[tex]\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{-14}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-14}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{(-2)}}}\implies \cfrac{-24}{10+2}\implies \cfrac{-24}{12}\implies -2[/tex]
[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{-2}[x-\stackrel{x_1}{(-2)}]\implies y-10=-2(x+2) \\\\\\ y-10=-2x-4\implies y=-2x+6[/tex]
A subway has good service 70% of the time and runs less frequently 30% of the time because of signal problems. When there are signal problems, the amount of time in minutes that you have to wait at the platform is described by the pdf probability density function with signal problems = pT|SP(t) = .1e^(−.1t). But when there is good service, the amount of time you have to wait at the platform is probability density function with good service = pT|Good(t) = .3e^(−.3t) You arrive at the subway platform and you do not know if the train has signal problems or running with good service, so there is a 30% chance the train is having signal problems. (a) After 1 minute of waiting on the platform, you decide to re-calculate the probability that the train is having signal problems based on the fact that your wait will be at least 1 minute long. What is that new probability? (b) After 5 minutes of waiting, still no train. You re-calculate again. What is the new probability? (c) After 10 minutes of waiting, still no train. You re-calculate again. What is the new probability?
Answer:
Part a: The probability is 0.3436
Part b: The probability is 0.5381
Part c: The probability is 0.7600
Step-by-step explanation:
As per the given data
[tex]p_{T | SP} = e^{0.1}\\p_{T | Good} = e^{0.3}\\[/tex]
(a)
Probability that train is delayed by more than 1 minute = P(T > 1) = P(SP) * P(T > 1 | SP) + P(Good) * P(T > 1 | Good)
[tex]= 0.3 e^{-.1 \times 1 }+ 0.7 * e^{-.3 \times 1}\\ = 0.3 e^{-.1} + 0.7 e^{-.3}[/tex]
Probability that after 1 minute of waiting, probability that train has signal problems = P(SP | T > 1)
= P(T > 1 | SP) * P(SP) / P(T > 1) (By Bayes theorem)
[tex]= \dfrac{0.3 e^{-.1 }}{ 0.3 e^{-.1 }+ 0.7 e^{-.3 }}\\\\= \dfrac{0.2714512}{0.790024}\\\\= 0.3435987[/tex]
(b)
Probability that train is delayed by more than 5 minutes = P(T > 5) = P(SP) * P(T > 5 | SP) + P(Good) * P(T > 5 | Good)
[tex]= 0.3 e^{-.1 \times 5 }+ 0.7 * e^{-.3 \times 5}\\ = 0.3 e^{-.5} + 0.7 e^{-1.5}[/tex]
Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 5)
= P(T > 5 | SP) * P(SP) / P(T > 5) (By Bayes theorem)
[tex]= \dfrac{0.3 e^{-.5 }}{ 0.3 e^{-.5 }+ 0.7 e^{-1.5 }}\\\\= \dfrac{0.1819592}{0.3381503}\\\\= 0.5381015[/tex]
(c)
Probability that train is delayed by more than 10 minutes = P(T > 10) = P(SP) * P(T > 10 | SP) + P(Good) * P(T > 10 | Good)
[tex]= 0.3 e^{-.1 \times 10 }+ 0.7 * e^{-.3 \times 10}\\ = 0.3 e^{-1.0} + 0.7 e^{-3.0}[/tex]
Probability that after 5 minute of waiting, probability that train has signal problems = P(SP | T > 10)
= P(T > 10 | SP) * P(SP) / P(T > 10) (By Bayes theorem)
[tex]= \dfrac{0.3 e^{-1.0 }}{ 0.3 e^{-1.0 }+ 0.7 e^{-3.0 }}\\\\= \dfrac{0.1103638}{0.1452148}\\\\= 0.7600038[/tex]
(a) The probabilities recalculated based on the waiting time are: 21.4% after 1 minute,
(b) 54.0% after 5 minutes, and
(c) 75.9% after 10 minutes.
These calculations use Bayes' theorem and the exponential distribution functions provided.
The question pertains to conditional probability and involves exponential distributions.
Let’s solve this step-by-step:
Part (a)
Given:
pT|SP(t) = [tex]0.1e^(-0.1t)[/tex]
pT|Good(t) = [tex]0.3e^(-0.3t)[/tex]
P(Signal Problems) = [tex]0.30[/tex]
P(Good Service) = [tex]0.70[/tex]
We need to find P(Signal Problems | T ≥ 1). Using Bayes’ theorem:
P(T ≥ 1 | Signal Problems) = [tex]\int_{1}^{\infty} 0.1 \, dx[/tex] [tex]e^(-0.1t)[/tex] [tex]dt =[/tex] [tex]e^(-0.1)[/tex]P(T ≥ 1 | Good Service) = [tex]\int_{1}^{\infty} 0.3 \, dx[/tex] [tex]e^(-0.3t)[/tex] [tex]dt =[/tex] [tex]e^(-0.3)[/tex]Using the law of total probability:
P(T ≥ 1) = P(T ≥ 1 | Signal Problems)P(Signal Problems) + P(T ≥ 1 | Good Service)P(Good Service)
= [tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70[/tex]
Now applying Bayes’ theorem:
P(Signal Problems | T ≥ 1) = [tex][[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.1)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-0.3)[/tex] [tex]* 0.70][/tex]
[tex]\approx [0.9048 * 0.30] / [0.9048 * 0.30 + 0.7408 * 0.70][/tex]
[tex]\approx 0.214[/tex]
Part (b)
Similarly for T ≥ 5, the probabilities become:
P(T ≥ 5 | Signal Problems) = [tex]e^(-0.5)[/tex]P(T ≥ 5 | Good Service) = [tex]e^(-1.5)[/tex]Using the law of total probability:
P(T ≥ 5) = [tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex][tex]* 0.70[/tex]
Applying Bayes’ theorem:
P(Signal Problems | T ≥ 5) = [tex][[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30] / [[/tex][tex]e^(-0.5)[/tex] [tex]* 0.30 +[/tex] [tex]e^(-1.5)[/tex] [tex]* 0.70][/tex]
≈ [tex][0.6065 * 0.30] / [0.6065 * 0.30 + 0.2231 * 0.70][/tex]
≈ [tex]0.540[/tex]
Part (c)
For T ≥ 10:
P(T ≥ 10 | Signal Problems) = [tex]e^{-1}[/tex]P(T ≥ 10 | Good Service) = [tex]e^{-3}[/tex]Using the law of total probability:
P(T ≥ 10) = [tex]e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70[/tex]
Applying Bayes’ theorem:
P(Signal Problems | T ≥ 10) = [tex]\frac{e^{-1} \cdot 0.30}{e^{-1} \cdot 0.30 + e^{-3} \cdot 0.70}[/tex]
≈ [tex][0.3679 * 0.30] / [0.3679 * 0.30 + 0.0498 * 0.70][/tex]
≈ [tex]0.759[/tex]
(1 point) Find the general solution to the homogeneous differential equation. ????2y????????2−20????y????????+136y=0 Use c1 and c2 in your answer to denote arbitrary constants, and enter them as c1 and c2. y(????)= ?
