Answer:
See steps below
Step-by-step explanation:
The function
[tex]\large f(x,y)=(1-x)^2+100(y-x^2)^2[/tex]
is a particular case of the general Rosenbrock’s Function.
a)
Since
[tex]\large ((1-x)^2\geq 0[/tex] for all the values of x and equals 0 when x=1 and
[tex]\large (y-x^2)^2=(y-1)^2\geq 0[/tex] for all the values of y and equals 0 only when y=1, we conclude that (1,1) is a minimum.
On the other hand,
f(x,y)>0 for (x,y) ≠ (1,1) so (1,1) is a global minimum.
b)
To confirm that this function is not convex, we will be using the following characterization of convexity
“f is convex if, and only if, the Laplace operator of f [tex]\large \nabla^2f \geq 0[/tex] for every (x,y) in the domain of f”
Given that the domain of f is the whole plane XY, in order to prove that f is not convex, we must find a point (x,y) at where the Laplace operator is < 0.
The Laplace operator is given by
[tex]\large \nabla^2f=\displaystyle\frac{\partial ^2f}{\partial x^2}+\displaystyle\frac{\partial ^2f}{\partial y^2}[/tex]
Let us compute the partial derivatives
[tex]\large \displaystyle\frac{\partial f}{\partial x}=-2(1-x)+200(y-x^2)(-2x)=-2+2x-400xy+400x^3\\\\\displaystyle\frac{\partial^2 f}{\partial x^2}=2-400y+1200x^2[/tex]
and
[tex]\large \displaystyle\frac{\partial f}{\partial y}=200(y-x^2)=200y-200x^2\\\\\displaystyle\frac{\partial^2 f}{\partial x^2}=200[/tex]
we have then
[tex]\large \nabla^2 f=2-400y+1200x^2+200[/tex]
if we take (x,y) = (0,1)
[tex]\large \nabla^2 f(0,1)=2-400+200=202-400=-198<0[/tex]
hence f is not convex.
The fox population in a certain region has a continuous growth rate of 4 percent per year. It is estimated that the population in the year 2000 was 27500.
Answer:
there is no question
but it grew 7500
over the time
Step-by-step explanation:
Question:
The fox population in a certain region has a continuous growth rate of 4 percent per year. It is estimated that the population in the year 2000 was 23700.
(a) Find a function that models the population (t) years after 2000 ( t=0 for 2000).
(b) Use the function from part (a) to estimate the fox population in the year 2008.
Answer:
A: P(t)=23700e^(.04t)
B: ≈32638
Step-by-step explanation:
A: The formula A=Pe^(rt) can be used. Initial value of 23700, e is the exponential constant, the rate is 4% which can be seen as 4/100, and t=time
B: Use the equation you created and substitute 8 for t since the years of difference between 2008-2000 is 8 years
23700e^(.04(8)) ≈ 32637.9280148
Rounded to the nearest fox is ≈32638
Find the x-coordinates of all critical points of the given function. Determine whether each critical point is a relative maximum, minimum, or neither by first applying the second derivative test, and, if the test fails, by some other method. f(x) = 5x4 − 2x3
Answer:
f''(0.3) > 0 therefore,
x = 0.3 is point of minima
and,
f''(0) = 0
thus,
x = 0 is point of neither maxima nor minima
Step-by-step explanation:
Given function:
f(x) = 5x⁴ − 2x³
Now,
To find the points of maxima or minima, put f'(x) = 0
thus,
f'(x) = (4)5x³ - (3)2x² = 0
or
20x³ - 6x² = 0
or
x(20x² - 6x) = 0
or
x = 0 and 20x² - 6x = 0
or
x = 0 and 2x(10x - 3) = 0
or
x = 0 and 2x = 0 and (10x - 3) = 0
or
x = 0 and x = 0 and x = [tex]\frac{3}{10}[/tex] = 0.3
thus,
condition for maxima or minima
f''(x) = (3 × 4)5x² - (2 × 3)2x
or
f''(x) = 60x² - 12x
at
x = 0
f''(0) = 60(0)² - 12(0) = 0
at x = 0.3
f''(0.3) = 60(0.3)² - 12(0.3)
= 5.4 - 3.6
= 1.8
since,
f''(0.3) > 0 therefore,
x = 0.3 is point of minima
and,
f''(0) = 0
thus,
x = 0 is point of neither maxima nor minima
a bridge is 38 feet above a river. how many seconds does it take a rock dropped from the bridge to pass by a tree limb that is 10 feet above the water?
The rock does not pass by the tree limb.
Explanation:To calculate the time it takes for the rock to pass by the tree limb, we can use the concept of free fall motion. The rock undergoes free fall motion from a height of 38 feet until it reaches a height of 10 feet. We can use the equation for free fall motion, h = ut + (1/2)gt^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
Plugging in the values, we have:
38 = 0 - (1/2)(9.8)t^2
Simplifying the equation, we get:
9.8t^2 = -38
t^2 = -38/9.8
t = sqrt(-38/9.8)
Since we cannot take the square root of a negative number, it means that the rock never reaches a height of 10 feet. Therefore, the rock does not pass by the tree limb.
Learn more about Free fall motion here:https://brainly.com/question/28806509
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Carleton Chemical claims that they can produce an average of more than 800 tons of meladone
per week. A random sample of 36 weeks of production yielded a sample mean of 823 tons, with
a standard deviation of 79.8 tons.
Does the sample data provide sufficient evidence to support the claim
made by Carleton Chemical? Use a significance level of α = .05.
