Answer:
21/64
Step-by-step explanation:
First, we need to note that the function f(x) = x² is increasing on (0, +∞), and it is decreasing on (-∞,0)
The first interval generated by the partition is [-1, -1/2], since f is decreasing for negative values, we have that f takes its minimum values at the right extreme of the interval, hence -1/2.
The second interval is [-1/2, 1/2]. Here f takes its minimum value at 0, because f(0) = 0, and f is positive otherwise.
Since f is increasing for positive values of x, then, on the remaining 2 intervals, f takes its minimum value at their respective left extremes, in other words, 1/2 and 3/4 respectively.
We obtain the lower Riemman sum by multiplying this values evaluated in f by the lenght of their respective intervals and summing the results, thus
LP(f) = f(-1/2) * ((-1/2) - (-1)) + f(0) * (1/2 - (-1/2)) + f(1/2)* (3/4 - 1/2) + f(3/4) * (1- 3/4)
= 1/4 * 1/2 + 0 * 1 + 1/4 * 1/4 + 9/16 * 1/4 = 1/8 + 0 + 1/16 + 9/64 = 21/64
As a result, the lower Riemann sum on the partition P is 21/64
Final answer:
To compute the lower Riemann sum for the given function f(x) = x² over the interval x ∈ [-1,1] with respect to the partition P = [-1, -1/2, 1/2, 3/4 ,1], we need to find the height of each subinterval and multiply it by the width of the subinterval. The lower Riemann sum is 5/8.
Explanation:
To compute the lower Riemann sum for the given function f(x) = x² over the interval x ∈ [-1,1] with respect to the partition P = [-1, -1/2, 1/2, 3/4 ,1], we need to find the height of each subinterval and multiply it by the width of the subinterval.
First, let's calculate the width of each subinterval:
width of subinterval 1: (-1) - (-1) = 0
width of subinterval 2: (-1/2) - (-1) = 1/2
width of subinterval 3: (1/2) - (-1/2) = 1
width of subinterval 4: (3/4) - (1/2) = 1/4
width of subinterval 5: (1) - (3/4) = 1/4
Next, let's calculate the height of each subinterval by substituting the left endpoint of each subinterval into the function:f((-1/2))^2 = 1/4, f(1/2)² = 1/4, f(3/4)² = 9/16, f(1)² = 1
Finally, we compute the lower Riemann sum by multiplying the height by the width for each subinterval and summing them up:
(0 * 0) + (1/2 * 1/4) + (1 * 9/16) + (1/4 * 1) = 5/8.
A researcher would like to estimate p, the proportion of U.S. adults who support recognizing civil unions between gay or lesbian couples. If the researcher would like to be 95% sure that the obtained sample proportion would be within 1.5% of p (the proportion in the entire population of U.S. adults), what sample size should be used?
(a) 17,778(b) 4,445(c) 1,112(d) 67(e) 45
Answer:
(b) 4,445
Step-by-step explanation:
If the researcher would like to be 95% sure that the obtained sample proportion would be within 1.5% of p (the proportion in the entire population of U.S. adults), what sample size should be used?
Given a=0.05, |Z(0.025)|=1.96 (check standard normal table)
So n=(Z/E)^2*p*(1-p)
=(1.96/0.015)^2*0.5*0.5
=4268.444
Take n=4269
Answer:(b) 4,445
Evaluate each of the following line integrals.
(a)
integral.gif
C
x dy − y dx, c(t) = (cos(t), sin(t)), 0 ≤ t ≤ 2π
(b)
integral.gif
C
x dy + y dx, c(t) = (2 cos(πt), 2 sin(πt)), 2 ≤ t ≤ 4
Answer:
a) 2π
b) 0
Step-by-step explanation:
Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]
[tex]\large \int_{C}[P(x,y)dx+Q(x,y)dy]=\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt[/tex]
Where P, Q are scalar functions
a)
C(t) = (x(t), y(t)) = (cos(t), sin(t)), 0 ≤ t ≤ 2π
P(x,y) = -y ==> P(x(t),y(t)) = -y(t) = -sin(t)
Q(x,y) = x ==> Q(x(t),y(t)) = x(t) = cos(t)
x'(t) = -sin(t)
y'(t) = cos(t)
[tex]\large \int_{C}[-ydx+xdy]=\int_{0}^{2\pi}[(-sin(t))(-sin(t))+cos(t)cos(t)]dt=\\\\\int_{0}^{2\pi}[sin^2(t)+cos^2(t)]dt=\int_{0}^{2\pi}dt=2\pi[/tex]
b)
C(t) = (x(t), y(t)) = (2cos(πt), 2sin(πt)), 2 ≤ t ≤ 4
P(x,y) = y ==> P(x(t),y(t)) = y(t) = 2sin(πt)
Q(x,y) = x ==> Q(x(t),y(t)) = x(t) = 2cos(πt)
x'(t) = -2πsin(πt)
y'(t) = 2πcos(πt)
[tex]\large \int_{C}[ydx+xdy]=\int_{2}^{4}[(2sin(\pi t))(-2\pi sin(\pi t))+2cos(\pi t)2\pi cos(\pi t)]dt=\\\\4\int_{2}^{4}[cos^2(\pi t)-sin^2(\pi t)]dt=4\int_{2}^{4}cos(2\pi t)dt=\\\\4\left[\frac{sin(2\pi t)}{2\pi}\right]_2^4=\frac{2}{\pi}(sin(8 \pi)-sin(4\pi))=0[/tex]
According to a report from a business intelligence company, smartphone owners are using an average of 20 apps per month Assume that number of apps used per month by smartphone owners is normally distributed and that the standard deviation is 4. Complete parts (a) through (d) below. a. If you select a random sample of 36 smartphone owners, what is the probability that the sample mean is between 19.5 and 20.5? (Round to three decimal places as needed.)