Answer:
Question is not clear please post question clearly lots of question marks.
Your differential equation is not displayed well. It though looks like this:
2d²y/dx² - 20dy/dx + 136y = 0
If this is not the differential equation, the method of solving this would still be used in solving the correct one.
We first write an auxiliary equation to the differential equation.
The auxiliary equation is:
2m² - 20m + 136 = 0
Dividing by 2, we have
m² - 10m + 68 = 0
Next, we solve the auxiliary equation to obtain the values of m.
Solving using the quadratic formula
m = [-b ± √(b² - 4ac)]/2a
Where a = 1, b = -10, and c = 68
m = [10 ± √(100 - 272)]/2
= 5 ± (1/2)√(-172)
= 5 ± (1/2)i√172
= 5 ± 6.6i
For solutions of the form a ± ib, the complimentary solution is
y = e^(ax)[C1cosbx + C2sinbx]
Therefore, the complimentary solution is
y = e^(5x)[C1cos(6.6x) + C2sin(6.6x)]
Define a random variable x = number of cups of coffee consumed on an average day. Let x = 4 represent four or more cups. Round your answers to four decimal places.
Answer:
E (X) = 6.4
Step-by-step explanation:
SOLUTION:
A random variable x = number of cups of coffee consumed on an average day.
∴Let x = 4 represent four or more cups. Round your answers to four decimal places.
X Probability (X)
0 0.1
1 0.15
2 0.3
3 0.75
4 0. 25
5 0.21
∴ E (X) = Ux(Mean)
0x.0.1 + 1 x.15 + 2 x 0.3 + 3 x 0.75 + 4 x 0.25 + 5 x 0.21 = 6.4
The least-squares regression model y =−3.4+5.2x and correlation coefficient r=−0.66 were calculated for a set of bivariate data with variables x and y . What is closest to the proportion of the variation in y that cannot be explained by the explanatory variable?
Answer:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For this case the value of r = -0.66
Now we can calculate the determination coeffcient:
[tex] r^2 = (-0.66)^2 = 0.4356[/tex]
And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable
And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable
Step-by-step explanation:
For this case we need to calculate the slope with the following formula:
[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]
[tex]\bar x= \frac{\sum x_i}{n}[/tex]
[tex]\bar y= \frac{\sum y_i}{n}[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x[/tex]
And the model obtained for this case is:
[tex] y = -3.4 +5.2 x[/tex]
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For this case the value of r = -0.66
Now we can calculate the determination coeffcient:
[tex] r^2 = (-0.66)^2 = 0.4356[/tex]
And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable
And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable
The Coefficient of determination gives the proportion of variation which can be explained by the regression line.
The proportion of variation that cannot be explained by the given regression line is 56.44%
From the question, the correlation Coefficient, R = - 0.66
The Coefficient of determination, R² can be calculated from the R value given. R² = - 0.66² = 0.4356This means that :
Percentage of variation that can be explained = (0.4356) × 100% = 43.56%The percentage of variation that cannot be explained = 100% - 43.56% = 56.44%Therefore, proportion of the variation in y that cannot be explained is 56.44%
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Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study27 has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is 0.183. The standard error for this estimate is 0.016.
a.Give notation for the quantity being estimated, and define any parameters used.
b.Give notation for the quantity that gives the best estimate, and give its value.
c.Give a 95% confidence interval for the quantity being estimated. Interpret the interval in context.
Answer:
a) [tex]\mu[/tex]
b) [tex]\bar{x}[/tex]
c) (0.152, 0.214)
Step-by-step explanation:
We are given the following in the question:
Sample mean = 0.183 ppm
Standard error = 0.016
a) quantity being estimated
We have to estimate the population mean.
Notation for population mean:
[tex]\mu[/tex]
b) Best estimate for population mean is the sample mean
Notation for sample mean:
[tex]\bar{x}[/tex]
The point estimate for population mean is
[tex]\mu = \bar{x} = 0.183[/tex]
c) 95% confidence interval
[tex]\bar{x}\pm z_{critical}(\text{Standard Error})[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting values, we get,
[tex]0.183 \pm 1.96(0.016)\\=0.183\pm 0.03136\\=(0.15164, 0.21436)\\\approx (0.152, 0.214)[/tex]
Thus, the 95% confidence interval is:
(0.152, 0.214)
Interpretation:
After treatment with moose drool, we are 95% certain or 95% confident that the interval (0.152, 0.214) contains the true mean of the population that is the mean level of the toxin ergovaline on the grass.
The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The best estimate for this quantity is 0.183 ppm. The 95% confidence interval for this estimate is (0.15164, 0.21436) ppm.