Answer:
t=1.729
[tex]p_v =P(t_{(35)}>1.729)=0.0463[/tex]
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=823[/tex] represent the mean production for the sample
[tex]s=79.8[/tex] represent the sample standard deviation for the sample
[tex]n=36[/tex] sample size
[tex]\mu_o =800[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
Part a: State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean production is higher than 800 tons, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 800[/tex]
Alternative hypothesis:[tex]\mu > 800[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Part b: Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{823-800}{\frac{79.8}{\sqrt{36}}}=1.729[/tex]
Part c: P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=36-1=35[/tex]
Since is a one side test the p value would be:
[tex]p_v =P(t_{(35)}>1.729)=0.0463[/tex]
Part d: Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the production its significant higher compared to the value of 800 tons at 5% of signficance.
use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 ), (15 comma 0 ), and (0 comma 15 )and whose cross sections perpendicular to the base and parallel to the y-axis are semicircles
Final answer:
The volume of the solid with a triangular base and semicircular cross sections is found by integrating the area of the semicircles along the height of the triangle. This approach involves calculating the area of semicircles using their radius which changes with y, and then integrating from 0 to 15 along the y-axis. The total volume is (56.25)π cubic units.
Explanation:
To find the volume of the solid with a triangular base and semicircular cross sections perpendicular to the base and parallel to the y-axis, we use the general slicing method. The base of the solid is a triangle with vertices at (0,0), (15,0), and (0,15), implying that it is a right-angled isosceles triangle on the xy-plane. The length of the legs are both 15 units.
For a slice at a particular y-value, the length of the base of the semicircle (which is also the diameter) is equal to the x-value at that y (since the triangle's equation is x + y = 15, x = 15 - y). The radius of the semicircle is thus ½(15 - y). The area of a semicircle is given by ½[tex]πr^{2}[/tex], substituting the expression for the radius we get ½π(½(15 - y))2.
To find the volume, we integrate this area from y = 0 to y = 15 along the y-axis. The integral of ½π(½(15 - y))2 dy from 0 to 15 is ½π × ½2 × ∫015 (15 - y)2 dy, which simplifies to ∑ volume = ⅔π(152) (⅓) = ½π(225)(⅓) = (56.25)π cubic units.
Information on a packet of seeds claims that 93% of them will germinate. Of the 200 seeds that I planted, only 175 germinated. (a) Find a 95% CI on the true proportion of seeds that germinate based on this sample. (b) Does this seem to provide evidence that the claim is wrong
Answer:
We reject H₀
we accept Hₐ seeds in the packet would germinate smaller than 93%
Step-by-step explanation:
Test of proportions
One tail-test (left side)
93 % = 0.93
p₀ = 0,93
1.- Hypothesis
H₀ ⇒ null hypothesis p₀ = 0.93Hₐ ⇒ Alternative hypothesis p = 0.8752.-Confidence interval 95 %α = 0,05 and z(c) = - 1.643.- Compute z(s)z(s) = (p - p₀)/√(p₀*q₀)/n z(s) = (0.875-0.93)/√0.93*0.07)200z(s) = - 0,055/ √0.0003255z(s) = - 0.055/ 0.018z(s) = - 3,064.-Compere z(c) and z(s)z(s) < z(c) -3.06 < -1.64z(s) is in rejection region, we reject H₀After checking to ensure that the sample data follows the necessary condition for using the t-procedure, we use Minitab to get a 90% confidence interval to estimate mu. We find it to be (118.51 ounces, 124.29 ounces). What does this interval tell us?
Answer:
the 90% confidence interval to estimate mu = (118.51 ounces, 124.29 ounces) means that out of all the data 90% of the sample has the weight between the range 118.51 ounces and 124.29 ounces.
Step-by-step explanation:
Data provided in the question:
Confidence level = 90%
90% confidence interval to estimate mu = (118.51 ounces, 124.29 ounces)
Now,
The confidence level tells us that out of all the value of the sample the given percentage of confidence level lies within the calculated interval.
Here in the given question,
the 90% confidence interval to estimate mu = (118.51 ounces, 124.29 ounces) means that out of all the data 90% of the sample has the weight between the range 118.51 ounces and 124.29 ounces.
Ryan boards a Ferris wheel at the 3-o'clock position and rides the Ferris wheel for multiple revolutions. The Ferris wheel rotates at a constant angular speed of 6.5 radians per minute and has a radius of 30 feet. The center of the Ferris wheel is 36 feet above the ground. Let t represent the number of minutes since the Ferris wheel started rotating.
a. Write an expression (in terms of t) to represent the varying number of radians 0 Cody has swept out since the ride started.
b. Write an expression (in terms of t) to represent Cody's height (in feet) above the center of the Ferris wheel.
c. Write an expression (in terms of t) to represent Cody's height (in feet) above the ground.
a. The expression for the varying number of radians Cody has swept out since the ride started is 6.5t radians.
b. Cody's height above the center of the Ferris wheel is[tex]\(30\cos(6.5t)\)[/tex] feet.
c. Cody's height above the ground is [tex]\(36 + 30\cos(6.5t)\)[/tex] feet.
a. To represent the varying number of radians Cody has swept out since the ride started, we use the formula for angular distance:
[tex]\[ \text{Angular distance} = \text{angular speed} \times \text{time} \][/tex]
Given that the Ferris wheel rotates at a constant angular speed of 6.5 radians per minute, the expression in terms of t is:
[tex]\[ \text{Angular distance} = 6.5t \][/tex]
b. To represent Cody's height above the center of the Ferris wheel, we use the relationship between angular displacement and height:
[tex]\[ \text{Height} = \text{radius} \times \cos(\text{angular distance}) \][/tex]
Substituting the expression for angular distance from part (a), we get:
[tex]\[ \text{Height} = 30 \times \cos(6.5t) \][/tex]
c. To represent Cody's height above the ground, we add the height of the center of the Ferris wheel to Cody's height above the center:
[tex]\[ \text{Total height} = \text{Center height} + \text{Height above center} \][/tex]
Given that the center of the Ferris wheel is 36 feet above the ground, the expression in terms of t becomes:
[tex]\[ \text{Total height} = 36 + 30 \times \cos(6.5t) \][/tex]
These expressions represent Cody's angular distance, height above the center of the Ferris wheel, and height above the ground as functions of time t.