Answer: 0.547
Step-by-step explanation:
As per given , we have
Population mean = [tex]\mu=20[/tex]
Population standard deviation= [tex]\sigma=4[/tex]
Sample size : n= 36
We assume that number of apps used per month by smartphone owners is normally distributed.
Let [tex]\overline{x}[/tex] be the sample mean.
Formula : [tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
The probability that the sample mean is between 19.5 and 20.5 :-
[tex]P(19.5<\overline{x}<20.5)\\\\=P(\dfrac{19.5-20}{\dfrac{4}{\sqrt{36}}}<\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{20.5-20}{\dfrac{4}{\sqrt{36}}})\\\\=P(-0.75<z<0.75)\\\\=P(z<0.75)-P(z<-0.75)\ \ [\because P(z_1<z<z_2)=P(z<z_2)-P(z<z_1)]\\\\=P(z<0.75)-(1-P(z<0.75))\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=2P(z<0.75)-1=2(0.7734)-1=0.5468\approx0.547[/tex]
[using standard normal distribution table for z]
Hence, the required probability = 0.547
To find the probability that the sample mean is between 19.5 and 20.5 for a random sample of 36 smartphone owners, calculate the z-scores for both values and find the area under the standard normal curve between those z-scores.
Explanation:To find the probability that the sample mean is between 19.5 and 20.5 for a random sample of 36 smartphone owners, we need to calculate the z-scores for both values and find the area under the standard normal curve between those z-scores.
The formula for calculating the z-score is: z = (x - μ) / (σ / √n), where x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.
Using the given information, the z-scores for 19.5 and 20.5 are: z1 = (19.5 - 20) / (4 / √36) = -0.75 and z2 = (20.5 - 20) / (4 / √36) = 0.75.
Now, we can use a standard normal table or a calculator to find the area under the curve between -0.75 and 0.75. The probability is approximately 0.467.
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Dan was trying to get faster at his multiplication tables. This table shows the time it took him to complete each of four tests containing 50 simple multiplication problems.
Which comparison of the time for each test is correct?
Test 1 < Test 4
Test 2 > Test 4
Test 1 > Test 3
Test 3 < Test 2
Answer:Test 1 < Test 4
Step-by-step explanation: Test 1 was easier so it took less time to complete the questions. As the tests continue they get harder meaning they will take more time to answer.
Answer:
the answer is a
Step-by-step explanation:
i just took the test
Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. Construct a 95% confidence interval for the population proportion who claim they always buckle up. What is the error bound?
Answer:
The 95% confidence interval would be given (0.761;0.839). The error bound is [tex]Me=\pm 0.0392[/tex]
Step-by-step explanation:
1) Data given and notation
n=400 represent the random sample taken
X=320 represent the people drivers claimed they always buckle up
[tex]\hat p=\frac{320}{400}=0.8[/tex] estimated proportion of people drivers claimed they always buckle up
[tex]\alpha=0.05[/tex] represent the significance level (no given, but is assumed)
Confidence =95% or 0.95
p= population proportion of people drivers claimed they always buckle up
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".
2) Calculating the interval for the proportion
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.8 - 1.96 \sqrt{\frac{0.8(1-0.8)}{400}}=0.761[/tex]
[tex]0.8 + 1.96 \sqrt{\frac{0.8(1-0.8)}{400}}=0.839[/tex]
And the 95% confidence interval would be given (0.761;0.839). The error bound is [tex]Me=\pm 0.0392[/tex]
We are confident that about 76.1% to 83.9% of people drivers that they always buckle up at 95% of confidence
A group of 24 people have found 7.2 kg of gold. Assuming the gold is divided evenly, how much gold will each one get in grams? Please someone help me out with this thank you
Answer:
300 g
Step-by-step explanation:
There are 1000 grams in one kilogram. Dividing 7200 grams into 24 equal parts makes each part ...
(7200 g)/(24 persons) = 300 g/person
Each person get 300 grams.
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minutes. We want to see if we can find significant evidence to prove that the mean waiting time of all customers is significantly more than 3 minutes. The test statistic is 2. What is the p-value? Round your answer to three decimal places.
Answer: The critical value is: insufficient.
Step-by-step explanation: Refer to Exhibit 9-2. The p-value is:
a. .0500
b. .0228
c. .0456
d. .0250
In order to test the hypotheses H0: μ ≤ 100 and Ha: μ > 100 at an α level of significance, the null hypothesis will be rejected if the test statistic z is:
The practice of concluding "do not reject H0" is preferred over "accept H0" when we:
a. have an insufficient sample size.
b. have not controlled for the Type II error.
c. are testing the validity of a claim.
d. are conducting a one-tailed test.
In hypothesis testing, the critical value is: insufficient
A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on three numbers pays 11 to 1 (that is, if you bet $1 and one of the three numbers you bet comes up, you get back your $1 plus another $11). How much do you expect to win with a $1 bet on three numbers?
Answer:
You are expected to lose $0.05 (or win -$0.05)
Step-by-step explanation:
Since the roulette wheel has the numbers 1 through 36, 0, and 00, there are 38 possible outcomes.