Explanation:a. The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The parameter used is the standard error, which measures the variability of the estimate.
b. The notation for the quantity that gives the best estimate is the mean level of the toxin ergovaline after treatment with moose drool, denoted as μ.
c. To calculate a 95% confidence interval, we need to determine the margin of error. Since the standard error is given as 0.016, the margin of error is 1.96 times the standard error, which is approximately 0.03136. The 95% confidence interval is then calculated by subtracting and adding the margin of error from the mean level of the toxin ergovaline after treatment with moose drool, resulting in an interval of (0.15164, 0.21436). This means that we are 95% confident that the true mean level of the toxin ergovaline after treatment with moose drool is between 0.15164 and 0.21436 ppm.
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I really need help! This is last minute and its 1 am. I'm tired so I'm going to leave it to you guys to solve my problems and provide statistical explanations. Please help me and answer these before 9: 15 in the morning EST on March 3rd, 2020.
Answer:
question 1[tex]n^{2} - 20n -96 = 0[/tex]
use product and sum method
product = -96
sum = -20
numbers needed = ( -24 , 4)
n - 24 = 0
n + 4 = 0
hence n = 24 and n = -4
Question 2
[tex]x^{2} + 12 x = 48[/tex]
in the form [tex]ax^{2} +bx +c = 0[/tex]
= [tex]x^{2} +12x - 48[/tex]
make use of the formula :
[tex]\frac{-b+-\sqrt{b^{2} -4ac} }{2a}[/tex]
replace values to make 2 equations :
1.[tex]\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = 3.17
2.[tex]\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = -15.2
hence x = 3.17 and x = -15.2
Question 3
[tex]x^{2} -14x+40=0[/tex]
use product and sum method
product = 40
sum = -14
numbers needed = (-10 , -4)
x - 10 = 0
x - 4 = 0
hence x = 10 and x = 4
Question 4
[tex]5b^{2} -20b-18 = 7[/tex]
in the form [tex]ax^{2} +bx +c = 0[/tex]
this becomes [tex]5b^{2} -20b-18-7[/tex]
= [tex]5b^{2} -20b-25[/tex]
can simplify by 5
= [tex]b^{2} -4b-5 =0\\[/tex]
use product and sum method
product = -5
sum = -4
numbers needed (-5 , 1)
b-5 = 0
b + 1 = 0
hence b = 5 and b = -1
Answer:
Step-by-step explanation:
1) n² - 20n - 96 = 0
n² - 20n + (- 20/2)² = 96 + (- 20/2)²
(n - 10)² = 96 + 100
(n - 10)² = 196
Taking square root of both sides
n - 10 = √196 = 14
n = 14 + 10
n = 24
2) x² + 12x = 48
x² + 12x + (12/2)² = 48 + (12/2)²
(x + 6)² = 48 + 36 = 84
Taking square root of both sides,
x + 6 = 9.2
x = 9.2 - 6
x = 3.2
3) x² - 14x + 40 = 0
x² - 14x = - 40
x² - 14x + (- 14/2)² = - 40 + (- 14/2)²
(x - 7)² = - 40 + 49 = 9
Taking square root of both sides,
x - 7 = 3
x = 3 + 7
x = 10
4) 5b² - 20b - 18 = 7
5b² - 20b = 7 + 18
5b² - 20b = 25
Dividing both sides by 5, it becomes
b² - 4b = 5
b² - 4b + (-4/2)² = 5 + (-4/2)²
(b - 2)² = 5 + 4 = 9
Taking square root of both sides
b - 2 = 3
b = 3 + 2
b = 5
Military radar and missile detection systems are designed to warn a country of an enemy attack. A reliability question is whether a detection system will be able to identify an attack and issue a warning. Assume that a particular detection system has a 0.80 probability of detecting a missile attack. Use the binomial probability distribution to answer the following questions. (a) What is the probability that a single detection system will detect an attack
Answer:
0.8 is the probability that a single detection system will detect a missile attack.
Step-by-step explanation:
We are given the following information:
We treat detection a missile attack as a success.
P(detecting a missile attack) = 80% = 0.8
Then the number of missile attack 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 = 1
We have to evaluate:
[tex]P(x = 1)\\= \binom{1}{1}(0.8)^1(1-0.8)^0\\= 0.8[/tex]
0.8 is the probability that a single detection system will detect a missile attack.
Answer:
(a) Probability that a single detection system will detect an attack is 0.80
Step-by-step explanation:
We are given that a reliability question is whether a detection system will be able to identify an attack and issue a warning. Assuming that a particular detection system has a 0.80 probability of detecting a missile attack.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials(samples) taken = 1 detection system
r = number of success
p = probability of success which in our question is probability of
detecting a missile attack, i.e., 80%
LET X = a particular detection system
Also, it is given that a single detection system is taken,
So, it means X ~ [tex]Binom(n=1,p= 0.80)[/tex]
Now, Probability that a single detection system will detect an attack is given by = P(X = 1)
P(X = 1) = [tex]\binom{1}{1}0.8^{1} (1-0.8)^{1-1}[/tex]
= [tex]1 \times 0.8 \times 1[/tex] = 0.80 .
Jeremy and Brenda drove their cars in
opposite directions. When they stopped
after some time, they were already
126 miles apart. If Jeremy drove twice
as far as Brenda, how many miles did
Jeremy drive?
Answer: Jeremy drove 84 miles.
Step-by-step explanation:
Let x represent the number of miles that Brenda drove.
If Jeremy drove twice
as far as Brenda, it means that the distance covered by Jeremy would be 2x miles
When they stopped after some time, they were already
126 miles apart. This means that the total distance covered by both of them is 126 miles. Therefore,
x + 2x = 126
3x = 126
x = 126/3
x = 42 miles
The number of miles that Jeremy drove is
42 × 2 = 84 miles
The given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem. y = c1 + c2 cos(x) + c3 sin(x), (−[infinity], [infinity]); y''' + y' = 0, y(π) = 0, y'(π) = 9, y''(π) = −1
To solve the initial-value problem for the differential equation y''' + y' = 0 with the general solution y = c1 + c2 cos(x) + c3 sin(x), we differentiate y to find y' and y'', apply the initial conditions at x = π, and solve for c1, c2, and c3. The specific solution is y = 1 + cos(x) - 9 sin(x).
