Answer:
[tex] \bar X = \frac{1435040}{80}=17938[/tex]
Step-by-step explanation:
Since we have a groued data for this case we can construct the following table to find the expected value.
Interval Frequency(fi) Midpoint(xi) xi*fi
5001-10000 16 7500.5 120008
10001-15000 14 12500.5 175007
15001-20000 15 17500.5 262507.5
20001-25000 17 22500.5 382508.5
25001-30000 18 27500.5 495009
Total 80 1435040
And we can calculate the mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n f_i x_i}{n}[/tex]
Where [tex] n=\sum_{i=1}^n f_i = 80[/tex]
And if we replace we got:
[tex] \bar X = \frac{1435040}{80}=17938[/tex]
A company buys computers and printers. Each computer costs $550 and each printer costs $390. If the company spends $8160 and buys a total of 16 machines, how many of each did it buy?
Answer:the company bought 12 computers and 4 printers.
Step-by-step explanation:
Let x represent the number of computers that the company bought.
Let y represent the number of printers that the company bought.
The company buys a total of 16 machines. It means that
x + y = 16
Each computer costs $550 and each printer costs $390. If the company spends $8160 for all the computers and printers that was bought, it means that
550x + 390y = 8160 - - - - - - - - - - 1
Substituting x = 16 - y into equation 1, it becomes
550(16 - y) + 390y = 8160
8800 - 550y + 390y = 8160
- 550y + 390y = 8160 - 8800
- 160y = - 640
y = - 640/ - 160
y = 4
Substituting y = 4 into x = 16 - y, it becomes
x = 16 - 4
x = 12
The company bought 12 computers and 4 printers.
Explanation:To solve this problem, we can set up a system of equations. Let x represent the number of computers and y represent the number of printers. We have two equations: x + y = 16 and 550x + 390y = 8160. We can solve this system by substitution or elimination. Let's use elimination.
Multiply the first equation by 390: 390x + 390y = 6240. Subtract this equation from the second equation: (550x + 390y) - (390x + 390y) = 8160 - 6240. Simplify: 160x = 1920. Divide both sides by 160: x = 12.
Substitute this value into the first equation: 12 + y = 16. Solve for y: y = 4. Therefore, the company bought 12 computers and 4 printers.
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An article reported the following data on oxygen consumption (mL/kg/min) for a sample of ten firefighters performing a fire-suppression simulation: 28.6 49.4 30.3 28.2 28.9 26.4 33.8 29.9 23.5 30.2Compute the following. (Round your answers to four decimal places.) a. The sample range mL/kg/minb. The sample variance s2 from the definition (i.e., by first computing deviations, then squaring them, etc.) mL2/kg2/min2c. The sample standard deviation mL/kg/mind. s2 using the shortcut method mL2/kg2/min
Answer:
a) The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]
b) The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]
c) The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]
Step-by-step explanation:
We are given the following data on oxygen consumption (mL/kg/min):
28.6, 49.4, 30.3, 28.2, 28.9, 26.4, 33.8, 29.9, 23.5, 30.2
a) The sample range
Range = Maximum - Minimum
[tex]\text{Range} = 49.4 - 23.5 = 25.9[/tex]
The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]
b) The sample variance
[tex]\text{Variance} = \displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{309.2}{10} = 30.92[/tex]
Sum of squares of differences =
5.3824 + 341.5104 + 0.3844 + 7.3984 + 4.0804 + 20.4304 + 8.2944 + 1.0404 + 55.0564 + 0.5184 = 444.096
[tex]s^2 = \dfrac{444.096}{9} = 49.344[/tex]
The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]
c) The sample standard deviation
It is the square root of sample variance.
[tex]s = \sqrt{s^2} = \sqrt{49.344} = 7.0245[/tex]
The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]
If angle X and angle Y are complementary angles which must be true?
Answer:
sinx = cosy
Step-by-step explanation:
option D is true because of the relationship between cos and sine
costheta = sine(90 - theta)
sine x = cos (90-x)
but x + y = 90
sine x = cos(x + y -x)
sineX = cos Y
The following confidence interval is obtained for a population proportion, p: 0.408 < p < 0.432
Use these confidence interval limits to find the margin of error, E.
A. 0.012
B. 0.013
C. 0.024
D. 0.420
Answer:
For this case the wisth of the interval represent 2ME and we have this:
[tex] 0.432-0.408 = 2ME[/tex]
And if we solve for ME we got:
[tex] ME = \frac{0.432-0.408}{2}=0.012[/tex]
So then the correct answer would be:
A. 0.012
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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 population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
The margin of error is given by:
[tex] ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
For this case the wisth of the interval represent 2ME and we have this:
[tex] 0.432-0.408 = 2ME[/tex]
And if we solve for ME we got:
[tex] ME = \frac{0.432-0.408}{2}=0.012[/tex]
So then the correct answer would be:
A. 0.012
On a circle of radius 5 feet, give the degree measure of the angle that would subtend an arc of length 1 feet. Round your answer to the nearest hundredth, or two decimal places.
Answer:
11.46°
Step-by-step explanation:
Let x be the angle that yields an arc length of 1 feet, if r= 5 feet, applying the circumference length equation and assuming that a full circumference has 360 degrees:
[tex]1=\frac{x}{360}*2\pi r \\1=\frac{x}{360}*2\pi 5\\x=\frac{360}{2 \pi 5}=11.46^o[/tex]
Rounding to the nearest hundredth, the angle should be 11.46°
Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3 ), t = time (min), k = the evaporation rate (mm/min), and A = surface area (mm2 ). Use Euler’s method to compute the volume of the droplet from t = 0 to 10 min using a step size of 0.25 min. Assume that k = 0.08 mm/min and that the droplet initially has a radius of 2.5 mm. Assess the validity of your results by determining the radius of your final computed volume and verifying that it is consistent with the evaporation rate.
