies to19/22 or roughly0.8636, so the chance you do not get an apple is0.8636 or86.36.
Four hundred people were asked whether gun laws should be more stringent. Three hundred said "yes," and 100 said "no". The point estimate of the proportion in the population who will respond "yes" is
Answer:
0.75
Step-by-step explanation:
We have been given that 400 people were asked whether gun laws should be more stringent. 300 hundred said "yes," and 100 said "no".
To find the point estimate of the proportion in the population who will respond "yes", we need to divide number of people who said yes by total number of people that is 300 by 400.
[tex]\text{People who will respond yes}=\frac{300}{400}[/tex]
[tex]\text{People who will respond yes}=\frac{3}{4}[/tex]
[tex]\text{People who will respond yes}=0.75[/tex]
Therefore, the point estimate of the proportion in the population who will respond "yes" is 0.75.
A recent article in a college newspaper stated that college students get an average of 5.5 hrs of sleep each night. A student who was skeptical about this value decided to conduct a survey by randomly sampling 25 students. On average, the sampled students slept 6.25 hours per night. Identify which value represents the sample mean and which value represents the claimed population mean.
Answer: Population mean = [tex]\mu=5.5[/tex]
Sample mean = [tex]\overline{x}=6.25[/tex]
Step-by-step explanation:
We know that the population mean [tex]\mu[/tex] is the average of the entire population.
The sample mean [tex]\overline{x}[/tex] is the mean of the sample which is derived from the whole population randomly.
Given : A recent article in a college newspaper stated that college students get an average of 5.5 hrs of sleep each night.
Thus , the population mean = 5.5 hrs
Also, On average, the sampled students slept 6.25 hours per night.
It implies , the sample mean = 6.25 hours
Final answer:
The sample mean is 6.25 hours of sleep per night from the student's survey of 25 students, while the claimed population mean is 5.5 hours as stated in the college newspaper article.
Explanation:
In this scenario, the sample mean is the average amount of sleep that the 25 randomly sampled students reported, which is 6.25 hours per night.
The claimed population mean is the value mentioned in the college newspaper article, stating that college students get an average of 5.5 hours of sleep each night.
The sample mean represents the average found from the sample taken by the student, while the claimed population mean represents the average that is supposedly true for the entire population of college students.
3. [6 Pts] Give an example of two uncountable sets A and B such that AnB is (a) finite, (b) countably infinite, (c) uncountably infinite.
Answer with explanation:
Let two uncountable sets A and B .
(a).Let A= [2,3]=uncountable
B= [3,4)=uncountable
[tex]A\cap B [/tex]={3}= finite
Hence, [tex] A\cap B [/tex] is finite set .
(b).Let A= Set of positive real numbers=Uncountable
B= Set of negative real numbers and positive integers= Uncountable
Now, [tex]A\cap B[/tex]=Set of positive integer numbers =countably infinite set.
Hence, [tex]A\cap B[/tex]is countably infinite .
(c). Let A= Set of real numbers=Uncountable
B= Set of irrational numbers =Uncountable
Then, [tex]A\cap B[/tex]= Set of irrational numbers= Uncountable.
Hence, [tex]A\cap B[/tex] is uncountable when A is set of real numbers and B is set of irrational numbers.
Real numbers A and B can be examples of the uncountable sets. By modifying these, their intersection could be finite, countably infinite, or uncountably infinite based on its definition. These sets and their intersections depict the distinct characteristics of uncountable sets.
Explanation:In Mathematics, an uncountable set is a set that cannot be put into one-to-one correspondence with the set of natural numbers or integers. In other words, it has more elements than the set of natural numbers.
Since the intersection of two uncountable sets could be any size (from empty to uncountably infinite), we can creatively define the sets to meet the requirements. Consider the real numbers A = [0,1] and B = [1,2] as our uncountable sets.
For (a) finite, let's modify B and set B to only include the single element {1}, their intersection, which is finite.
For (b) countably infinite, instead B = {x in Q | 1 ≤ x ≤ 2}, where Q is the set of rational numbers. They intersect at countably infinite set of rational numbers between 1 and 2.
For (c) uncountably infinite. If both A and B are intervals on the real number line that have nonempty intersection, their intersection is uncountably infinite.
Learn more about Uncountable Sets here:
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