An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 100100 engines and the mean pressure was 7.27.2 lbs/square inch. Assume the variance is known to be 0.810.81. If the valve was designed to produce a mean pressure of 7.37.3 lbs/square inch, is there sufficient evidence at the 0.010.01 level that the valve does not perform to the specifications? State the null and alternative hypotheses for the above scenario.
Answer:
Null hypothesis:[tex]\mu \geq 7.3[/tex]
Alternative hypothesis:[tex]\mu < 7.3[/tex]
[tex]z=\frac{7.2-7.3}{\frac{0.81}{\sqrt{100}}}=-1.235[/tex]
[tex]p_v =P(z<-1.235)=0.1084[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so then we can conclude that the mean pressure for the automobile engines is not significantly less than 7.3 at 1% of significance.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=7.2[/tex] represent the sample mean
[tex]\sigma=0.81[/tex] represent the population standard deviation
[tex]n=100[/tex] sample size
[tex]\mu_o =7.3[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean pressure is lower than the specificated value 7.3, the system of hypothesis are :
Null hypothesis:[tex]\mu \geq 7.3[/tex]
Alternative hypothesis:[tex]\mu < 7.3[/tex]
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{7.2-7.3}{\frac{0.81}{\sqrt{100}}}=-1.235[/tex]
4) P-value
Since is a one-side lower test the p value would given by:
[tex]p_v =P(z<-1.235)=0.1084[/tex]
5) Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean pressure for the automobile engines is not significantly less than 7.3 at 1% of significance.
A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained physical activity. What sample size should be obtained if he wishes the estimate to be within two percentage points with 90% confidence, assuming that
(a) he uses the estimates of 21.6% male and 19.2% female from a previous year?
(b) he does not use any prior estimates?
Answer:
Step-by-step explanation:
The standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean(μ) and standard deviation(σ)
The general formula for the sample size is given below:
[tex]n=p^{'}(1-p^{'})(\frac{Z_{\frac{a}{3} } }{E} )^{2}[/tex]
The formular for finding sample size is given as:
[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]
a.)
it is given that [tex]E=±0.02, p^{'}_{1}=0.216, p^{'}_{2}=0.192[/tex]
The confidence level is 0.90
For (1 - ∝) = 0.90
∝=0.10; ∝/2 = 0.05
frm the standard normal table, the required [tex]Z_{0.05}[/tex] value for 90% confidence is 1.645. The sample size is as shown:
[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]
=[tex]n=(\frac{1.645}{0.05} )^{2} * (0.216(1-0.216)+0.192(1-0.192))\\=351.22≅352[/tex]
The required sample size is 352 (nearest whole number)
b.)
it is given that [tex]E=±0.02, p^{'}_{1}=0.5, p^{'}_{2}=0.5[/tex]
The confidence level is 0.90
For (1 - ∝) = 0.90
∝=0.10; ∝/2 = 0.05
frm the standard normal table, the required [tex]Z_{0.05}[/tex] value for 90% confidence is 1.645. The sample size is as shown:
[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]
=[tex]n=(\frac{1.645}{0.05} )^{2} * (0.5(1-0.5)+0.5(1-0.5))\\=541.205≅542[/tex]
The required sample size is 542 (nearest whole number)
Using the estimates from a previous year (21.6% male and 19.2% female), a sample size of 811 should be obtained. If no prior estimates are used, a sample size of 848 is needed.
Explanation:To determine the sample size needed for estimating the difference in the proportion of men and women who participate in regular sustained physical activity, we can use the formula:
[tex]n = (Z^2 * p * (1-p)) / E^2[/tex]
Where:
n is the required sample sizeZ is the Z-score corresponding to the desired confidence level (90%)p is the estimated proportion in the population (either the estimates from a previous year or a hypothesized value)E is the desired margin of error (two percentage points)(a) If the therapist uses the estimates of 21.6% male and 19.2% female from a previous year, we can assume that the sample proportion for both genders is the same (20.4%).
Plugging in the values:
[tex]n = ((1.645^2) * 0.204 * (1-0.204)) / (0.02^2)[/tex]
= 810.611
Therefore, a sample size of 811 should be obtained.
(b) If the therapist does not use any prior estimates, we can assume that the sample proportion for both genders is 0.5 (maximum variability).
Plugging in the values:
[tex]n = ((1.645^2) * 0.5 * (1-0.5)) / (0.02^2)[/tex]
= 847.075
Therefore, a sample size of 848 should be obtained.
A large jetliner with a wingspan of 40 m flies horizontally and due north at a speed of 300 m/s in a region where the magnetic field of earth is 60 uT directed 50 degrees below the horizontal. What is the magnitude of the induced emf between the ends of the wing?
a) 250 mV
b)350 mV
c) 550 mV
d) 750 mV
Using Faraday's law of electromagnetic induction, the induced emf on a jetliner with a 40 m wingspan flying at 300 m/s through a 60 uT magnetic field directed 50 degrees below the horizontal is calculated to be 550 mV.
Explanation:The question involves calculating the induced electromotive force (emf) in the wings of a jetliner flying through the Earth's magnetic field, a concept from physics. By applying Faraday's law of electromagnetic induction, we can determine the induced emf when a conductor moves through a magnetic field.
To solve for the induced emf, we use the equation emf = B * v * l * sin(θ), where B is the magnetic field strength, v is the velocity of the conductor, l is the length of the conductor, and θ is the angle between the direction of velocity and the magnetic field. Since the magnetic field is directed 50 degrees below the horizontal and the plane is flying horizontally due north, the angle θ will be 90 degrees (because the movement is perpendicular to the direction of the magnetic field's horizontal component).