Answer:
V = 20.2969 mm^3 @ t = 10
r = 1.692 mm @ t = 10
Step-by-step explanation:
The solution to the first order ordinary differential equation:
[tex]\frac{dV}{dt} = -kA[/tex]
Using Euler's method
[tex]\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\[/tex]
Where initial droplet volume is:
[tex]V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3[/tex]
Hence, the iterative solution will be as next:
i = 1, ti = 0, Vi = 65.45[tex]V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88[/tex]
i = 2, ti = 0.5, Vi = 63.88[tex]V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33[/tex]
i = 3, ti = 1, Vi = 62.33[tex]V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813[/tex]
We compute the next iterations in MATLAB (see attachment)
Volume @ t = 10 is = 20.2969
The droplet radius at t=10 mins
[tex]r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\[/tex]
The average change of droplet radius with time is:
Δr/Δt = [tex]\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min[/tex]
The value of the evaporation rate is close the value of k = 0.08 mm/min
Hence, the results are accurate and consistent!
Using Euler's method, we approximated the volume and surface area of the object over a period of 10 minutes. Starting with an initial radius of 2.5 mm, and given a decay constant of 0.08 mm/min, we computed the final volume to be approximately 26.19 mm³ and the final surface area to be about 31.70 mm². This resulted in a final radius of approximately 2.06 mm.
To solve this problem using Euler's method, we'll use the following formulas:
dV/dt = -kA
A = 4πr²
V = (4/3)πr³
Given that the initial radius r₀ = 2.5 mm, we can compute the initial volume V₀ and surface area A₀. Then, we'll iterate through time steps using Euler's method:
Vₙ₊₁ = Vₙ - kA * Δt
Aₙ₊₁ = 4π(rₙ - (k/3) * Δt)²
Using k = 0.08 mm/min and a step size of Δt = 0.25 min, we perform the calculations:
V₀ = (4/3)π(2.5)³ ≈ 65.45 mm³
A₀ = 4π(2.5)² ≈ 98.17 mm²
Iterating:
t = 0.25 min:
V₁ ≈ 65.45 - 0.08 * 98.17 * 0.25 ≈ 61.44 mm³
A₁ ≈ 4π(2.44)² ≈ 59.17 mm²
t = 0.5 min:
V₂ ≈ 61.44 - 0.08 * 59.17 * 0.25 ≈ 59.58 mm³
A₂ ≈ 4π(2.42)² ≈ 58.01 mm²
Continuing this process until t = 10 min, we obtain:
V₄₀ ≈ 26.19 mm³
A₄₀ ≈ 31.70 mm²
Finally, calculating the radius r₄₀ corresponding to V₄₀:
r₄₀ = ((3V₄₀)/(4π))^(1/3) ≈ 2.06 mm
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Of all the luggage handled by the airlines at JFK in the last 10 years, 5% was lost or stolen or damaged. For a randomly selected sample of 400 pieces of outgoing luggage at JFK International Airporta) Would it be unusual for 10 pieces of luggage to be lost?b) The mean and standard deviation of the binomial distribution is given byc) Would 40 pieces of luggage lost be an unusual occurrence?d) Find the probability that less than 6 pieces of luggage are lost or stolen.
Answer:
a) 10 pieces of luggage to be lost is not likely because its z-score is -2.29
b) mean=20 and standard deviation=4.359
c) 40 pieces of luggage lost is very extreme given that in average 5% of the luggage is lost.
d) The probability that less than 6 pieces of luggage are lost or stolen is 0.0007
Step-by-step explanation:
Let p be the proportion of stolen or damaged luggages of all the luggage handled by the airlines at JFK in the last 10 years.
p=0.05 (5%)
b) Mean and standard deviation of the binomial distribution is given as:
Mean= n×pStandard Deviation = [tex]\sqrt{n*p*(1-p)}[/tex] where n is the sample sizeThus,
Mean=400×0.05=20 and
Standard Deviation=[tex]\sqrt{400*0.05*0.95}[/tex] ≈ 4.359
a) We should calculate z-score of 10 pieces of luggage lost to decide if it is unusual.
Z-score can be calculated as follows:
[tex]z= \frac{X-M}{s}}[/tex] where
X = 10 pieces of luggage lost M is the mean lost luggage (20 pieces)s is the standard deviation (4.359)Thus, [tex]z= \frac{10-20}{4.359}[/tex] ≈ -2.29
This is unusual because it is in the 1st percentile and 99% of the possible number of lost luggage are higher than this score.
c) 40 pieces of lost luggage is also unusual because its z score is:
[tex]z= \frac{40-20}{4.359}[/tex] ≈ 4.59, in the 100th percentile, that is very extreme.
d) the probability that less than 6 pieces of luggage are lost or stolen
= P(z<z*) where z* is the z-score of 6 pieces of luggage are lost or stolen
That is [tex]z= \frac{6-20}{4.359}[/tex] ≈ -3.21
And P(z<3.21)= 0.0007
A digital scale that provides weights to the nearest gram is used. a. What is the sample space for this experiment? Let A denote the event that a weight exceeds 11 grams, let B denote the event that a weight is less than or equal to 15 grams, and let C denote the event that a weight is greater than or equal to 8 grams and less than 12 grams. Describe the following events. b. A ∪ B c. A ∩ B d. A' e. A ∪ B ∪ C f. (A ∪ C)' g. A ∩ B ∩ C h. B' ∩ C i. A ∪ (B ∩ C)
Answer:
a) [tex] S= [x \in x \geq 0][/tex]
Because the weigth can't be negative.
b) AUB = "a weight exceeds 11 grams OR is less than or equal to 15 grams" and that is represent by all the sample space S.