In this bet, you are allowed to pick 3 out of the 38 numbers. Thus, your chances of winning (P(W)) and losing (P(L)) are:
[tex]P(W)=\frac{3}{38}\\P(L) = 1 - P(W)\\P(L) = \frac{35}{38}\\[/tex]
The expected value of the bet is given by the sum of the product of each outcome pay by its probability. Winning the bet means winning $11 while losing the bet means losing $1. The expected value is:
[tex]EV = (11*\frac{3}{38}) -(1*\frac{35}{38})\\EV = -\$0.0525[/tex]
Therefore, with a $1 bet, you are expected to lose roughly $0.05
The time in seconds that it takes for a sled to slide down a hillside inclined at an angle θ is given by the formula below, where d is the length of the slope in feet. Find the time it takes to slide down a 2000 ft slope inclined at 30°. (Round your answer to one decimal place.) t = d 16 sin θ
Answer: It takes 15.8 seconds to slide down a 2000 ft slope.
Step-by-step explanation:
Since we have given that
[tex]t=\sqrt{\dfrac{d}{16\sin\theta}}[/tex]
where, t is the time taken,
d is the length to slide down a slope
sin θ is the angle at which it inclined.
So, we have d = 2000 ft
θ = 30°
So, the time taken is given by
[tex]t=\sqrt{\dfrac{2000}{16\sin 30^\circ}}\\\\t=\sqrt{\dfrac{2000}{8}}\\\\t=\sqrt{250}\\\\t=15.81\ seconds[/tex]
Hence, it takes 15.8 seconds to slide down a 2000 ft slope.
The time it takes to slide down a 2000 ft slope inclined at 30° is 250 seconds, calculated using the equation t = d / (16 sin θ).
Explanation:We're given the equation for calculating time of a sled sliding down a slope as t = d / (16 sin θ). Here, d is the length of the slope, and θ is the incline of the hill slope.
To calculate the time it takes to slide down a 2000 ft slope inclined at 30°, we substitute the respective values into the formula: t = 2000 / (16 sin 30). The sin value for 30° is 0.5, so the formula simplifies to t = 2000 / (16 * 0.5), which yields t = 2000 / 8. By performing that division, we see that t = 250 seconds. Therefore, it takes 250 seconds to slide down a 2000 ft slope inclined at 30°.
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I need help with this challenge question pls someone help asap
Answer:
Step-by-step explanation:
You were given the following joint probability function for
Y1 = { 0, if the child survived, 1, if not,
and
Y2 = { 0 if no belt used, 1 if adult belt used, and 2 if car seat belt used
Notice that Y1 is the number of fatalities per child and, since children's car seats usually utilize two belts, Y2 is the number of seat belts in use at the time of accident.
Given:
Y1
y2 0 1 total
0 0.38 0.17 0.55
1 0.14 0.02 0.16
2 0.24 0.05 0.29
Total 0.76 0.24 1
Are Y1 and Y2 independent? Why or why not?
Answer: No, Y1 and Y2 are not independent
Step-by-step explanation:
Because they don't satisfy this condition:
FsubscriptY1Y2(Y1,Y2) = FsubscriptY1(y1) × FsubscriptY2(y2)
... for all given values of Y1 and Y2
This is the condition for independence.
How do we know that Y1 and Y2 don't satisfy this condition?
We use the information in the Joint Probability Distribution Table.
Let's see if the condition stands when Y1 is zero and Y2 is zero
FsubscriptY1Y2(0,0) = 0.38
FsubscriptY1(0) × FsubscriptY2(0) = 0.76×0.55 = 0.418
We can see that 0.38 is not equal to 0.418
Doing the test for any other combination of Y1 and Y2 values will give unequal figures as well.
Evaluate ModifyingBelow Integral from nothing to nothing With Upper C xy dx plus (x plus y )dy along the curve y equals 2 x squared from (1 comma 2 )to (2 comma 8 ).
[tex]y=2x^2\implies\mathrm dy=4x\,\mathrm dx[/tex]
Then in the integral we have
[tex]\displaystyle\int_Cxy\,\mathrm dx+(x+y)\,\mathrm dy=\int_1^2(2x^3+4x(x+2x^2))\,\mathrm dx=\int_1^210x^3+4x^2\,\mathrm dx=\boxed{\frac{281}6}[/tex]
A survey of 1010 college seniors working towards an undergraduate degree was conducted. Each student was asked, "Are you planning or not planning to pursue a graduate degree?" Of the 1010 surveyed, 658 stated that they were planning to pursue a graduate degree. Construct and interpret a 98% confidence interval for the proportion of college seniors who are planning to pursue a graduate degree. Round to the nearest thousandth.
Answer:
The 98% confidence interval would be given (0.616;0.686).
We are confident at 98% that the true proportion of people that they were planning to pursue a graduate degree is between (0.616;0.686).
Step-by-step explanation:
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".
Description in words of the parameter p
[tex]p[/tex] represent the real population proportion of people that they were planning to pursue a graduate degree
[tex]\hat p[/tex] represent the estimated proportion of people that they were planning to pursue a graduate degree
n=1010 is the sample size required
[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Numerical estimate for p
In order to estimate a proportion we use this formula:
[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.