Thus, the induced emf is calculated as follows:
B = 60 uT = 60 x 10-6 Tv = 300 m/sl = 40 msin(θ) = sin(90°) = 1emf = 60 x 10-6 T * 300 m/s * 40 m * 1 = 0.72 V = 720 mV
However, since this value is not an option in the multiple-choice answers provided, we might consider only the vertical component of the magnetic field, which contributes to the induction effect:
Bvertical = B * sin(50°)sin(50°) = 0.77 (approximately)Bvertical = 60 x 10-6 T * 0.77 = 46.2 x 10-6 Temf = 46.2 x 10-6 T * 300 m/s * 40 m * 1 = 0.55 V = 550 mVTherefore, the correct answer is 550 mV, which corresponds to option (c).
At an insurance sales conference, exactly 100 people are selected at random from conference registration records. Under "specialty", 25 people indicated that they specialize in health insurance, 36 said they focused on life insurance, and 24 said "Other". The remaining individuals specified either Car insurance or Home Insurance. a) If you were to pick 2 individuals at random, what is the probability that the first one specializes in life insurance, and the second one also specializes in life insurance
Answer:
P(1st health insurance)*p(2nd health insurance) = 0.1296
Step-by-step explanation:
p(health insurance) = =25/100 = 0.25
p(life insurance) = =36/100 = 0.36
p(other) = =24/100 = 0.24
p(car or home) = =1-0.25-0.36-0.24 = 0.15
a) P(1st health insurance)*p(2nd health insurance)
= 0.36*0.36
= 0.1296
The following hypotheses are given. H0 : σ1² ≤ σ2² H1 : σ1² > σ2² A random sample of five observations from the first population resulted in a standard deviation of 12. A random sample of seven observations from the second population showed a standard deviation of 7. At the 0.01 significance level, is there more variation in the first population?
Final answer:
To assess whether there is more variation in the first population than in the second, calculate the F-statistic using the ratio of the squares of the sample standard deviations and compare it to the critical F-value at the 0.01 significance level. A calculated F-value significantly greater than 1 suggests greater variation in the first population, but the final decision depends on comparing the F-value to the F-distribution table.
Explanation:
To determine whether there is more variation in the first population at the 0.01 significance level, we can use an F-test for the comparison of two variances given that the populations are normally distributed. We begin by calculating the F-statistic, which is the ratio of the squares of the two sample standard deviations. The formula we use is:
F = (S1)2 / (S2)2
For the given data, the first population standard deviation is 12 and the second is 7, so we calculate the F-statistic as:
F = (122) / (72) = 144 / 49 = 2.9388
An F-value significantly greater than 1 suggests that the variance in the first population is greater than that in the second population. To determine if this observed F-value is statistically significant, we compare it to the critical value of F for the given degrees of freedom at the chosen significance level. However, critical F-values are obtained from statistical tables or software, and as we do not have the specific value for the degrees of freedom here, we cannot make the final comparison.
If the calculated F-value exceeds the critical value from an F-distribution table, we would reject the null hypothesis and conclude that there is more variation in the first population. If it does not exceed the critical value, we would not reject the null hypothesis.
In a health club, research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard deviation of 5.4 minutes. It is assumed that this is a normally distributed variable.
Find the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill.
a)0.38
b)0.69
c)0.99
d)0.31
Answer: d) 0.31
Step-by-step explanation:
Given : In a health club, research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard deviation of 5.4 minutes.
i.e. [tex]\mu=42.5[/tex] and [tex]\sigma= 5.4[/tex]
It is assumed that this is a normally distributed variable.
Let x denotes the time spend by a person on treadmill.
Then, the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill.
[tex]P(30<x<40)=P(\dfrac{30-42.5}{5.4}<\dfrac{x-\mu}{\sigma}<\dfrac{40-42.5}{5.4})\\\\\approxP(-2.31<z<0.46)\\\\=P(z<-0.46)-P(z<-2.31)\ \ [\because\ P(z_1<z<z_2)=P(z<z_2)-P(z<z_1)]\\\\=1-P(z<0.46)-(1-P(z\leq2.31))\ \ [\because P(Z>z)=1-P(Z\leq z)]\\\\=1-0.6772419-(1-0.9895559)\ \ \text{[By using z-table or calculator]}\\\\=0.3227581-0.0104441=0.312314\approx0.31[/tex]
Hence, the required probability = 0.31
Thus , the correct answer = d) 0.31
A rectangle has a perimeter of 52 inches. The length of the rectangle is 4 inches more than its width. What is the length of the rectangle?
Answer: The length of the rectangle is 15 inches
Step-by-step explanation:
Let L represent the length of the rectangle.
Let W represent the width of the rectangle.
The perimeter of a rectangle is expressed as 2 length + 2 width
Perimeter = 2(L + W)
The rectangle has a perimeter of 52 inches. It means that
2(L + W) = 52
L + W = 52/2 = 26 - - - - - - - 1
The length of the rectangle is 4 inches more than its width. It means that
L = W + 4
Substituting L = W + 4 into equation 1, it becomes
W + 4 + W = 26
2W = 26 - 4 = 22
W = 22/2 = 11
L = W + 4 = 11 + 4 = 15
For the function below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subintervals. Then take a limit of these sums as n right arrow infinityn → [infinity] to calculate the area under the curve over [a comma b ][a,b]. f (x )equals x squared plus 3f(x)=x2+3 over the interval [0 comma 4 ]
Answer:
See below
Step-by-step explanation:
We start by dividing the interval [0,4] into n sub-intervals of length 4/n
[tex][0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4][/tex]
Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.
Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)
[tex]\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12[/tex]
but
[tex]1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}[/tex]
so the upper sum equals
[tex]\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12[/tex]
When [tex]n\rightarrow \infty[/tex] both [tex]\displaystyle\frac{3}{n}[/tex] and [tex]\displaystyle\frac{1}{n^2}[/tex] tend to zero and the upper sum tends to
[tex]\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}[/tex]
The average age for a person getting married for the first time is 26 years.