c) A ∩ B ="a weight exceeds 11 grams AND is less than or equal to 15 grams" and that is represent by [tex] 11 < X \leq 15 [/tex] who is the same as [tex] 12 \leq X \leq 15[/tex] .
d) A' = "a weight NOT exceeds 11 grams" [tex] X \leq 11[/tex] that's the complement of the event A
e) A ∪ B ∪ C = "represent all the possib;e values for the sample space or S"
f) (A ∪ C)'="for this case (AUC) represent the weigths that exceeds 11 gr OR are between 8 and less than 11, so on this case values [tex] X \geq 8[/tex], so then the complement (AUC)' woudl be all the values [tex] X <8[/tex]"
g) A ∩ B ∩ C =[tex]\emptyset[/tex] since we don't have a common interval for the 3 events at the same time
h) B' ∩ C = The complement of B are the [tex] X>15[/tex] and for C we have values [tex] 8 \leq X <12[/tex] and the intersection between these two events is the [tex]\emptyset[/tex].
i) A ∪ (B ∩ C) = For this case (B ∩ C) represent the values between [tex] 8\leq X <12[/tex] and if we do the union A ∪ (B ∩ C) we got [tex] X \geq 8[/tex]
Step-by-step explanation:
For this case we have defined the following events, assuming that X represent the weight:
A= "a weight exceeds 11 grams" [tex]X>11[/tex]
B= " a weight is less than or equal to 15 grams" [tex] X \leq 15[/tex]
C= "a weight is greater than or equal to 8 grams and less than 12 grams" [tex] 8 \leq X < 12[/tex]
Part a
The sample space is given by:
[tex] S= [x \in x \geq 0][/tex]
Because the weigth can't be negative.
Part b
AUB = "a weight exceeds 11 grams OR is less than or equal to 15 grams" and that is represent by all the sample space S.
Part c
A ∩ B ="a weight exceeds 11 grams AND is less than or equal to 15 grams" and that is represent by [tex] 11 < X \leq 15 [/tex] who is the same as [tex] 12 \leq X \leq 15[/tex] .
Part d
A' = "a weight NOT exceeds 11 grams" [tex] X \leq 11[/tex] that's the complement of the event A
Part e
A ∪ B ∪ C = "represent all the possib;e values for the sample space or S"
Part f
(A ∪ C)'="for this case (AUC) represent the weigths that exceeds 11 gr OR are between 8 and less than 11, so on this case values [tex] X \geq 8[/tex], so then the complement (AUC)' woudl be all the values [tex] X <8[/tex]"
Part g
A ∩ B ∩ C =[tex]\emptyset[/tex] since we don't have a common interval for the 3 events at the same time
Part h
B' ∩ C = The complement of B are the values in[tex] X>15[/tex] and for C we have values [tex] 8 \leq X <12[/tex] and the intersection between these two events is the [tex]\emptyset[/tex].
Part i
A ∪ (B ∩ C) = For this case (B ∩ C) represent the values between [tex] 8\leq X <12[/tex] and if we do the union A ∪ (B ∩ C) we got [tex] X \geq 8[/tex]
The question refers to the mathematical concept of events and their union and intersection. In this context, A, B, and C represent specific weight ranges. Various combinations of these, such as A ∪ B or A ∩ B ∩ C, refer to different sets of possible weights.
Explanation:Sample space in this experiment would basically consist of all possible weights that the digital scale can measure to the nearest gram. In Principle, it's unlimited but practically it will depend on the maximum limit of the scale.
A is the event that a weight exceeds 11 grams. B is the event that a weight is less than or equal to 15 grams. C is the event that a weight is greater than or equal to 8 grams and less than 12 grams.
A ∪ B represents weights that exceed 11 grams or are less than or equal to 15 grams.A ∩ B refers to weights that are both more than 11 grams and less than or equal 15 grams.A' is the event of weights that do not exceed 11 grams.A ∪ B ∪ C comprises weights that are either more than 11 grams, less than or equal to 15 grams, or between 8 and 12 grams.(A ∪ C)' refers to weights that are not more than 11 grams and not between 8 and 12 grams.A ∩ B ∩ C represents weights that are more than 11 grams, less than or equal to 15 grams, and simultaneously between 8 and 12 grams.B' ∩ C indicates weights that are more than 15 grams and at the same time between 8 and 12 grams.A ∪ (B ∩ C) signifies weights that are more than 11 grams, or are less than or equal to 15 grams and between 8 and 12 grams.Learn more about Sets and Probability here:https://brainly.com/question/11179193
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Which equation shows this relationship?
Answer: y=x+2
Step-by-step explanation:
Find a unit vector orthogonal to the plane containing the points A= 1,0,0 , B= 3,−1,−3 , and C= 1,3,−2 .
Answer:
unit normal vector n will be n=(a,b,c) = (4/√171,11/√171,6/√171)
Step-by-step explanation:
There are several ways to solve this problem
1) build 2 vectors AB and BC such that the vectorial product ABxBC is the orthogonal vector to the plane , then find unit vector
2) since the 3 points belongs to the plane solve a linear system of 4 equation with 4 variables
for the second solution , the equation of the plane with normal vector n=(a,b,c) and containing the point (x₀,y₀,z₀) is
a*(x-x₀)+b*(y-y₀)+c*(z-z₀) =0
and
a²+b²+c² = 1 (unit vector)
then choosing A=(x₀,y₀,z₀)=(1,0,0)
a*(x-1)+b*(y-0)+c*(z-0) =0
for B
a*(3-1)+b*(-1-0)+c*(-3-0) =0
1) 2*a - b - 3*c =0
for C
a*(1-1)+b*(3-0)+c*(-2-0) =0
2) 3*b - 2*c=0 → b= 2/3*c
replacing in 1)
2*a - 2/3*c - 3*c =0
2*a-11/3*c=0 → a=11/6*c
thus
a²+b²+c² = 1
(11/6*c)²+(2/3*c)²+c² = 1
(121/36+4/9+1)*c² = 1
171/36*c²=1 → c= 6/√171
therefore
a=11/6*c = 11/6*6/√171= 11/√171
b=2/3*c= 2/3*6/√171= 4/√171
then the unit normal vector n will be
n=(a,b,c) = (4/√171,11/√171,6/√171)
Students are working on a project which requires toothpicks. Each student receives the same number of toothpicks, t, from the
teacher. Which equation can be used to find the total number of toothpicks, N, given out by the teacher for the project to
s students?