We're given the general solution of the third-order differential equation y''' + y' = 0 as y = c1 + c2 cos(x) + c3 sin(x), and we need to find specific constants c1, c2, and c3 that satisfy the initial conditions y(π) = 0, y'(π) = 9, and y''(π) = -1.
First, differentiate y = c1 + c2 cos(x) + c3 sin(x) to find y' and y''.
Substitute x = π into the resulting equations to apply the initial conditions.
Solve the system of equations to find the values of c1, c2, and c3.
Let's carry out these steps:
Differentiation gives us:
y' = -c2 sin(x) + c3 cos(x)
y'' = -c2 cos(x) - c3 sin(x)
Applying initial conditions at x = π:
y(π) = c1 - c2 = 0
y'(π) = -c3 = 9
y''(π) = -c2 = -1
Solve the system:
c1 = c2
c2 = 1
c3 = -9
Therefore, the specific solution to the given initial-value problem is y = 1 + cos(x) - 9 sin(x).
This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 100 customer orders to fill. Each order requires one component part that is purchased froma supplier. However, typically, 2% of the components are identifiedas defective, and the components can be assumed to beindependent.a)If the manufacturer stocks 100 components, what is theprobability that the 100 orders can be filled without reorderingcomponents?b) If the manufacturer stocks 102 components, what is theprobability that the 100 orders can be filled without reorderingcomponents?c) If the manufacturer stocks 105 components, what is theprobability that the 100 orders can be filled without reorderingcomponents?
Answer:
(a) 0.1326
(b) 0.2732
(c) 0.0410
Step-by-step explanation:
Let X = number of defective components.
The probability of X is, P (X) = p = 0.02.
The random variable X follows a Binomial distribution with parameters n and p. The probability mass function of a Binomial distribution is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3,...[/tex]
(a)
Compute the probability that the 100 orders can be filled without reordering components as follows:
n = 100
[tex]P(X=0)={100\choose 0}0.02^{0}(1-0.02)^{100-0}=1\times1\times0.13262=0.1326[/tex]
Thus, the probability that the 100 orders can be filled without reordering components is 0.1326.
(b)
Compute the probability that out of 102 orders 2 orders needs reordering as follows:
n = 102
[tex]P(X=2)={102\choose 2}0.02^{2}(1-0.02)^{102-2}=5151\times0.0004\times0.13262=0.2732[/tex]
Thus, the probability that out of 102 orders 2 orders needs reordering is 0.2732.
(c)
Compute the probability that out of 105 orders 2 orders needs reordering as follows:
n = 105
[tex]P(X=5)={105\choose 5}0.02^{5}(1-0.02)^{105-5}=96560646\times0.0000000032\times0.13262=0.0410[/tex]
Thus, the probability that out of 105 orders 5 orders needs reordering is 0.0410.
An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 6 cm, and the height is 12 cm. Find the rate at which the water level is rising when the water level is 5 cm.
Answer:
The volume of a pyramid is 1/3 Sh, where S is the area of the base and h is the height. Since the area of base (Square) is , S = [tex]s^{2}[/tex], where s is the side of the base. So the volume is
V = 1/3 [tex]s^{2}[/tex]h.
Further solution is on paper (Pictures attached)
In a certain small dorm, the dorm council will consist of 8 students, ofwhom 3 must be women and 5 must be men, because there are a total of 15 women and 25 men currently living there. What is the total possible number of complete councils that could be selected
Answer:
The total possible number of complete councils that could be selected is 24,174,150.
Step-by-step explanation:
The order of the men and the women is not important. So we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
What is the total possible number of complete councils that could be selected
3 women from a set of 15
5 men from a set of 25
[tex]T = C_{15,3}*C_{25,5} = \frac{15!}{3!12!}*\frac{25!}{5!20!} = 455*53130 = 24174150[/tex]
The total possible number of complete councils that could be selected is 24,174,150.
A researcher would like to evaluate the claim that large doses of vitamin C can help prevent the common cold. One group of participants is given a large dose of the vitamin (500 mg per day), and a second group is given a placebo (sugar pill). The researcher records the number of colds each individual experiences during the 3-month winter season. a. Identify the dependent variable for this study.b. Is the dependent variable discrete or continuous?c. What scale of measurement (nominal, ordinal, or interval/ratio) is used to measure the dependent variable?d. What is the independent variable?e. What research method is being used (experimental or correlational?
Answer:
(a) the dependent variable here are the Participants.
(b) the dependent variable is discrete.
(c) The scale measurement of measurement used is interval/ratio to measure the dependent variable.
(d) the independent variable is Vitamin C
(e) The research method being used is experimental.
Step-by-step explanation:
The dependent variable (sometimes known as the responding variable) is what is being studied and measured in the experiment.
Examples of continuous dependent variable may include costs, profits and sales.But some dependent variables are discrete – that is, they take on a relatively small number of integer values.
Interval scale and ratio scale are the two variable measurement scales where they define the attributes of the variables quantitatively.A ratio scale is a measurement scale which has more or less all the properties of an interval scale. Ratio data on this scale has measurable intervals.
Experimental research is a study that strictly adheres to a scientific research design. It includes a hypothesis, a variable that can be manipulated by the researcher, and variables that can be measured, calculated and compared.
Answer: 1. The dependent variable is the common cold.
2. It is a discrete variable.
3. The interval/ratio scale.
4. The independent value is the large doses of Vitamin C administered.
5. The research method used is Experimental.
Step-by-step explanation:
1. The dependent variable is the factor that we are trying to understand. In this case, it is the common cold and how it is affected by large doses of vitamins.
2. A discrete value is computed by counting. So the dependent variable - the common cold was computed by counting the number of occurrences.
3. The Interval/ratio scale is used because both the order of measurement and the differences between them are observed.
4. The Independent variable is the control in the experiment which can be compared to changes in the dependent variable. The large doses of vitamin C affects the common cold.
5. Correlational research observes patterns in variables that occur naturally while Experimental research introduces a change and monitors its effect. The research is Experimental because a change in the form of the placebo is introduced and observed.
he purpose of an x-bar chart is to determine whether there has been a: change in the AOQ. change in the number of defects in a sample. change in the percent defective in a sample. change in the dispersion of the process output. change in the central tendency of the process output.