Assume that the ages for the first marriages have a normal distribution with a standard deviation of about 4 years.90% of people getting married for the first time get married before what age?
Answer:
90% of people getting married for the first time get married before 32 years.
Step-by-step explanation:
The average age for a person getting married for the first time is 26 years i.e. [tex]\mu=26[/tex]
The ages for the first marriages have a normal distribution with a standard deviation of about 4 years i.e. [tex]\sigma = 4[/tex]
90% of people getting married for the first time.
The z-value at 90% confidence interval is z=1.64.
The z-score formula is given by,
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Where, x is the required sample mean or age.
Substitute the values,
[tex]1.64=\frac{x-26}{4}[/tex]
[tex]1.64\times 4=x-26[/tex]
[tex]6.56=x-26[/tex]
[tex]x=6.56+26[/tex]
[tex]x=32.56[/tex]
Therefore, 90% of people getting married for the first time get married before 32 years.
A veterinary researcher takes an SRS of 60 horses presenting with colic whose average age is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of horses coming to the veterinary clinic is 8 years. The probability that a sample mean is 12 or larger for a sample from the horse population is:
The probability that a sample mean is 12 or larger for a sample from the horse population is approximately 0.0264, which matches the correct answer.
Given:
Population mean (μ): 10 years
Sample mean ([tex]\bar{x}[/tex]): 12 years
Sample size (n): 60 horses
Population standard deviation (σ): 8 years
First, let's calculate the standard error of the sample mean (SE) using the formula:
SE = σ / √n
SE = 8 / √60 ≈ 1.032
Now, let's find the z-score using the formula:
z = ([tex]\bar{x}[/tex] - μ) / SE
z = (12 - 10) / 1.032 ≈ 1.938
Next, we find the probability that a sample mean is 12 or larger by finding the area under the standard normal curve to the right of z = 1.938.
Using a standard normal distribution table or calculator, we find that P(Z > 1.938) ≈ 0.0264.
The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 21 who smoke. Step 1 of 2 : Suppose a sample of 292 Americans over 21 is drawn. Of these people, 225 don't smoke. Using the data, estimate the proportion of Americans over 21 who smoke. Enter your answer as a fraction or a decimal number rounded to three decimal places
Answer:
[tex]\frac{67}{292} = 0.229[/tex]
Step-by-step explanation:
Out of 292 drawn Americans over 21, 225 of them don't smoke. That means the rest of them are smokers:
So there are 292 - 225 = 67 smokers here.
We can then calculate the portion of Americans over 21 who smoke:
[tex]\frac{67}{292} = 0.229[/tex]
Which of the following is generally true about a distribution that is highly skewed to the right?
A) The median is less than the mean
B) The median is greater than the mean
C) The median is approximately equal to the mean
D) Whether the mean or median is larger depends on the value of the standard deviation of the distribution
Step-by-step explanation:
The answer is A because since the median is resistant to extreme values but the mean is not, the mean tends to move toward extreme values in the distribution.
Sorry if I'm wrong but i tried my best sorry :) :(
The statement which is generally true about a distribution that is highly skewed to the right is A) The median is less than the mean.
What is Mean?Mean of a set of data is defined as the average of all the values. It gives the exact middle point of the data set.
In a right skewed distribution, the data values are more concentrated on the right side.
So the values of the data will be larger, which in turn makes the value of the mean to be higher.
Since median is the exact middle value in the set, it does not depend on the right values.
So mean will be larger than median.
Hence the correct option is A.
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Determine whether Rolle's Theorem can be applied to the function f(x) = (x + 3)(x - 2)^2 on the closed interval {-3, 2}. If Rolle's Theorem can be applied, find all numbers c in the open interval (-3,2) such that f'(c) = 0.
Rolle's Theorem applies, -(8/3)
Rolle's Theorem applies, -(4/5)
Rolle's Theorem applies, -(4/3)
Rolle's Theorem does not apply.
Answer:
Option C) Rolle's theorem applies
[tex]c = \displaystyle\frac{-4}{3}[/tex]
Step-by-step explanation:
We are given that:
[tex]f(x) = (x + 3)(x - 2)^2[/tex]
Closed interval: [-3,2]
Rolle's Theorem:
According to this theorem if the given function
continuous in [a,b]differentiable in (a,b)f(a) = f(b)the, there exist c in (a,b) such that
[tex]f'(c) = 0[/tex]
Continuity of function:
Since the given function is a continuous function, it is continuous everywhere. Therefore, f(x) is continuous in [-3,2]
Differentiability of function:
A polynomial function is differentiable For all arguments. Therefore, f(x) is differentiable in (-3,2)
Now, we evaluate f(-3) and f(2)
[tex]f(x) = (x + 3)(x - 2)^2\\f(-3) = (-3+3)(-3-2)^2 = 0\\f(2) - (2+3)(2-2)^2 = 0\\\Rightarrow f(-3) = f(2) = 0[/tex]
Thus, Rolle's theorem applies on the given function f(x).
According to Rolle's theorem there exist c in (a,b) such that f'(c) = 0
[tex]f(x) = (x + 3)(x - 2)^2\\f'(x) = (x-2)^2 + 2(x+3)(x-2) = 3x^2-2x-8\\f'(c) = 0\\f'(c) = 3c^2-2c-8 = 0\\\Rightarrow (c-2)(3c+4) = 0\\\Rightarrow c = 2, c = \displaystyle\frac{-4}{3}[/tex]
c should lie in (-3,2)
Thus,
[tex]c = \displaystyle\frac{-4}{3}[/tex]
Final answer:
Rolle's Theorem can be applied to the function f(x) = (x + 3)(x - 2)^2 on the closed interval {-3, 2}, and the number c in the open interval (-3, 2) where f'(c) = 0 is c = -4/3.