OA. N =
OB. N = st
OC. N = s +t
OD. N =
Reset
Submit
Answer:
The correct answer is B. N = st
Step-by-step explanation:
1. Let's review the information given to us to answer the question correctly:
t = number of toothpicks received by each student
s = number of students
N = total number of students
2. Which equation can be used to find the total number of toothpicks, N, given out by the teacher for the project?
Total number of toothpicks for the project = number of toothpicks received by each student * number of students, replacing with the variables:
N = t * s = s * t
The correct answer is B. N = st
Answer:
B. N =st
Step-by-step explanation:
there are 21 members of the soccer team on the us. if each player carries on 4 pieces of equipment, how many pieces of equipment are on the bus?
Answer:
84
Step-by-step explanation:
21x4=84
Answer:the total number of pieces of equipment in the bus is 84
Step-by-step explanation:
The total number of members of the soccer team on the bus is 21.
if each player carries on 4 pieces of equipment, then the total number of pieces of equipment in the bus would be
21 × 4 = 84
To estimate the mean score μ μ of those who took the Medical College Admission Test on your campus, you will obtain the scores of an SRS of students. From published information you know that the scores are approximately Normal with standard deviation about 6.2 6.2 . You want your sample mean ¯ x x¯ to estimate μ μ with an error of no more than 1.4 1.4 point in either direction. (a) What standard deviation must ¯ x x¯ have so that 99.7 % 99.7% of all samples give an ¯ x x¯ within 1.4 1.4 point of μ μ ? Use the 68 – 95 – 99.7 68–95–99.7 rule. (Enter your answer rounded to four decimal places.
Answer: 0.4667
Step-by-step explanation:
According to 68–95–99.7 rule , About 99.7% of all data values lies with in 3 standard deviations from population mean ([tex]\mu[/tex]).
Here , margin of error = 3s , where s is standard deviation.
As per given , we have want our sample mean [tex]\overline{x}[/tex] to estimate μ μ with an error of no more than 1.4 point in either direction.
If 99.7% of all samples give an [tex]\overline{x}[/tex] within 1.4 , it means that
[tex]3s=1.4[/tex]
Divide boths ides by 3 , we get
[tex]s=0.466666666667\approx0.4667[/tex]
Hence, So [tex]\overline{x}[/tex] must have 0.4667 as standard deviation so that 99.7 % 99.7% of all samples give an [tex]\overline{x}[/tex] within 1.4 point of μ .
To estimate the mean score of an MCAT with an error of no more than 1.4 points, one needs to calculate the standard deviation for the sample mean that allows 99.7% of samples to fall within this range. After determining this standard deviation, apply the Central Limit Theorem to increase sample size and approach the population mean.
Explanation:To estimate the mean score μ of those who took the Medical College Admission Test on your campus with an error of no more than 1.4 points in either direction, you are seeking a standard deviation for the sample mean ¯x that will allow for 99.7% of samples to fall within this range. This relates to the 68 – 95 – 99.7 rule, which in this context will be interpreted as within three standard deviations of the mean, hence, you need ¯x to have a standard deviation of 1.4/3.
Using the formula for ¯x standard deviation which is σ/√n (where σ is population standard deviation and n is the sample size), you can rewrite the formula as 1.4/3 = 6.2/√n and solve for n to get the desired sample size.
This use of the empirical rule and standard deviations is part of a bigger concept known as the Central Limit Theorem, which states that as sample size increases, the sample mean gets nearer to the population mean.
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Find an autonomous differential equation with all of the following properties:Equilibrium solutions at y=0 and y=3, y' > 0 for 0 < y < 3; and y' < 0 for -inf < y < 0 and 3 < y < inf.
The differential equation y' = y(y - 3) satisfies all the given conditions. It has equilibrium solutions at y = 0 and y = 3, y' > 0 for 0 < y < 3, and y' < 0 for -inf < y < 0 and 3 < y < inf.
Explanation:The autonomous differential equation that meets the given conditions is
y' = y(y-3). To confirm this, we can check each condition:
Equilibrium solutions at y=0 and y=3: Setting y' = 0 and solving for y in the differential equation gives y = 0 and y = 3, which satisfies this condition.y' > 0 for 0 < y < 3: The expression y(y-3) is positive in this interval, satisfying this condition.y' < 0 for -inf < y < 0 and 3 < y < inf: The expression y(y-3) is negative in these intervals, satisfying this condition.Learn more about Differential Equations here:https://brainly.com/question/33814182
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A research team conducted a study showing that approximately 15% of all businessmen who wear ties wear them so tightly that they actually reduce blood flow to the brain, diminishing cerebral functions. At a board meeting of 15 businessmen, all of whom wear ties, what are the following probabilities (Round your answers to three decimal places.) (a) at least one tie is too tight (b) more than two ties are too tight (c) no tie is too tight (d) at least 18 ties are not too tight
Answer:
a) 0.913
b) 0.397
c) 0.087
Step-by-step explanation:
We are given the following information:
We treat wearing tie too tight as a success.