[tex]\hat p=\frac{658}{1010}=0.651[/tex] represent the estimated proportion of people that they were planning to pursue a graduate degree
Confidence interval
The confidence interval for a proportion is given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 98% confidence interval the value of [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2=0.01[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.33[/tex]
And replacing into the confidence interval formula we got:
[tex]0.651 - 2.33 \sqrt{\frac{0.651(1-0.651)}{1010}}=0.616[/tex]
[tex]0.651 + 2.33 \sqrt{\frac{0.651(1-0.651)}{1010}}=0.686[/tex]
And the 98% confidence interval would be given (0.616;0.686).
We are confident at 98% that the true proportion of people that they were planning to pursue a graduate degree is between (0.616;0.686).
What is the value of X?
Answer:
x = 55°
Step-by-step explanation:
The sum of the angles in a triangle is 180°, so you have ...
x + 50° +75° = 180°
x = 55° . . . . . . . . . . . . subtract 125°
Pedro thinks that he has a special relationship with the number 3. In particular, Pedro thinks that he would roll a 3 with a fair 6-sided die more often than you'd expect by chance alone. Suppose p is the true proportion of the time Pedro will roll a 3.
(a) State the null and alternative hypotheses for testing Pedro's claim. (Type the symbol "p" for the population proportion, whichever symbols you need of "<", ">", "=", "not =" and express any values as a fraction e.g. p = 1/3)
H0 = _______
Ha = _______
(b) Now suppose Pedro makes n = 30 rolls, and a 3 comes up 6 times out of the 30 rolls. Determine the P-value of the test:
P-value =________
Answer:
p value = 0.3122
Step-by-step explanation:
Given that Pedro thinks that he has a special relationship with the number 3. In particular,
Normally for a die to show 3, probability p = [tex]\frac{1}{6} =0.1667[/tex]
Or proportion p = 0.1667
Pedro claims that this probability is more than 0.1667
[tex]H_0 = P =0.1667______H_a = P>0.1667_____[/tex]
where P is the sample proportion.
b) n=30 and [tex]P = 6/30 =0.20[/tex]
Mean difference = [tex]0.2-0.1667=0.0333[/tex]
Std error for proportion = [tex]\sqrt{\frac{p(1-p)}{n} } \\=0.0681[/tex]
Test statistic Z = p difference/std error = 0.4894
p value = 0.3122
p >0.05
So not significant difference between the two proportions.
The null hypothesis is that Pedro's claim is not true, while the alternative hypothesis is that Pedro's claim is true. The P-value is approximately 0.082.
Explanation:
(a) The null hypothesis states that Pedro's claim is not true, so the null hypothesis is H0: p = 1/6. The alternative hypothesis states that Pedro's claim is true and he rolls a 3 more often than expected, so the alternative hypothesis is Ha: p > 1/6.
(b) To determine the P-value, we can use the binomial distribution. The probability of rolling a 3 with a fair 6-sided die is 1/6. Using the binomial distribution, we can calculate the probability of rolling a 3 six times out of 30 rolls. This gives a P-value of approximately 0.082, which is the probability of observing a result as extreme as the one obtained or more extreme, assuming the null hypothesis is true.
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Two teams play a series of games, the first team to win 4 games is the winner overall. Suppose that one of the teams is stronger and has probability 0.6 to win each game, independent of any other games. What is the probability that the stronger team wins the series in exactly i games. Do it for i = 4,5,6,7. Compare the probability that the stronger team wins with the probability that it would win a 2 out of 3 series.
Answer:
Step-by-step explanation:
Figuring it out rn
You are designing a 1000 cm^3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore beA = 8r^2 + 2pi rhWhat is the ratio now of h to r for the most economical can?
Answer:
h /r = 2.55
Step-by-step explanation:
Area of a can:
Total area of the can = area of (top + bottom) + lateral area
lateral area 2πrh without waste
area of base (considering that you use 2r square) is 4r²
area of bottom ( for same reason ) 4r²
Then Total area = 8r² + 2πrh
Now can volume is 1000 = πr²h h = 1000/πr²
And A(r) = 8r² + 2πr(1000)/πr²
A(r) = 8r² + 2000/r
Taking derivatives both sides
A´(r) = 16 r - 2000/r²
If A´(r) = 0 16 r - 2000/r² = 0
(16r³ - 2000)/ r² = 0 16r³ - 2000 = 0
r³ = 125
r = 5 cm and h = ( 1000)/ πr² h = 1000/ 3.14* 25
h = 12,74 cm
ratio h /r = 12.74/5 h /r = 2.55
To find the ratio of h to r for the most economical can, we need to minimize the amount of aluminum used. The ratio is 0.
Explanation:To find the ratio of h to r for the most economical can, we need to minimize the amount of aluminum used. The total amount of aluminum used is given by the equation A = 8r^2 + 2πrh. To minimize A, we can take the derivative of A with respect to h and set it equal to 0. Solving this equation will give us the value of h in terms of r, and we can then find the ratio of h to r.
Taking the derivative of A with respect to h, we get dA/dh = 2πr. Setting this equal to 0 and solving for h, we find that h = 0. To find the ratio of h to r, we divide both sides of the equation by r, giving us h/r = 0.
Therefore, the ratio of h to r for the most economical can is 0.
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The probability that an individual without a college education earns more than $100,000 is 0.4, whereas the probability that a person with a B.S. or higher degree earns more than $100,000 is 0.6. The probability that a person chosen at random has a B.S. degree is 0.5. What is the probability that a person has at least a B.S. degree if it is known that he or she earns more than $100,000?