Explanation:
To determine whether Rolle's Theorem can be applied to the function f(x) = (x + 3)(x - 2)^2 on the closed interval {-3, 2}, we need to check if the function satisfies the three conditions for Rolle's Theorem:
f(x) is continuous on the closed interval [-3, 2].f(x) is differentiable on the open interval (-3, 2).f(-3) = f(2).First, note that the function f(x) = (x + 3)(x - 2)^2 is a polynomial function, and all polynomial functions are continuous and differentiable for all real numbers. Therefore, condition 1 is satisfied. Next, we find the derivative of f(x):
f'(x) = (2x-4)(x-2)+(x+3)(2(x-2)) = 4x^2 - 12x + 8 + 2x^2 - 2x - 12 = 6x^2 - 14x - 4.
To find the values of c in the open interval (-3, 2) where f'(c) = 0, we set f'(x) = 0:
6x^2 - 14x - 4 = 0.
We can solve this quadratic equation using factoring, the quadratic formula, or by completing the square. Solving the equation, we find the roots as x = -4/3 and x = 2/3. However, only the root x = -4/3 is in the interval (-3, 2). So, Rolle's Theorem can be applied, and the number c in the open interval (-3, 2) where f'(c) = 0 is c = -4/3.
Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places. 64 0 sin( x ) dx, n = 4
The Midpoint Rule is used to approximate the value of an integral. In this case, we are approximating the integral ∫ sin(x) dx on the interval [0, 64] using the Midpoint Rule with n = 4. The approximate value of the integral is approximately 39.6007.
Explanation:The Midpoint Rule is used to approximate the value of an integral by dividing the interval into equal subintervals and evaluating the function at the midpoint of each subinterval. In this case, the integral is ∫ sin(x) dx on the interval [0, 64] and n = 4.
We can calculate the width of each subinterval by dividing the total interval length by the number of subintervals: (64-0)/4 = 16.
Next, we evaluate the function sin(x) at the midpoint of each subinterval and sum up the results, multiplying by the width of each subinterval:
Approximation = 16 * [sin(8) + sin(24) + sin(40) + sin(56)]
Calculating the values of sin(8), sin(24), sin(40), and sin(56) using a calculator, we get:
Approximation ≈ 16 * [0.1392 + 0.4121 + 0.7451 + 0.9309] ≈ 39.6007
Therefore, the approximate value of the integral ∫ sin(x) dx on the interval [0, 64] using the Midpoint Rule with n = 4 is approximately 39.6007.
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To approximate the integral, we divide the interval into subintervals, find the midpoints, evaluate the function at the midpoints, and multiply by the width of the subintervals.
Explanation:To approximate the integral using the Midpoint Rule, we divide the interval [0, 64] into n subintervals of equal width. In this case, n = 4, so each subinterval has width (64-0)/4 = 16.
The midpoint of each subinterval is used to estimate the value of the function sin(x) within that subinterval.
The midpoints of the subintervals are: 8, 24, 40, and 56.
We evaluate sin(x) at these midpoints and multiply each value by the width of the subintervals (16).
Finally, we sum these values to approximate the integral.
Using a calculator or computer, we find that sin(8) = 0.989, sin(24) = -0.905, sin(40) = 0.745, and sin(56) = -0.521.
Therefore, the approximation of the integral is (0.989 + (-0.905) + 0.745 + (-0.521)) * 16 = 3.296.
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The mean monthly expenditure on gasoline per household in Middletown is determined by selecting a random sample of 100 households. The sample mean is $128, with a sample standard deviation of $38, what is the upper bound of a 90% confidence interval for the mean monthly expenditure on gasoline per household in Middletown? a. $134.66 b. $162.20 c. $129.38 d. $131.42 e. $13287
Answer:
a. $134.66
Step-by-step explanation:
1) 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".
[tex]\bar X =128[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=38 represent the sample standard deviation
n=100 represent the sample size
2) Calculate the confidence interval
Since the sample size is large enough n>30. The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]
Now we have everything in order to replace into formula (1):
[tex]128-1.64\frac{38}{\sqrt{100}}=121.768[/tex]
[tex]128+1.64\frac{38}{\sqrt{100}}=134.232[/tex]
The closest value would be $134.66 and that would be the answer for this case.
The upper bound of a 90% confidence interval for the mean monthly gasoline expenditure per household, based on the sample provided, is approximately $134.66. This is derived using the z-score for a 90% confidence level and the given sample mean and standard deviation.
Explanation:To calculate the upper bound of a 90% confidence interval for the mean monthly expenditure on gasoline per household, we can use the formula for the confidence interval of the mean, which is sample mean ± (critical value * (sample standard deviation / sqrt(sample size))). Since the sample size is 100, the sample mean is $128, and the sample standard deviation is $38, we need to find the critical value for a 90% confidence level. For a 90% confidence interval and a sample size of 100, which results in a degrees of freedom of 99, we could use a t-table to find the critical value; however, given the large sample size, the critical value will approximate the z-score, which is about 1.645 for a 90% confidence level.
The calculation is:
$128 + (1.645 * ($38 / sqrt(100)))
$128 + (1.645 * ($38 / 10))
$128 + (1.645 * 3.8)
$128 + 6.251
$134.251
The closest answer choice to our calculation is a. $134.66, which we can reasonably conclude to be a slight rounding difference in the critical value used.
sequences: If tn = 4n−1, find t1, t2, t3 and tn+1 . Express tn+1−tn in its simplest form.
To solve the question, we find the specific terms of the sequence using the given formula and illustrate that the difference between any two consecutive terms of this sequence is consistently 4, displaying the sequence's linear progression.