P(Tight tie) = 15% = 0.15
Then the number of businessmen follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 15
We have to evaluate:
a) at least one tie is too tight
[tex]P(x \geq 1) = P(x = 1) +....+ P(x = 15)\\=1 - P(x = 0)\\= 1 - \binom{15}{0}(0.15)^0(1-0.15)^{15}\\=1 - 0.087\\= 0.913[/tex]
b) more than two ties are too tight
[tex]P(x > 2) = P(x = 3) +....+ P(x = 15)\\=1 - P(x = 0) - P(x=1) - P(x=2)\\= 1 - \binom{15}{0}(0.15)^0(1-0.15)^{15}-\binom{15}{1}(0.15)^1(1-0.15)^{14}-\binom{15}{0}(0.15)^2(1-0.15)^{13}\\=1 - 0.087 - 0.231 - 0.285\\= 0.397[/tex]
c) no tie is too tight
[tex]P(x = 0)\\=\binom{15}{0}(0.15)^0(1-0.15)^{15}\\=0.087[/tex]
d) at least 18 ties are not too tight
This probability cannot be evaluated as the number of success or the failures exceeds the number of trials given which is 15.
The probability is asked for 18 failures which cannot be evaluated.
To calculate the probabilities, we can use the concept of binomial probability. Using appropriate formulas, we can find the probabilities (a) at least one tie is too tight (b) more than two ties are too tight (c) no tie is too tight (d) at least 18 ties are not too tight.
Explanation:To calculate the probabilities, we can use the concept of binomial probability. In this case, the probability that a businessman wears the tie too tightly is 15% or 0.15. The total number of businessmen at the board meeting is 15.
(a) To calculate the probability that at least one tie is too tight, we need to find the complement of the probability that no ties are too tight. Using the formula 1 - P(no tight ties), we calculate 1 - (0.85^15) = 0.999 or 99.9%.
(b) To calculate the probability that more than two ties are too tight, we need to find the sum of probabilities of having 3 or more ties too tight. Using the binomial probability formula, we calculate P(X > 2) = 1 - (P(X=0) + P(X=1) + P(X=2)) = 1 - (0.85^15 + 15*0.15*(0.85^14) + 105*(0.15^2)*(0.85^13)) = 0.014 or 1.4%.
(c) To calculate the probability that no ties are too tight, we use the probability P(X=0) = 0.85^15 = 0.203 or 20.3%.
(d) To calculate the probability that at least 18 ties are not too tight, we need to find the sum of probabilities of having 18 or more ties not too tight. Using the binomial probability formula, we calculate P(X >= 18) = P(X=18) + P(X=19) + ... + P(X=15) = (15 choose 18)*(0.85^18)*(0.15^2) + (15 choose 19)*(0.85^19)*(0.15^1) + (15 choose 20)*(0.85^20)*(0.15^0) = 0.999 or 99.9%.
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A box contains 15 resistors. Ten of them are labeled 50 Ω and the other five are labeled 100 Ω.
(a) What is the probability that the first resistor is 100Ω?
(b) What is the probability that the second resistor is 100 Ω, given that the first resistor is 50 Ω?
(c) What is the probability that the second resistor is 100 Ω, given that the first resistor is 100 Ω?
The probability that the first resistor is 100Ω is 1/3. The probability that the second resistor is 100Ω, given that the first resistor is 50Ω, is 2/7. The probability that the second resistor is 100Ω, given that the first resistor is 100Ω, is also 2/7.
Explanation:(a) Probability that the first resistor is 100Ω:
The total number of resistors is 15, with 5 of them labeled 100Ω.
So, the probability is 5/15 or 1/3.
(b) Probability that the second resistor is 100Ω, given that the first resistor is 50Ω:
If the first resistor is 50Ω, there are still 4 resistors labeled 100Ω out of the remaining 14 resistors.
So, the probability is 4/14 or 2/7.
(c) Probability that the second resistor is 100Ω, given that the first resistor is 100Ω:
If the first resistor is 100Ω, there are still 4 resistors labeled 100Ω out of the remaining 14 resistors.
So, the probability is 4/14 or 2/7.
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In a certain country, the life expectancy for women in 1900 was 4949 years and in 2000 it was 8181 years. Assuming that life expectancy between 2000 and 2100 increases by the same percentage as it did between 1900 and 2000, what will the life expectancy be for women in 2100?
Answer:
The life expectancy for 2100 will be 134 years.
Step-by-step explanation:
Consider the provided information.
In 1900 women life expectancy was 49 years and in 2000 it was 81 years.
First calculate the difference of life expectancy between 1900 to 2000.
81-49=32 years
Percentage change is:
[tex]\text{Percentage change}=\frac{32}{49}\times 100\approx65.31[/tex]
It is given that the increase in percentage remains same.
Therefore, next year life expectancy will be:
[tex]\begin{aligned}81+65.31\% \times81&=81(1+\frac{65.31}{100})\\&=81(1.6531)\\&=133.9011\approx134\end{aligned}[/tex]
Hence, the life expectancy for 2100 will be 134 years.
To find the life expectancy for women in 2100 with the same percentage increase as from 1900 to 2000, we calculate the percentage increase between these years and apply it to the year 2000 figure, resulting in an estimated life expectancy of approximately 134 years in 2100.
Explanation:In 1900, the life expectancy for women was 49 years, and in 2000 it was 81 years. To calculate the life expectancy for women in 2100, assuming it increases by the same percentage as from 1900 to 2000, we must first determine the percentage increase from 1900 to 2000.
The percentage increase is calculated as follows:
Percentage Increase = ((Life Expectancy in 2000 - Life Expectancy in 1900) / Life Expectancy in 1900) * 100
This gives us:
Percentage Increase = ((81 - 49) / 49) * 100 = (32 / 49) * 100 ≈ 65.31%
Now, we apply this percentage increase to the life expectancy in 2000 to estimate the life expectancy for women in 2100:
Estimated Life Expectancy in 2100 = Life Expectancy in 2000 * (1 + Percentage Increase)
Estimated Life Expectancy in 2100 = 81 * (1 + 0.6531) ≈ 133.9
Therefore, if the life expectancy for women continues to increase by the same percentage, it will be approximately 134 years in 2100.