Answer:
E. Change in the central tendency of the process output.
Step-by-step explanation:
The x-bar chart is a control chart for the central tendency of the process output. Therefore it's purpose is to determine whether there has been a change in the central tendency of the process output.
Answer:
The change in the central tendency of the process output.
Step-by-step explanation:
Lets examine each of the listed options
The change in the percent defective in a sample is associated with the change in average outgoing quality (AOQ).
The change in the dispersion of the process output is associated with R-chart since the function of R-chart is to detect changes in the dispersion.
The change in the number of defects in a sample is associated with C-chart since the function of a C-chart is to show the number of flaws per unit in a sample.
The change in the central tendency of the process output is associated with X-bar chart since the function of X-bar chart is to check whether the values are within the appropriate limits (central tendency) of the process output or not.
Therefore, the purpose of an X-bar chart is to determine whether there has been a change in the central tendency of the process output.
Since the ratios for sides of triangles are the same, when the angles are the same, ? have been developed which show these ratios for every angle encountered in a triangle.
Answer:
Step-by-step explanation:
Mathematically, if two triangles are similar, then all the three of their angles are congruent to each other and their corresponding sides are in the same proportion. This means that the ratio of their corresponding sides are equal to each other.
Answer: A trigonometric table will provide all the angles in respect to the ratios of the sides.
Step-by-step explanation:
A trigonometric table would provide all of the ratios of the sides in respect to the angles. Sin, cos, and tan are parts of trigonometric table.
Trigonometry table, tabulated values for some or all of the six trigonometric functions for various angular values.
Since the ratios for sides of triangles are the same, when the angles are the same, trigonometric tables have been developed which show these ratios for every angle.
what is the recursive formula for this geometric sequence? -4,-24,-144,-864,...
[tex]\bf -4~~,~~\stackrel{-4\cdot 6}{-24}~~,~~\stackrel{-24\cdot 6}{-144}~~,~~\stackrel{-144\cdot 6}{-864}~\hfill \impliedby \begin{array}{llll} \textit{we're multiplying by 6 the previous}\\ \textit{term(n-1) to get the current one} \end{array} \\\\[-0.35em] ~\dotfill\\\\ a_n = a_{n-1}(6)\qquad for~~a_1=-4~~,~~n>2\qquad \impliedby \textit{recursive formula}[/tex]
Below are the jersey numbers of 11 players randomly selected from a football team. Find the range, variance, and standard deviation for the given sample data. What do the results tell us? 59 19 30 75 42 49 57 81 11 87 91
Answer:
Range = The difference between the highest and lowest which is
The mean is the sum of all values divided by the number of values
= 59+19+30+75+42+49+57+81+11+87+91 divided by n (which is 11)
601÷11
= 54.6
Variance = sum of squared deviations from the mean divided by n-1
=679.65
Standard Deviation = This is the square root of variance which gives us;
26.07
The length and width of a rectangle are measured as 45 cm and 24 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle.
Answer:
dA(r) = 6.9 cm²
Step-by-step explanation:
Area of a rectangle is
A(r) = L * W (1)
Where L stands for length and W for width.
Taking differentials on both sides f equation (1)
dA(r) = W*dL + L*dW
As error in measurments are at most 0,1 cm ( in each side), then the maximum error in calculating the area is
dA(r) = 24* ( 0,1) (cm²) + 45*(0,1)(cm²) ⇒ dA(r) = 2.4 + 4.5 (cm²)
dA(r) = 6.9 cm²
A factorial experiment was designed to test for any significant differences in the time needed to perform English to foreign language translations with two computerized language translators. Because the type of language translated was also considered a significant factor, translations were made with both systems for three different languages: Spanish, French, and German. Use the following data for translation time in hours.
Language
Spanish French German
System 1 9 12 13
13 16 17
System 2 9 15 19
13 17 25
Test for any significant differences due to language translator system (Factor A), type of language (Factor B), and interaction. Use = .05.
Answer:
See the attached pictures for answer.
Step-by-step explanation:
See the attached pictures for explanation.
Answer:
Step-by-step explanation:
We have a factorial experiment with Factors A and B, levels a=2, b=3, and n=2 observations.
It's necessary to apply a Two-Factor with replication ANOVA by using Excel or calculating by hand.
1. Sum of Squares could be calculated with the following formulas:
[tex]SS_{T}=\sum\limits^a_{i=1}\sum\limits^b_{j=1}\sum\limits^n_{k=1} {(y_{ijk}-y_{...} ^{2} )}\\SS_{A}=bn\sum\limits^a_{i=1} {(y_{i..}^{2}) -Ny_{...} ^{2} }\\SS_{B}=an\sum\limits^b_{j=1} {(y_{.j.}^{2}) -Ny_{...} ^{2} }\\SS_{AB}=n\sum\limits^a_{i=1}\sum\limits^b_{j=1} {(y_{ij.}^{2}) -Ny_{...} ^{2} -SC_{A} -SC_{B}-SC_{AB}}[/tex]
2. Calculate Degrees of Freedom
Factor A: [tex]a-1[/tex]
Factor B: [tex]b-1[/tex]
Interaction: [tex](a-1)(b-1)[/tex]
Within: [tex]ab(n-1)[/tex]
Total: [tex]abn-1[/tex]
3. Mean Square:
Factor A: [tex]MS_{A}=\frac{SS_{A}}{a-1}[/tex]
Factor B: [tex]MS_{B}=\frac{SS_{B}}{b-1}[/tex]
Interaction: [tex]MS_{AB}=\frac{SS_{AB}}{(a-1)(b-1)}[/tex]
Within:[tex]MS_{E}=\frac{SS_{E}}{ab(n-1)}[/tex]
4. Estimate F and P-value
[tex]F_{A}=\frac{SS_{A} }{SS_{E}}[/tex]
[tex]F_{B}=\frac{SS_{B} }{SS_{E}}[/tex]
[tex]F_{AB}=\frac{SS_{AB} }{SS_{E}}[/tex]
Open the attachment to see the results.