Explanation:The question requires us to first find the terms t1, t2, t3, and tn+1 of a sequence described by the formula tn = 4n - 1, and then to find the difference tn+1 - tn in its simplest form. To solve this problem, we substitute the appropriate values of n into the given formula and simplify.
t1 = 4(1) - 1 = 3t2 = 4(2) - 1 = 7t3 = 4(3) - 1 = 11To find tn+1, substitute n+1 for n: tn+1 = 4(n+1) - 1 = 4n + 3To find the difference tn+1 - tn in its simplest form, we subtract tn from tn+1:
tn+1 - tn = (4n + 3) - (4n - 1) = 4
The difference between any two consecutive terms in this sequence is 4. This reveals a consistent increment pattern in the sequence, illustrating its linear nature.
A manufacturer claims that the average ten- sile strength of thread A exceeds the average tensile strength of thread B by at least 12 kilograms. To test this claim, 50 pieces of each type of thread were tested under similar conditions. Type A thread had an aver- age tensile strength of 86.7 kilograms with a standard deviation of 6.28 kilograms, while type B thread had an average tensile strength of 77.8 kilograms with a standard deviation of 5.61 kilograms. Test the manufacturer’s claim using a 0.05 level of significance
The question asks for a hypothesis test comparing the average tensile strength of two types of threads. We formulate a null and an alternative hypothesis, calculate the test statistic, and finally determine the p-value. If the p-value is less than the significance level, we reject the null hypothesis.
Explanation:The subject of the question is about testing a manufacturer's claim about the average tensile strength of two different types of threads using statistical methods. We can test this claim using a hypothesis test for the difference between two means.
First, we set up our hypotheses. The null hypothesis (H0) could be that the average tensile strength of thread A is greater than or equal to the average tensile strength of thread B by 12kg. The alternative hypothesis (Ha) would be that the average tensile strength of thread A is not greater by at least 12kg.
The next step is to calculate the test statistic. This requires finding the difference between the two sample means, subtracting the hypothesized difference (12kg), and then dividing by the standard error of the difference. To find the standard error, we need to combine the standard deviations of the two samples.
Once we have the test statistic, we can find the p-value, which is the probability of observing a test statistic as extreme as the one calculated if the null hypothesis is true. If the p-value is less than the significance level (0.05 in this case), we reject the null hypothesis and support the manufacturer's claim.
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We perform a hypothesis test comparing the mean tensile strength of thread A and B. We formulate our null and alternative hypotheses, calculate the pooled standard deviation, standard error, and test statistic. Based on the p-value compared to the significance level, we conclude whether the data supports the manufacturer's claim.
Explanation:To test the manufacturer's claim, we will perform a hypothesis test for the difference of means, which can be conducted since there are two independent samples: thread type A and thread type B. We set up our null and alternative hypotheses as follows:
First, we calculate the pooled standard deviation and the standard error. Then, we compute the test statistic (z-score) using this formula: z = (sample difference - hypothesized difference) / standard error. The sample difference is 86.7 - 77.8 = 8.9 kg, and we are hypothesizing a difference of 12 kg.
Next, we find our p-value by looking this z-score up in a standard normal distribution table, and compare it with our significance level (0.05). If the p-value is less than 0.05, we reject the null hypothesis, thus supporting the manufacturer's claim. If the p-value is greater than 0.05, we do not reject the null hypothesis, indicating that the data does not provide strong evidence to support the manufacturer's claim that average tensile strength of thread A exceeds that of thread B by at least 12 kilograms.
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When you sample the first individuals you can find, you call it a _____; it's cheap and easy to do, but statistically not a very strong method.
A.
cluster
B.
stratified random sample
C.
convenience sample
D.
cluster sample
E.
simple random sample
Answer:
E.
simple random sample
Step-by-step explanation:
Simple random sample: Every member and set of members has an equal chance of being included in the sample. Random samples are usually fairly representative since they don't favor certain members.
A line is parameterized by x=8+6t x = 8 + 6 t and y=3+2t (a) Which of the following points are on the section of the line obtained by restricting t to nonnegative numbers (for each, enter Y if the point is on the section, and N if not)? (−4,−1): ___________ (26 , 9) :___________ (32,11): ___________ Then, give one more point that is on the section of the line obtained by this restriction:____________ (b) What are the endpoints of the line segment obtained by restricting t to − 5 ≤ t ≤ − 2 ? left endpoint: __________ right endpoint: _________
Answer:
We have the line parametrized by
[tex]x=8+6t\\y=3+2t[/tex]
Solving for t in each equation we have that
[tex]t=\frac{x-8}{6}\\t=\frac{y-3}{2}[/tex]
The point (a,b) lies in the line if when we replace a in the first equation and b in the second equation, the values of t coincide.
a)
1. (-4,-1)
[tex]t=\frac{-4-8}{6}=-2\\t=\frac{-1-3}{2}=-2[/tex]
Then, (-4,-1) lies in the line but no lies in the section of the line obtained by restricting t to nonnegative numbers.
2. (26,9)
[tex]t=\frac{26-8}{6}=3\\t=\frac{9-3}{2}=3[/tex]
Since t is positive then (26,9) lies in the line and lies in the section of the line obtained by restricting t to nonnegative numbers.
3. (32,11)
[tex]t=\frac{32-8}{6}=4\\t=\frac{11-3}{2}=4[/tex]
Since t is positive then (32,11) lies in the line and lies in the section of the line obtained by restricting t to nonnegative numbers.
4. If we take t=2 we obtain the point
[tex]x=8+6(2)=20\\y=3+2(2)=7[/tex]
(20,7) that lies in the section of the line obtained by restricting t to nonnegative numbers.
b)
When t=-5,
[tex]x=8+6(-5)=-22\\y=3+2(-5)=-7[/tex]
correspond to the point (-22,-7).
when t=-2
[tex]x=8+6(-2)=-4\\y=3+2(-2)=-1[/tex]
correspond to the point (-4,-1).