A clothing business finds there is a linear relationship between the number of shirts, n, it can sell and the price, p, it can charge per shirt. In particular, historical data shows that 1000 shirts can be sold at a price of $30 , while 3000 shirts can be sold at a price of $22 . Find a linear equation in the form p=mn+b that gives the price p they can charge for n shirts.
Answer:
p = -0.004n+34
Step-by-step explanation:
The slope of the linear equation can be found by using the two given points (30, 1000) and (22, 3000):
[tex]m = \frac{p_1-p_0}{n_1-n_0}= \frac{30-22}{1000-3000}\\m=-0.004[/tex]
Applying the point (30, 1000) to the general form of a linear equation with m =-0.004, gives us the linear relationship between price (p) and number of shirts (n):
[tex](p-p_0) = m(n-n_0)\\p-30=-0.004(n-1000)\\p=-0.004n+34[/tex]
Answer: p = -0.004n+34
Step-by-step explanation:
Given that we need to derived the linear equation of the form;
p = mn+b ....1
Where p is the p is the price and n is the number of shirts they can sell
We need to substitute the values of p and n for the two cases to determine the slope m and constant b.
Case 1:
n = 1000 and p = $30
Substituting into the equation 1, we have:
30 = 1000m +b .....2
Case 2:
n = 3000 and p = $22
Substituting into the equation 1
22 = 3000m + b ......3
Substracting equation 2 from 3, we have
22-30 = 3000m-1000m +b-b
-8 = 2000m
m = -8/2000
m = -0.004
Substituting the value of m into equation 2
30 = 1000(-0.004) + b
b = 30 + 1000(0.004)
b = 30 + 4 = 34
b = 34
Therefore substituting the values of m and b into equation 1, we have our linear equation:
p = -0.004n+34
Suppose that the number of a certain type of computer that can be sold when its price is P (in dollars) is given by a linear function N(P). (a) Determine N(P) if N(1000) = 10000 and N(1700) = 5800. (Use symbolic notation and fractions where needed.) N(P) = (b) Select the statement that gives the slope of the graph of N(P), including units and describes what the slope represents. 6 computers per dollar -6 dollars per computer computers per dollar -6 computers per dollar (c) What is the change AN in the number of computers sold if the price is increased by AP = 110 dollars? (Give your answer as a whole number.) AN = computers
Answer:
a) N(P) = -6P + 16000
b) slope = -6 computers per dollar
That means the number of computer sold reduce by 6 per dollar increase in price.
c) ∆N = -660 computers
Step-by-step explanation:
Since N(P) is a linear function
N(P) = mP + C
Where m is the slope and C is the intercept.
Case 1
N(1000) = 10000
10000 = 1000m + C ....1
Case 2
N(1700) = 5800
5800 = 1700m + C ....2
Subtracting equation 1 from 2
700m = 5800 - 10000
m = -4200/700
m = -6
Substituting m = -6 into eqn 1
10000 = (-6)1000 + C
C = 10000+ 6000 = 16000
N(P) = -6P + 16000
b) slope = -6 computers per dollar
That means the number of computer sold reduce by 6 per dollar increase in price.
Slope is the change in number of computer sold per unit Change in price.
c) since slope m = -6 computers per dollar
∆P = 110 dollars
∆N = m × ∆P
Substituting the values,
∆N = -6 computers/dollar × 110 dollars
∆N = -660 computers.
The number of computer sold reduce by 660 when the price increase by 110 dollars
Answer:
110
Step-by-step explanation:
Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 4.5 millimeters (mm) and a standard deviation of 1.7 mm. For a randomly found shard, find the following probabilities. (Round your answers to four decimal places.)
(a) the thickness is less than 3.0 mm.
(b) the thickness is more than 7.0 mm.
(c) the thickness is between 3.0 mm and 7.0 mm.
Answer:
a) [tex]P(X<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]
[tex]P(z<-0.882)=0.189[/tex]
b) [tex]P(X>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]
[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]
c) [tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]
[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/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 thickness of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(4.5,1.7)[/tex]
Where [tex]\mu=4.5[/tex] and [tex]\sigma=1.7[/tex]
We are interested on this probability
[tex]P(X<3.0)[/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(X<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]
And we can find this probability using excel or the normal standard table:
[tex]P(z<-0.882)=0.189[/tex]
Part b
We are interested on this probability
[tex]P(X>3.0)[/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(X>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]
And we can find this probability using excel or the normal standard table:
[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]
Part c
[tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]
And we can find this probability using excel or the normal standard table liek this:
[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/tex]
Calculate the fuel economy of your family or personal car in terms of gallons per mile and the total cost for one fill-up git hub.
Answer:
0.028 Gallons/mile
Step-by-step explanation:
1 Gallon = 3.78541 Litres
1 Mile = 1.60934 Km
Total Fuel Expenditure (Dec 2019) = PKR 7,000
Fuel cost / Litre = 114Litre/Km
Total Fuel consumption = (7000/114)
= 61.40 Litres or 16.22 Gallons
Total Distance Travelled in Dec 2019 = 921 Km or 572.28 miles
Fuel Economy = 16.22/572.28
= 0.028 Gallons/mile
One time tank fill up cost = PKR 1,500
An engineer in charge of water rationing for the U.S. Army wants to determine if the average male soldier spends less time in the shower than the average female soldier. Let μm represent the average time in the shower of male soldiers and μf represent the average time in the shower of female soldiers.
a) What are the appropriate hypotheses for the engineer?