Answer:
0.6
Step-by-step explanation:
Given:
P( Person chosen at random has a B.S. degree), P(C) = 0.5
P( Person chosen at random does not have a B.S. degree), P(C') = 1 - 0.5 = 0.5
P(Student earns more than $100,000) = P(E)
P(Student earns more than $100,000, without going college) = P(E | C') = 0.4
P(Student earns more than $100,000, with college degree) = P(E | C) = 0.6
Now,
P(at least a B.S. degree | earns more than $100,000), P(C | E)
using Baye's theorem
we have
P(C | E) = [tex]\frac{P(C)\timesP(E | C)}{P(C)\timesP(E | C)+P(C')\timesP(E | C')}[/tex]
or
P(C | E) = [tex]\frac{0.5\times0.6}{0.5\times0.6+0.5\times0.4}[/tex]
or
P(C | E) = [tex]\frac{0.3}{0.5}[/tex]
or
P(C | E) = 0.6
To find the probability that a person has at least a B.S. degree if it is known that he or she earns more than $100,000, we can use Bayes' Theorem and the given probabilities.
Explanation:To find the probability that a person has at least a B.S. degree if it is known that he or she earns more than $100,000, we can use Bayes' Theorem. Let A be the event that the person has at least a B.S. degree, and let B be the event that the person earns more than $100,000. The probability of A given B can be calculated as:
P(A|B) = (P(B|A) * P(A)) / P(B)
Given that P(B|A) = 0.6, P(A) = 0.5, and P(B) = 0.4, we can substitute the values into the formula:
P(A|B) = (0.6 * 0.5) / 0.4 = 0.75
Therefore, the probability that a person has at least a B.S. degree if it is known that he or she earns more than $100,000 is 0.75.
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Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
a. The line segment x=4 y=0 z=[0;4]
Find a scalar potential function f for F, such that
F=∇f.
The work done by F over the line segment is?
b. The helix r(t)= (4cost)i +(4sint)j +(2t/pi)k t=[0;2pi]
Find df/dt for F?
The Work done by F over the Helix?
c. The x axis from (4,0,0) to (0,0,0) followed by the line z=x, y=0 from (0,0,0) to (4,0,4)
What is the integral to comput the work done by F along the x-axis from followed by the line z=x?
What is the work done by F over the 2 curves
[tex]\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k[/tex]
We want to find [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. This means
[tex]\dfrac{\partial f}{\partial x}=x^2+y[/tex]
[tex]\dfrac{\partial f}{\partial y}=y^2+x[/tex]
[tex]\dfrac{\partial f}{\partial z}=ze^z[/tex]
Integrating both sides of the latter equation with respect to [tex]z[/tex] tells us
[tex]f(x,y,z)=e^z(z-1)+g(x,y)[/tex]
and differentiating with respect to [tex]x[/tex] gives
[tex]x^2+y=\dfrac{\partial g}{\partial x}[/tex]
Integrating both sides with respect to [tex]x[/tex] gives
[tex]g(x,y)=\dfrac{x^3}3+xy+h(y)[/tex]
Then
[tex]f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)[/tex]
and differentiating both sides with respect to [tex]y[/tex] gives
[tex]y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C[/tex]
So the scalar potential function is
[tex]\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}[/tex]
By the fundamental theorem of calculus, the work done by [tex]\vec F[/tex] along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it [tex]L[/tex]) in part (a) is
[tex]\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}[/tex]
and [tex]\vec F[/tex] does the same amount of work over both of the other paths.
In part (b), I don't know what is meant by "df/dt for F"...
In part (c), you're asked to find the work over the 2 parts (call them [tex]L_1[/tex] and [tex]L_2[/tex]) of the given path. Using the fundamental theorem makes this trivial:
[tex]\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3[/tex]
[tex]\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4[/tex]
The article "Plugged In, but Tuned Out" (USA Today, January 20, 2010) summarized data from two surveys of randomly selected kids ages 8 to 18. One survey was conducted in 1999 and the other was conducted in 2009. Data on the number of hours per day spent using electronic media, consistent with summary quantities given in the article, are below.time1999<-c(4, 5, 7, 7, 5, 7, 5, 6, 5, 6, 7, 8, 5, 6, 6)time2009<-c(5, 9, 5, 8, 7, 6, 7, 9, 7, 9, 6, 9, 10, 9, 8)Find the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999. Show all steps of the confidence interval. You may use the formula or R for calculations.
Answer:
(-3.0486, -0.2848)
Step-by-step explanation:
Let the number of hours per day spent using electronic media from 1999 be the first population and the number of hours per day spent using electronic media from 1999 the second population.
We have small sample sizes [tex]n_{1} = 15[/tex] and
[tex]n_{2} = 15[/tex].
[tex]\bar{x}_{1} = 5.9333[/tex] and [tex]\bar{x}_{2} = 7.6[/tex]; [tex]s_{1} = 1.0998[/tex] and [tex]s_{2} = 1.5946[/tex].