-22<-4 and -7<-1
then the left endpoint (-22,-7) and right endpoint (-4,-1)
We are 95% confident that the population proportion in 2000 that supported preferential hiring of women is about 0.0509 and the population proportion in 2010 is 0.04723. We are 95% confidence that the population proportion in 2000 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2010. We are 95% confidence that the population proportion in 2010 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2000. We are 95% confident that the population proportion in 2010 that supported preferential hiring of women is about 0.0509 and the population proportion in 2000 is 0.04723.
Answer:
What do Americans think about preferential hiring of women? Has there been a change in the past decade? In 2000(group 1) and 2010(Group 2), the General Social Survey asked participants if they would favor or oppose preferential hiring of women. In 2000, out of 849 respondents, 271 said yes. In 2010, out of 696 respondents, 225 said yes. The 95% confidence interval for p1-p2 is (-0.0509, 0.04273). What would be an appropriate conclusion? O We are 95% confident that the population proportion in 2010 that supported preferential hiring of women is about 0.0509 and the population proportion in 2000 is 0.04723. We are 95% confident that the pulation proportion in 2000 that supported preferential hiring of women is about 0.0509 and the population proportion in 2010 is 0.04723. We are 95% confidence that the population proportion in 2010 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2000. O We are 95% confidence that the population proportion in 2000 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2010.
Step-by-step explanation:
"How much do students pay, on average, for textbooks during the first semester in college? From a random sample of 400 students the mean cost was found to be $357.75 ,and the sample standard deviation was $37.89. Assuming that the population is normally distributed, find the margin of error of a 95% confidence interval for the population mean. Compute the confidence interval and describe in words what it means."
Answer: The margin of error = 3.71, confidence interval = (354.04, 361.46) and it means that mean cost is lies within the confidence interval.
Step-by-step explanation:
Since we have given that
Sample size = 400
Mean = $357.75
Standard deviation = $37.89
At 95% confidence level, z = 1.96
We first find the margin of error.
Margin of error is given by
[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{37.89}{\sqrt{400}}\\\\=3.71[/tex]
95% confidence interval would be
[tex]\bar{x}\pm \text{margin of error}\\\\=357.75\pm 3.71\\\\=(357.75-3.71,357.75+3.71)\\\\=(354.04,361.46)[/tex]
Hence, the margin of error = 3.71, confidence interval = (354.04, 361.46) and it means that mean cost is lies within the confidence interval.
Final answer:
The margin of error for a 95% confidence interval of the population mean cost of textbooks is approximately $3.71, resulting in a confidence interval of ($354.04, $361.46). This indicates a 95% level of confidence that the true average cost of textbooks lies within this range.
Explanation:
To calculate the margin of error for the 95% confidence interval for the population mean, we use the formula for margin of error (ME): ME = z * (s/√n), where z is the z-score corresponding to the confidence level, s is the sample standard deviation, and n is the sample size. Given that the student sample size is 400, the sample mean cost is $357.75, and the sample standard deviation is $37.89, we first find the z-score for a 95% confidence level, which is approximately 1.96. We then compute the margin of error.
Margin of Error calculation:
ME = 1.96 * ($37.89/√400) = $1.96 * ($37.89/20) = $1.96 * 1.8945 ≈ $3.71
Now that we have the margin of error, we can calculate the 95% confidence interval for the population mean, which is the range where we expect the true population mean to lie. The confidence interval is: ($357.75 - ME, $357.75 + ME), or ($357.75 - $3.71, $357.75 + $3.71), giving us ($354.04, $361.46). This means that we are 95% confident that the true average cost of textbooks during the first semester in college is between $354.04 and $361.46.
During 2012, global oil consumption grew by 0.7%, to reach 88 million barrels per day. Assume that it continues to increase at this rate, (a) Write the first four terms of the sequence a_n giving daily oil consumption n years after 2011; give a formula for the general term a_n. Round your answers for a_1, a_2, a_3, and a_4 to three decimal places.
a_1 =_____________.
a_2 = _____________.
a-3 = _____________.
a_4 = _____________.
a_n = _____________.
(b) In what year is consumption first expected to exceed 195 million barrels a day? Consumption exceeds 195 million barrels per day during the year
Answer:
(a)
[tex]a_{1} = 88.000 \ million\\a_{2} = 94.160 \ million\\a_{3} = 100.751 \ million\\a_{4} = 107.804 \ million\\a_{n} = 82.243*1.07^n\\[/tex]
(b) 2024
Step-by-step explanation:
Global oil consumption in 2011 is given by:
[tex]C_{2011} =\frac{88}{1.07}=82.243 \ million[/tex]
(a) Assuming a constant growth of 0.7% per year, the formula for the daily oil consumption n years after 2011 is:
[tex]a_{n} = a_{0}*1.07^n\\a_{0} = C_{2011} = 82.243\\a_{n} = 82.243*1.07^n[/tex]
The terms a_1, a_2,a_3 and a_4, corresponding to the global oil consumption in the years of 2012, 2013, 2014 and 2015, respectively, are given by:
[tex]a_{1} = 82.243*1.07^1\\a_{1} = 88.000 \ million\\a_{2} = 82.243*1.07^2\\a_{2} = 94.160 \ million\\a_{3} = 82.243*1.07^3\\a_{3} = 100.751 \ million\\a_{4} = 82.243*1.07^4\\a_{4} = 107.804 \ million\\[/tex]
(b) To find the in year in which consumption reaches 195 million barrels a day, apply logarithmic properties:
[tex]a_{n} = 82.243*1.07^n\\ln(a_{n}) = ln(82.243)+ n*ln(1.07)\\\\n=\frac{ln(195)-ln(82.243)}{ln(1.07)} \\n= 12.760[/tex]
Consumption will reach 195 million barrels, 12.7 years after 2011, round it to the next whole year to find when consumption exceeds 195 million:
[tex]Y = 2011+13 = 2024[/tex]