H0: μm = μf versus Ha: μm > μf
H0: σm = σf versus Ha: σm > σf
H0: μm = μf versus Ha: μm ≠ μf
H0: μm = μf versus Ha: μm < μf
b) Among a sample of 66 male soldiers the average shower time was found to be 2.68 minutes and the standard deviation was found to be 0.65 minutes. Among a sample of 69 female soldiers the average shower time was found to be 2.7 minutes and the standard deviation was found to be 0.5 minutes. What is the test statistic? Give your answer to three decimal places.
c) What is the P-value for the test? Give your answer to four decimal places.
d) Using a 0.1 level of significance, what is the appropriate conclusion?
Reject the claim that the average shower times are different for male and female soldiers because the P-value is greater than 0.1.
Conclude that the average shower time for males is less than the average shower time for females because the P-value is less than 0.1.
Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.
Conclude that the average shower time for males is equal to the average shower time for females because the P-value is less than 0.1.
Answer:
a) H0: μm = μf versus Ha: μm < μf
b) [tex]t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200[/tex]
c) [tex]p_v =P(t_{133}<-0.200)=0.421[/tex]
d) Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.
Step-by-step explanation:
Data given and notation
[tex]\bar X_{m}=2.68[/tex] represent the mean for the sample male
[tex]\bar X_{f}=2.7[/tex] represent the mean for the sample female
[tex]s_{m}=0.65[/tex] represent the sample standard deviation for the males
[tex]s_{f}=0.5[/tex] represent the sample standard deviation for the females
[tex]n_{m}=66[/tex] sample size for the group male
[tex]n_{f}=69[/tex] sample size for the group female
t would represent the statistic (variable of interest)
Part a
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the average male soldier spends less time in the shower than the average female soldier, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{m}-\mu_{f}\geq 0[/tex]
Alternative hypothesis:[tex]\mu_{m} - \mu_{f}< 0[/tex]
Or equivalently:
Null hypothesis:[tex]\mu_{m}-\mu_{f}= 0[/tex]
Alternative hypothesis:[tex]\mu_{m} - \mu_{f}< 0[/tex]
And the best option is:
H0: μm = μf versus Ha: μm < μf
Part b
We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{(\bar X_{m}-\bar X_{f})-\Delta}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{f}}{n_{f}}}}[/tex] (1)
And the degrees of freedom are given by [tex]df=n_m +n_f -2=66+69-2=133[/tex]
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
What is the test statistic?
With the info given we can replace in formula (1) like this:
[tex]t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200[/tex]
Part c What is the p-value?
Since is a left tailed test the p value would be:
[tex]p_v =P(t_{133}<-0.200)=0.421[/tex]
Part d
The significance level given is [tex] \alpha =0.1[/tex] since the p value is higher than the significance level we can conclude:
Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.
The question relates to hypothesis testing in statistics, insights into average shower times for male and female soldiers. After formulating the hypotheses, we calculate the test statistic using the provided sample data, and then find the corresponding P-value. If the P-value is less than our significance level, we reject the null hypothesis and side with the alternative hypothesis.
Explanation:The subject of this question falls under Mathematics, specifically it deals with hypothesis testing statistics.
a) The appropriate hypotheses for the engineer to consider would be: H0: μm = μf versus Ha: μm > μf
b) To calculate the test statistic, we use the formula for the test statistic in a independent two-sample t-test which incorporates the sample sizes, means, and standard deviations from the two groups. The formula is (avg(male soldiers)-avg(female soldiers))/sqrt(((sd(male soldiers))^2/number(male soldiers))+((sd(female soldiers))^2/number(female soldiers))). Plug in given values, we can obtain the test statistic.
c) The P-value can be obtained by looking up the test statistic in the T distribution table.
d) If the P-value is greater than the level of significance (0.1), we would fail to reject the null hypothesis. If the P-value is less than the level of significance, we would reject the null hypothesis. The conclusion is based on the specific P-value we computed.
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A study examined the alertness benefits of higher caffeine intake. In the study, the researchers investigated the link between daily caffeine intake and alertness at work (measured on a 6-point scale where 1 equals"not alert" and 6 equals"very alert"). Participants for the study were those who used a popular social networking Web site. Of the respondents, those who had very high or very low intakes were excluded, leaving a sample size of 49 respondents.
Do the data represent a population or a sample?
A. The data represent a population since the data are all the units that are the subject of the study
B. The data represent a sample since the data are a subset of units.
C. The data represent a population since the data are a subset of units.
D. The data represent a sample since the data are all the units that are the subject of the study.
Answer:
B. The data represent a sample since the data are a subset of units.
True, we have a sample and all the elements on this sample are subset for the units of the population.
Step-by-step explanation:
We need to remember that the population represent all the possible elements or individuals of interest and by the other hand the sample is a subset of the population that is used to analyze patterns in the population of interest
Let's analyze one by one the options.
A. The data represent a population since the data are all the units that are the subject of the study
False, we have a sample size n =49 not all the possible elements of the population for this case we don't have a population.
B. The data represent a sample since the data are a subset of units.
True, we have a sample and all the elements on this sample are subset for the units of the population.
C. The data represent a population since the data are a subset of units.
False, by definition the population CAN'T be a subset of the sample size, and that's not the case.
D. The data represent a sample since the data are all the units that are the subject of the study.
False, the data is a sample BUT are not all the units for the study because that only occurs when we have a population.
On an indoor circular track of circumference 50 feet, Joneal starts at point $S$, runs in a counterclockwise direction, and then stops when he has run exactly one mile (5280 feet). On which quarter of the circle, $A$, $B$, $C$ or $D$, did Joneal stop
Answer:
He stoped on 3th quarter,i.e, $C$.
Step-by-step explanation:
He ran 105 full circles ( 5280/50=105 ( rst= 30ft) ). So in the last circles he started from point S to run 30ft more.
The quarter of the circle is long 50ft/4= 12,5ft. So for 30 feets he must run 2 quarters, its 25ft. The last 5ft he ran on the 3th quarter, so he stoped on C.