5. Analysis of P-values
Factor A: P-value (0.1280 )>(0.05)
Factor B: P-value (0.0315 )<(0.05)
Interaction: P-value (0.2963 )<(0.05)
P-value of Factor B is less than 0.05, then type of language is insignifficant.
P-value of Factor A and interaction are greater than 0.05, then the translator system and the interaction of both factors are signifficant.
2. A canoe requires 8 hours of fabrication. A row boat requires 5 hours of fabrication. The fabrication department has at most 110 hours to labor each week. Write the equation and solve the problems.
Answer:
Let x is the number of the cannon and y is the number of the row boat.Thus the equation will become
8x+5y=110
Answer: 8C+5R≤110
Step-by-step explanation: Let hours of fabrication of canoes = C
Hours of fabrication of rowboat = R
If a canoe requires 8 hours of fabrication. And a row boat requires 5 hours of fabrication. Then
8C + 5R
The fabrication department has at most 110 hours to labor each week.
Therefore
8C+5R≤110
At most means less than or equal to.
In this question, f : R → R is a differentiable function. We will use linear algebra try to find a good linear approximation to f near a point x = a. Our linear approximation will be of the form y = c + mx. Use the values of f at the point a and the nearby point a h to find a good approximation. (You will need to set up a linear system involving c, m,a, h, f(a), f(a+ h), where c and m are the variables. You should also be able to solve this linear system exactly for the vector.
Answer:
See step by step explanations to get answer.
Step-by-step explanation:
Given that:
Our linear approximation will be of the form y = c + mx. Use the values of f at the point a and the nearby point a h to find a good approximation. (You will need to set up a linear system involving c, m,a, h, f(a), f(a+ h), where c and m are the variables. You should also be able to solve this linear system exactly for the vector.
One method used to distinguish between Granitic (G) and Basaltic (B) rocks is to examine a portion of the infrared spectrum of the sun’s energy reflected from the rock surface. Let R1, R2, and R3 denote measures of spectrum intensities at three different wave lengths; typically, for granite R1 < R2 < R3, whereas for basalt R3 < R1 < R2. When measurements are made remotely (using aircraft), various orderings of the Ri′s may arise whether the rock is basalt of granite. Flights over regions of know composition have yielded the following information: Reading R_1 P(basalt|R1 < R2 < R3). If measurements yielded R1 < R2 < R3, would you classify the rock as granite? b. If measurements yielded R1 < R3 < R2, how would you classify the rock? Answer the same question forR3
Answer:
Step-by-step explanation:
so from the question i have here, i will be giving a step by step analysis of the question.
(a). Here we are showing a relationship, i.e.
P(Granite ║ R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║ R₁ ∠ R₂ ∠ R₃)
from the LHS;
P(Granite ║ R₁ ∠ R₂ ∠ R₃) = P(Granite ║ R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)
= P(Granite)P(R₁ ∠ R₂ ∠ R₃ ║ Granite) / [ P(Granite R₁ ∠ R₂ ∠ R₃) + P(Basaltic R₁ ∠ R₂ ∠ R₃) ]
= 0.25 × 0.6 / [(0.25×0.6)+(0.75×0.1)] = 0.667
from the RHS;
P (Basalt ║ R₁ ∠ R₂ ∠ R₃) = P(Basalt R₁ ∠ R₂ ∠ R₃) / P(R₁ ∠ R₂ ∠ R₃)
= P(Basalt)P(R₁ ∠ R₂ ∠ R₃ ║ Basalt) / [ P(Basalt R₁ ∠ R₂ ∠ R₃) + P(Granite R₁ ∠ R₂ ∠ R₃) ]
= 0.75 × 0.1 / [(0.25 × 0.6)+(0.75 × 0.1)] = 0.333
Therefore from this we can infer that;
P(Granite ║ R₁ ∠ R₂ ∠ R₃) ˃ P (Basalt ║ R₁ ∠ R₂ ∠ R₃)
(b). here we are asked to classify the rocks considering the measurement yield.
Measurement yielded R1 < R3 < R2
P(Granite ║ R₁ ∠ R₃ ∠ R₂) = P(Granite R₁ ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)
= P(Granite)P(R₁ ∠ R₃ ∠ R₂ ║ Granite) / [ P(Granite R₁ ∠ R₃ ∠ R₂) + P(Basaltic R₁ ∠ R₃ ∠ R₂) ]
= 0.25 × 0.25 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.294
also for RHS;
P (Basalt ║ R₁ ∠ R₃ ∠ R₂) = P(Basalt ║ ∠ R₃ ∠ R₂) / P(R₁ ∠ R₃ ∠ R₂)
= P(Basalt)P(R₁ ∠ R₃ ∠ R₂ ║ Basalt) / [ P(Basalt R₁ ∠ R₃∠ R₂) + P(Granite R₁ ∠ R₃ ∠ R₂) ]
= 0.75 × 0.2 / [(0.25 × 0.25)+(0.75 × 0.2)] = 0.706
from this we can infer that;
P(Granite ║ R₁ ∠ R₃ ∠ R₂) ∠ P (Basalt ║ R₁ ∠ R₃ ∠ R₂)
Also considering measurements yielded R₃ ∠ R₁ ∠ R₂
P(Granite ║ R₃ ∠ R₁ ∠ R₂) = P(Granite R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)
= P(Granite)P(R₃ ∠ R₁ ∠ R₂ ║ Granite) / [ P(Granite R₃ ∠ R₁ ∠ R₂) + P(Basaltic R₃ ∠ R₁ ∠ R₂) ]
= 0.25 × 0.15 / [(0.25×0.15)+(0.75×0.7)] = 0.067
from the RHS;
P (Basalt ║ R₃ ∠ R₁ ∠ R₂) = P(Basalt R₃ ∠ R₁ ∠ R₂) / P(R₃ ∠ R₁ ∠ R₂)
= P(Basalt)P(R₃ ∠ R₁ ∠ R₂ ║ Basalt) / [ P(Basalt R₃ ∠ R₁ ∠ R₂) + P(Granite R₃ ∠ R₁ ∠ R₂) ]
1 - 0.067 = 0.933
from this we can infer that;
P(Granite ║ R₃ ∠ R₁ ∠ R₂) ∠ P (Basalt ║ R₃ ∠ R₁ ∠ R₂)
cheers i hope this helps
Final answer:
Measurements yielding R1 < R2 < R3 suggest a granitic composition, while the pattern R1 < R3 < R2 leans granitic, and R3 < R1 < R2 lean towards basaltic. The rock type can be further supported by the appearance of the rock, with granitic rocks being coarse and light-colored due to feldspar and quartz, while basaltic rocks are fine-grained and dark with ferromagnesian minerals. If granite contains basaltic inclusions, the granite is younger.