The pooled estimate is given by
[tex]s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(15-1)(1.0998)^{2}+(15-1)(1.5946)^{2}}{15+15-2} = 1.8762[/tex]
The 99% confidence interval for the true mean difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is given by
[tex](\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{15}+\frac{1}{15}}[/tex], i.e.,
[tex](5.9333-7.6)\pm t_{0.005}1.3697\sqrt{\frac{1}{15}+\frac{1}{15}}[/tex]
where [tex]t_{0.005}[/tex] is the 0.5th quantile of the t distribution with (15+15-2) = 28 degrees of freedom. So
[tex]-1.6667\pm(-2.7633)(1.3697)(0.3651)[/tex], i.e.,
(-3.0486, -0.2848)
######
Using R
time1999 <- c(4, 5, 7, 7, 5, 7, 5, 6, 5, 6, 7, 8, 5, 6, 6)
time2009 <- c(5, 9, 5, 8, 7, 6, 7, 9, 7, 9, 6, 9, 10, 9, 8)
n1 <- length(time1999)
n2 <- length(time2009)
(mean(time1999)-mean(time2009))+qt(0.005, df = 28)*sqrt(((n1-1)*var(time1999)+(n2-1)*var(time2009))/(n1+n2-2))*sqrt(1/n1+1/n2)
(mean(time1999)-mean(time2009))-qt(0.005, df = 28)*sqrt(((n1-1)*var(time1999)+(n2-1)*var(time2009))/(n1+n2-2))*sqrt(1/n1+1/n2)
To find the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999, we can use the formula: CI = (x1 - x2) ± Z * sqrt((s1^2/n1) + (s2^2/n2)). Using R, the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is (0.4733, 2.1933) hours.
Explanation:To find the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999, we can use the formula:
CI = (x1 - x2) ± Z * sqrt((s1^2/n1) + (s2^2/n2))
where x1 and x2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and Z is the critical value for a 99% confidence level.
Using R, we can calculate the confidence interval as follows:
time1999 <- c(4, 5, 7, 7, 5, 7, 5, 6, 5, 6, 7, 8, 5, 6, 6)The 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is (0.4733, 2.1933) hours. This means we are 99% confident that the true difference between the means falls within this interval.
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The government of Preon (a small island nation) was voted in at the last election with 68% of the votes. That was 2 years ago, and ever since then the government has assumed that their approval rating has been the same. Some recent events have affected public opinion and the government suspects that their approval rating might have changed. They decide to run a hypothesis test for the proportion of people who would still vote for them.The null and alternative hypotheses are:H0: Pi symbol = 0.68HA: Pi symbol ≠ 0.68The level of significance used in the test is α = 0.1. A random sample of 102 people are asked whether or not they would still vote for the government. The proportion of people that would is equal to 0.745. You may find this standard normal table useful throughout this question.Calculate the test statistic (z) for this hypothesis test.
Answer: 1.41
Step-by-step explanation:
Test statistic(z) for proportion is given by :-
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
, where p=population proportion.
[tex]\hat{p}[/tex]= sample proportion
n= sample size.
As per given , we have
[tex]H_0:\mu=0.68\\\\ H_a: \mu\neq0.68[/tex]
n= 102
[tex]\hat{p}=0.745[/tex]
Then, the test statistic (z) for this hypothesis test will be :-
[tex]z=\dfrac{0.745-0.68}{\sqrt{\dfrac{0.68(1-0.68)}{102}}}\\\\=\dfrac{0.065}{\sqrt{\dfrac{0.2176}{102}}}\\\\=\dfrac{0.065}{\sqrt{0.0021333}}\\\\=\dfrac{0.065}{0.04618802}=1.40729132792\approx1.41[/tex]
[Rounded to the two decimal places]
Hence, the test statistic (z) for this hypothesis test = 1.41
The populations, P, of six towns with time t in years are given by
1) P=2400(0.8)^t
2) P=900(0.77)^t
3)P=2100(0.98)^t
4)P=600(1.18)^t
5)P=1100(1.08)^t
6)P=1700(1.191)^t
Answer the following questions regarding the populations of the six towns above. Whenever you need to enter several towns in one answer, enter your answer as a comma separated list of numbers. For example if town 1, town 2, town 3, and town 4, are all growing you could enter 1, 2, 3, 4 ; or 2, 4, 1, 3 ; or any other order of these four numerals separated by commas.
(a) Which of the towns are growing?
(b) Which of the towns are shrinking?
(c) Which town is growing the fastest?
What is the annual percentage growth RATE of that town? %
(d) Which town is shrinking the fastest?
What is the annual percentage decay RATE of that town? %
(e) Which town has the largest initial population?
(f) Which town has the smallest initial population?
Answer:
Since, in the population function,
[tex]P = ab^t[/tex]
a = initial population,
b = population change factor,
If 0 < b < 1, then population will shrink,
While, if b > 1, then the population will grow,
(a) Since, 1.18, 1.08 and 1.191 is greater than 1,
Thus, town 4), 5) and 6) are growing.
(b) Since, 0.8, 0.77 and 0.98 are less than 1,
Thus, town 1), 2) and 3) are shrinking.
(c) An exponential growth function with highest change factor grows fastest.
∵ 1.191 > 1.18 > 1.08
⇒ town 6) is growing fastest.
(d) An exponential decay function with lowest change factor shrinks fastest,
∵ 0.77 < 0.8 < 0.98 < 1.08 < 1.18 < 1.191,
⇒ Town 2) shrinks fastest.
(e) Since,
2400 > 2100 > 1700 > 1100 > 900 > 600
⇒ town 1) has the largest initial population.
(f) Similarly,
Town 4) has the smallest initial population.
Answer:
a.
i, ii, and iv
Step-by-step explanation:
Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?
Answer:
92
Step-by-step explanation:
She is taking four tests. She wants to average a 95 on those tests. Therefore, she must score 95*4=380 total. She scored 97 and 91, or 188 total. Therefore, she must score 380-188 = 192 total on her next two tests. Her maximum score possible on the fourth test is 100. Therefore, the lowest score possible for the third test would be 192-100=92
The lowest possible score Isabella could have made on the third test
is 92 marks.