This answer $C$, if $C$ is the 3th quadrant, i don't see the picture of the track.
Answer:
C
Step-by-step explanation:
Between which two whole numbers does the product of 9 and 7 1/8 lie?
Answer:
We can answer that the product of 9 and 7 1/8 lies between 64 and 65, the two whole numbers asked.
Step-by-step explanation:
Let's find the product of 9 and 7 1/8, this way:
9 * 7 1/8 = 9 * 7.125
9 * 7.125 = (9 * 7) + (9 * .125)
63 + 1.125 = 64.125
We can answer that the product of 9 and 7 1/8 lies between 64 and 65, the two whole numbers asked.
Answer:
64 and 65
Step-by-step explanation:
9 X 7 1/8
=64.125
Consider the angle -6 radians. Determine the quadrant in which the terminal side of the angle is found and find the corresponding reference angle ¯ θ . Round the reference angle to 4 decimal places.
Final answer:
The angle -6 radians places the terminal side in the third quadrant with a reference angle of 0.28 radians when expressed in the positive acute form and rounded to four decimal places.
Explanation:
The question involves determining the quadrant of an angle measured in radians and finding its corresponding reference angle, which is a common task in Mathematics, specifically in the study of trigonometry.
To find the quadrant in which the terminal side of the angle -6 radians lies, we need to recall that one full revolution around the unit circle is 2π or approximately 6.28 radians. Since the given angle is negative, we move in the clockwise direction from the positive x-axis. Dividing -6 by 6.28, we realize that it is just shy of a full negative revolution, thereby placing the terminal side in the third quadrant.
To find the reference angle ¯θ, which is the positive acute angle the terminal side makes with the x-axis, we need to subtract the given angle from one full revolution (if necessary) and find the absolute value. Hence, ¯θ = |2×π - (-6)| = |6.28 - (-6)| = |0.28|, which is the same as 0.28 radians when rounded to four decimal places.
Suppose that we generate a pseudo-random number U = 0.128. Use this to generate an Exponential ( λ = 1 / 3 ) random variate.
Answer:
The generated random variate is X = 6.173
Explanation:
To generate an exponential random variate (with parameter λ) from a pseudo-random number U, we use the formula:
X = − ln(U)/λ
Fro the given problem, we have U = 0.128 and λ = 3.
So, X = -In(0.128) / (1/3) = -In(0.128)/ 0.333 = 2.056/0.333 = 6.173
To generate an Exponential random variate with a given pseudo-random number, we use the formula -ln(U) / λ, where U is the pseudo-random number and λ is the rate parameter. In this case, plugging in U = 0.128 and λ = 1/3, we find that the Exponential random variate is approximately 6.147.
Explanation:To generate an Exponential random variate, we use the formula:
X = -ln(U) / λ
where U is a pseudo-random number between 0 and 1, and λ is the rate parameter for the Exponential distribution.
In this case, U = 0.128 and λ = 1/3. Plugging in these values, we have:
X = -ln(0.128) / (1/3)
X ≈ -ln(0.128) / (1/3) ≈ -(-2.049) / (1/3) ≈ 2.049 / (1/3) ≈ 6.147
Therefore, the Exponential random variate is approximately 6.147.
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What is next in this sequence of numbers: 1, 11, 21, 1211, 111221, 312211, ...?
Answer:
13112221
Step-by-step explanation:
Each sequence of numbers is a verbal representation of the sequence before it. Thus, starting with 1, the next sequence would be "one one," or "11." That sequence is followed by "two one," or "21," and so on and so forth.
This may also be a good explanation:
The first number is just ONE (amount) "1" (0-9 numeral). So if you say there's ONE "1" (seriously just say it aloud) the next number would be an 11. Then there are TWO "1's", creating 21. Then ONE "2" and ONE "1" which creates 1,211. Then ONE "1", ONE "2", and TWO "1's" creating 111,221 ... and so on.
The first number 1 is read as one one, so the second number is written as 11, this is read as two ones, so the next number is written as 21 ( two ones)
This continues throughout the sequence.
The last number written is 312211 which is read as one three, one one, two twos, two ones
This gets written as 13112221
In a given population, the percent of people with blue eyes is 32%. If 17 people from that population are randomly selected, what is the probability that exactly 9 of them will have blue eyes?
Answer:
The required probability is 0.0391.
Step-by-step explanation:
Consider the provided information.
The percent of people with blue eyes is 32%
Thus, p = 32% = 0.32
The percent of people with non blue eyes is 100%-32%=68% = 0.68
q = 68% = 0.68
We need to determine the probability that exactly 9 of them will have blue eyes if 17 are selected.
Thus, n=17 and r=9
Use the formula of binomial distribution: [tex]P(r) = ^nC_r p^r q^{n-r}[/tex]
Substitute the respective values in the above formula.
[tex]P(r=9) = ^{17}C_9 \times(0.32)^9 \times 0.68^{17-9}[/tex]
[tex]P(r=9) = \frac{17!}{9!8!} \times(0.32)^9 \times 0.68^{8}[/tex]
[tex]P(r=9) \approx0.0391[/tex]
Hence, the required probability is 0.0391.
Evaluate the integral by changing the order of integration in an appropriate way. Integral from 0 to 1 Integral from 0 to 1 Integral from x squared to 1 3 x font size decreased by 7 z e Superscript z font size decreased by 5 y squared Baseline dy dx dz
Answer:
The question is not clear, but here is a similar question with the same approach.
Integral from 0 to 1 Integral from 1 to 2 Integral from 2 to 3, { (x+y+z) } dx dy dz by changing the order of integration in an appropriate way.
Step-by-step explanation:
The approach is that of multiple integral where changing the order of integration is done appropriately
The step by step with detailed workings are shown in the attachment below.