Explanation:
In analyzing rock compositions from infrared spectrum measurements, if measurements yield R1 < R2 < R3, based on typical patterns, the rock would more likely be classified as granitic because, for granite, the intensity measurements usually increase in that order. Conversely, when measurements produce R1 < R3 < R2, it does not fit neatly into the categories provided for basalt or granite, but it leans closer to a granitic composition given that R3 is not the least among the three, which is a characteristic of basaltic rocks. Similarly, if R3 presents a lower intensity compared to R1 and R2, and you have to classify without further information, it might suggest a basaltic nature following the typical pattern for basalt (R3 < R1 < R2), although it is important to note that real-world applications may require additional contextual information to accurately determine the rock type.
When studying rock samples, you should also consider differences in mineral size (coarse-textured like granite versus fine-textured like basalt), mineral content (dark, mafic minerals vs light, felsic minerals), and elemental content of the igneous rock types. For instance, granitic rock contains feldspar and quartz, which is reflected in its coarse and light-colored appearance. In contrast, basaltic rock, which is fine-grained and dark-colored, includes ferromagnesian minerals and feldspars. Additionally, if a granitic rock has inclusions of basalt (xenoliths), this implies that the granitic rock would typically be younger than the basalt since the inclusions must have been present already for the granite to incorporate them during its formation process.
Consider a value to be significantly low if its z score less than or equal to minus 2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent year, the mean test score was 21.6 and the standard deviation was 5.4. Identify the test scores that are significantly low or significantly high. What test scores are significantly low?
Answer:
The scores that are less than or equal to 10.8 are considered significantly low.
The scores that are greater than or equal to 32.4 are considered significantly high.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 21.6
Standard Deviation, σ = 5.4
We are given that the distribution of score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Significantly low score:
[tex]z \leq -2\\z = \displaystyle\frac{x-21.6}{5.4} \leq -2\\\\\displaystyle\frac{x-21.6}{5.4} \leq -2\\\\x\leq -2(5.4) + 21.6\\\Rightarrow x \leq 10.8[/tex]
Thus, scores that are less than or equal to 10.8 are considered significantly low.
Significantly high score:
[tex]z \geq 2\\z = \displaystyle\frac{x-21.6}{5.4} \geq 2\\\\\displaystyle\frac{x-21.6}{5.4} \geq 2\\\\x\geq 2(5.4) + 21.6\\\Rightarrow x \geq 32.4[/tex]
Thus, scores that are greater than or equal to 32.4 are considered significantly high.
A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2. Calculate how far the car travels in the time it takes to stop. Round your answer to one decimal place.
Answer:
80 feet
Step-by-step explanation:
Given:
Initial speed of the car ([tex]v_0[/tex]) = 40 ft/sec
Deceleration of the car ([tex]\frac{dv}{dt}[/tex]) = -10 ft/sec²
Final speed of the car ([tex]v_x[/tex]) = 0 ft/sec
Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.
Now, deceleration means rate of decrease of velocity.
So, [tex]\frac{dv}{dt}=-10\ ft/sec^2[/tex]
Negative sign means the velocity is decreasing with time.
Now, [tex]\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})[/tex] using chain rule of differentiation. Therefore,
[tex]\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx[/tex]
Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,
[tex]\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft[/tex]
Therefore, the car travels a distance of 80 feet before stopping.
Is there more wood in a 60-foot-high tree trunk with a radius of 2.2 feet or in a 50-foot-high tree trunk with a radius of 2.6 feet? Assume that the trees can be regarded as right circular cylinders.
Answer:
V₂=1061.85 ft³
Step-by-step explanation:
To determine where more wood is found, just find the volume of each log and see which one has the largest volume
knowing that the volume of a straight cylinder is
V=πR²h, where R=radius and h=height
[tex]V_{1}=\pi (2.2)^{2}60= 912.31ft^{3}\\V_{2}=\pi (2.6)^{2}50= 1061.85ft^{3}[/tex]
As we can observe
V₂>V₁
1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260. Use this sample information to construct a 95% confidence interval for the average daily revenue. The store manager believes that the coffee and pastry strategy would lead to an average daily revenue of $1,200. Is the manager correct based on the 95% confidence interval?
Answer:
No, the manager is not correct based on the 95% confidence interval.
Step-by-step explanation:
We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .
The Pivotal quantity for 95% confidence interval is given by;
[tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, X bar = sample mean = $1080
s = sample standard deviation = $260
n = sample size = 35 {five-week}
So, 95% confidence interval for average daily revenue, [tex]\mu[/tex] is given by;
P(-2.032 < [tex]t_3_4[/tex] < 2.032) = 0.95
P(-2.032 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.032) = 0.95
P(-2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95
P(X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] , X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ]
= [ 1080 - 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] , 1080 + 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] ]
= [ 990.70 , 1169.30 ]
No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.
Therefore, the store manager believe is not correct.
The 95% confidence interval for the store's average daily revenue is calculated to be approximately ($993.97, $1166.03). Since $1200 is outside this interval, the manager's belief that the coffee and pastry strategy will lead to an average daily revenue of $1200 is not backed by this confidence level.