What is a numerical expression?A numerical expression is a mathematical statement written in the form of numbers and unknown variables. We can form numerical expressions from statements.
Given, Isabella must take four 100-point tests in her math class and her goal is to achieve an average grade of 95 on the tests.
So, The total marks she needs to score in four games is (4×95).
= 380.
Now, She scores 97 and 91 in her first and second tests.
So, She needs to score
380 - (97 + 91)
= 380 - 188.
= 192 in her two tests.
Now, The highest marks she can score in the fourth test is 100.
Therefore, The lowest possible marks she can score is,
= 192 - 100.
= 92.
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Let D be the region bounded below by the plane zequals=0, above by the sphere x squared plus y squared plus z squared equals 900x2+y2+z2=900, and on the sides by the cylinder x squared plus y squared equals 100x2+y2=100. Set up the triple integrals in cylindrical coordinates that give the volume of D using the following orders of integration.
a. dz dr dthetaθ
b. dr dz dthetaθ
c. dthetaθ dz dr
Answer:
The cylinder is given straight away by x^2+y^2=r^2=16\implies r=4. To get the cylinder, we complete one revolution, so that 0\le\theta\le2\pi. The upper limit in z is a spherical cap determined by
x^2+y^2+z^2=144\iff z^2=144-r^2\implies z=\sqrt{144-r^2}
So the volume is given by
\displaystyle\iiint_{\mathcal D}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}\int_{z=0}^{z=\sqrt{144-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
and has a value of \dfrac{128(27-16\sqrt2)\pi}3 (not that we care)
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Step-by-step explanation:
A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 418 gram setting. It is believed that the machine is underfilling the bags. A 9 bag sample had a mean of 411 grams with a standard deviation of 20 . A level of significance of 0.025 will be used. Assume the population distribution is approximately normal. Is there sufficient evidence to support the claim that the bags are underfilled?
Answer: There is sufficient evidence to support the claim that the bags are under-filled.
Step-by-step explanation:
Since we have given that
[tex]H_0:\mu=418\\\\H_a:\mu<418[/tex]
Sample mean = 411
Standard deviation = 20
n = 9
So, the test statistic value is given by
[tex]z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\\z=\dfrac{411-418}{\dfrac{20}{\sqrt{9}}}\\\\\\z=\dfrac{-7}{\dfrac{20}{3}}\\\\\\z=-1.05[/tex]
At 0.025 level of significance,
critical value z = -2.306
since -2.306<-1.05
so, we will reject the null hypothesis.
Yes, there is sufficient evidence to support the claim that the bags are underfilled.
A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. Assuming a population standard deviation of six hours, what is the required sample size if the error should be less than a half hour with a 95% level of confidence?
Answer:
Sample size should be atleast 55320
Step-by-step explanation:
Given that a group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week.
population standard deviation [tex]\sigma = 6 hrs[/tex]
Std error = [tex]\frac{6}{\sqrt{n} }[/tex]
Margin of error for 95%
= [tex]1.96*\frac{6}{\sqrt{n} } <0.5\\\sqrt{n} >235.2\\n \geq 55319.04[/tex]
Since sample size is number of items it cannot be indecimal.
So we can round off to the next high integer on a safer side.
The required sample size for a 95% level of confidence, with a population standard deviation of six hours and error less than a half hour, is approximately 554 students.
Explanation:
The question is about finding the required sample size for a statistics project. This project involves figuring out the average amount of time students spend studying per week with a given population standard deviation, desired error, and level of confidence.
To solve this, we use the formula for sample size: n = [(Z_α/2*σ)/E]^2. Here, Z_α/2 is the Z-score corresponding to the desired level of confidence, σ is the population standard deviation, and E is the desired error.
For a 95% level of confidence, the Z-score is about 1.96 (you can find this value in Z-tables). Given σ = 6 (the population standard deviation stated in the question) and E = 0.5 (the desired error less than a half hour), we substitute these into the formula to get:
n = [(1.96*6)/0.5]^2
So, n ≈ 553.29. Because we can't have a fraction of a student, we should round this number up to the nearest whole number.
Therefore, "the required sample size is 554 students".
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Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z ≥ 0, and f(x, y, z) = 0 otherwise. (a) Find the value of the constant C. (b) Find P(X ≤ 1.375 , Y ≤ 1.5). (Round answer to five decimal places). (c) Find P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1). (Round answer to six decimal places).
a.
[tex]f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}[/tex]
is a proper joint density function if, over its support, [tex]f[/tex] is non-negative and the integral of [tex]f[/tex] is 1. The first condition is easily met as long as [tex]C\ge0[/tex]. To meet the second condition, we require
[tex]\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}[/tex]
b. Find the marginal joint density of [tex]X[/tex] and [tex]Y[/tex] by integrating the joint density with respect to [tex]z[/tex]:
[tex]f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz[/tex]
[tex]\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}[/tex]
Then
[tex]\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy[/tex]
[tex]\approx\boxed{0.12886}[/tex]
c. This probability can be found by simply integrating the joint density:
[tex]\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]
[tex]\approx\boxed{0.012262}[/tex]
(a) We determined the constant C by integrating the joint density function over the entire space, finding = 1/10.
(b) We calculated P(X≤1.375,Y≤1.5) by integrating the joint density function over the specified region, resulting in approximately 0.1286.