Explanation:In the field of statistics, a confidence interval (CI) is a type of interval estimate that is used to indicate the reliability of an estimate. The method for calculating a 95% confidence interval for the average daily revenue involves the sample mean, the standard deviation, and the z-score associated with a 95% confidence level, which is approximately 1.96. Let's use the provided data to calculate:
Calculate the standard error by dividing the standard deviation by the square root of the sample size. Here, the standard deviation is $260, and the sample size is 5 weeks * 7 days/week = 35 days. So, the standard error is $260 / sqrt(35) = $43.89.Multiply the standard error by the z-score to get the margin of error. So, $43.89 * 1.96 = $86.03.Calculate the lower and upper bounds of the 95% confidence interval by subtracting and adding the margin of error from/to the sample mean. So, ($1080 - $86.03, $1080 + $86.03) = ($993.97, $1166.03).The range of this 95% confidence interval is from $993.97 to $1166.03. This means we are 95% confident that the true average daily revenue lies within this interval. Since $1200 lies outside this interval, the manager's belief is not supported by this confidence interval.
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A 8-inch tall sunflower is planted in a garden and the height of the sunflower increases exponentially. 5 days after being planted the sunflower is 13.4805 inches tall. What is the 5-day growth factor for the height of the sunflower
Answer:
The growth factor is approximately 0.11 or 11%.
Step-by-step explanation:
We have been given that a 8-inch tall sunflower is planted in a garden and the height of the sunflower increases exponentially. 5 days after being planted the sunflower is 13.4805 inches tall.
We will use exponential growth formula to solve our given problem.
[tex]y=a\cdot (1+r)^x[/tex], where,
y = Final value,
a = Initial value,
r = Growth rate in decimal form,
x = Time.
Upon substituting initial value [tex]a=8[/tex], [tex]x=5[/tex] and [tex]y=13.4805[/tex], we will get:
[tex]13.4805=8\cdot(1+r)^5[/tex]
[tex]8\cdot(1+r)^5=13.4805[/tex]
[tex]\frac{8\cdot(1+r)^5}{8}=\frac{13.4805}{8}[/tex]
[tex](1+r)^5=\frac{13.4805}{8}[/tex]
Now, we will take 5th root of both sides of equation as:
[tex]\sqrt[5]{(1+r)^5} =\sqrt[5]{\frac{13.4805}{8}}[/tex]
[tex]1+r =\sqrt[5]{1.6850625}[/tex]
[tex]1+r =1.1100005724234515[/tex]
[tex]1-1+r =1.1100005724234515-1[/tex]
[tex]r=0.1100005724234515[/tex]
[tex]r\approx 0.11[/tex]
Therefore, the growth factor is approximately 0.11 or 11%.
The 5-day growth factor of the sunflower, which grows exponentially, is calculated by dividing the final height (13.4805 inches) by the initial height (8 inches) giving a growth factor of approximately 1.6850625.
Explanation:The 5-day growth factor of the sunflower that increases exponentially can be calculated by dividing the final height by the initial height. The formula is:
5-day growth factor = final height / initial height
To calculate the 5-day growth factor, we substitute the heights into the formula. We have:
5-day growth factor = 13.4805 inches (final height) / 8 inches (initial height)
So, 5-day growth factor = 1.6850625. This means that the height of the sunflower is multiplied by approximately 1.685 each day for 5 days, which accounts for the exponential growth.
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Ask Your Teacher What was the age distribution of prehistoric Native Americans? Extensive anthropological studies in the southwestern United States gave the following information about a prehistoric extended family group of 86 members on what is now a Native American reservation. For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the class 31 and over, use 35.5 as the class midpoint. (Round your answers to one decimal place.)
Answer:
[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]
Now in order to calculate the variance we can use the following formula:
[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]
and replacing we got:
[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]
And the deviation would be the square root of the variance:
[tex] s = \sqrt{111.33} = 10.6[/tex]
Step-by-step explanation:
We assume the following dataset
Age range (years) 1-10 11-20 21-30 >31
Number of individuals 30 18 23 10
Solution to the problem
We can solve the problem creating the following table:
Class Midpoint(xi) fi xi*fi xi^2 *fi
1-10 5.5 30 165 907.5
11-20 15.5 18 279 4324.5
21-30 25.5 23 586.5 14955.75
>31 35.5 10 355 12602.5
___________________________________
Total 81 1385.5 32790.25
The midpoint is calculated as the average between the lower and the upper interval values.
We can calculate the mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]
Now in order to calculate the variance we can use the following formula:
[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]
and replacing we got:
[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]
And the deviation would be the square root of the variance:
[tex] s = \sqrt{111.33} = 10.6[/tex]
The numerical course grades in a statistics course can be approximated by a normal model with a mean of 70 and a standard deviation of 10. The professor must convert the numerical grades to letter grades. She decides that she wants 10% A's, 30% B's, 40% C's, 15% D's, and 5% F's. a. What is the cutoff for an A grade?
Answer:
The cutoff for an A grade is 82.8.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 70, \sigma = 10[/tex]
a. What is the cutoff for an A grade?
The top 10% of the class get an A grade. So the cutoff is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 70}{10}[/tex]
[tex]X - 70 = 1.28*10[/tex]
[tex]X = 82.8[/tex]
The cutoff for an A grade is 82.8.
A college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section was 460 with a standard deviation of 105, and the mean score for the Quantitative Reasoning was 429 with a standard deviation of 148. Suppose that both distributions are nearly normal. Round calculated answers to 4 decimal places unless directed otherwise.
Answer:
Step-by-step explanation:
Given that a college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section.
Quantitative reasons scores are N (429, 148)
Hence Z score for the college senior = [tex]\frac{530-429}{148} \\=0.6824[/tex]
The mean score for Verbal Reasoning section was 460 with a standard deviation of 105
Verbal reasoning score was N(460, 105)
The score of college senior = 530
Z score for verbal reasoning =[tex]\frac{530-460}{105} \\=0.6667[/tex]
comparing we can say he scored more on quantiative reasoning.
Thus comparison was possible only by converting to Z scores.