(c) For P(X≤1.375,Y≤1.5,Z≤1), we utilized the results from part (b) and integrated over, yielding approximately 0.1163.
The problem involves determining the constant C and calculating certain probabilities for the given joint density function f(x, y, z).
Below is the step-by-step solution:
Part (a): Finding the Constant C
First, we need to find the value of C such that the total probability is 1.
This means we need to evaluate the integral of the joint density function over the entire space.
We solve the integral:[tex]\int \int \int C e^{-(0.5x + 0.2y + 0.1z)} \, dx \, dy \, dz = 1[/tex], where x, y, z ≥ 0
Breaking it down, we integrate one variable at a time:[tex]\int_{0}^{\infty} Ce^{-0.5x} \, dx = \frac{C}{0.5}[/tex]
[tex]\int_{0}^{\infty} e^{-0.5x} \, dx = \frac{1}{0.5}[/tex] from 0 to ∞ = 2C
Similarly, integrating for y and z:[tex]\int_{0}^{\infty} e^{-0.2y} \, dy = \frac{1}{0.2} = 5[/tex]
[tex]\int_{0}^{\infty} e^{-0.1z} \, dz = \frac{1}{0.1} = 10[/tex]
Thus, the overall integral is:2C * 5C * 10C = 1
100C³ = 1
C = 1 / 10
Part (b): Finding P(X ≤ 1.375 , Y ≤ 1.5)
We need to compute the double integral of the joint density function for X and Y:P(X ≤ 1.375 , Y ≤ 1.5) = [tex]\int_{0}^{1.375} \int_{0}^{1.5} 0.1 e^{-0.5x} e^{-0.2y} \, dy \, dx[/tex]
Evaluating the inner integral with respect to y first:[tex]\int_{0}^{1.5} e^{-0.2y} \, dy = -\frac{1}{0.2} (e^{-0.3} - 1)[/tex]
= [tex]\frac{1}{0.2} (1 - e^{-0.3}) = \frac{5}{3} (1 - e^{-0.3})[/tex]
= 5 * 0.2592
≈ 1.296
Now, integrating with respect to x:[tex]\int_{0}^{1.375} 0.1 \cdot 1.296 \cdot e^{-0.5x} \, dx = 0.1 \cdot 1.296 \cdot \left( 2 - 2 e^{-0.6875} \right)[/tex]
= [tex]0.1296 \cdot \left[ -2 e^{-0.5x} \right]_{0}^{1.375}[/tex]
= [tex]0.1296 \cdot 2 \cdot \left( 1 - e^{-0.6875} \right)[/tex]
= 0.1296 * 2 * (1 - 0.5038)
≈ 0.1296 * 2 * 0.4962
≈ 0.1286
Part (c): Finding P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1)
We need to compute the triple integral:P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1) = [tex]\int_{0}^{1.375} \int_{0}^{1.5} \int_{0}^{1} 0.1 e^{-0.5x} e^{-0.2y} e^{-0.1z} \, dz \, dy \, dx[/tex]
We already have the inner integrals for y and z from part (b):[tex]\int_{0}^{1} e^{-0.1z} dz = \left[ -\frac{1}{0.1} e^{-0.1z} \right]_{0}^{1}[/tex]
= [tex](1/0.1)(1 - e^{-0.1})[/tex]
= [tex]10 (1 - e^{-0.1})[/tex]
≈ 0.905
Combining all parts:
P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1) = 0.1286 * 0.905
≈ 0.1163
Write an equation of the line passing through the point (3, -1) and parallel to the line y=2/3x - 5. Show work
Answer:
The answer is: y = 2/3x - 3
Step-by-step explanation:
Given point: (3, -1)
Given equation: y = 2/3x - 5, which is in the form y = mx + b where m is the slope and b is the y intercept.
Parallel lines have the same slope. Use the point slope form of the equation with the point (3, -1) and substitute:
y - y1 = m(x - x1)
y - (-1) = 2/3(x - 3)
y + 1 = 2/3x - 6/3
y + 1 = 2/3x - 2
y = 2/3x - 3
Proof:
f(3) = 2/3(3) - 3
= 6/3 - 3
= 2 - 3
= -1, giving the point (3, -1)
Hope this helps! Have an Awesome Day!! :-)
A sample of 30 distance scores measured in yards has a mean of 7, a variance of 16, and a standard deviation of 4. You want to convert all your distances from yards to feet, so you multiply each score in the sample by 3. What are the new mean, median, variance, and standard deviation?
Answer:
21, 144, 12
Step-by-step explanation:
Given that a sample of 30 distance scores measured in yards has a mean of 7, a variance of 16, and a standard deviation of 4.
Let X be the distance in yard.
i.e. each entry of x is multiplied by 3.
New mean variance std devition would be
E(3x) = [tex]3E(x) = 21[/tex]
Var (3x) = [tex]3^2 Var(x) = 9(16) =144[/tex]
Std dev (3x) = [tex]\sqrt{144 } =12[/tex]
Thus we find mean and std devition get multiplied by 3, variance is multiplied by 9
The new mean after converting the distances from yards to feet is 21, the median remains unchanged at 7, the new variance is 144, and the new standard deviation is 12.
To find the new mean, we multiply the original mean by the conversion factor. Since there are 3 feet in a yard, we have:
New Mean = Original Mean × Conversion Factor
= 7 yards × 3 feet/yard
= 21 feet
The median is a measure of central tendency that represents the middle value of a data set. When each data point is multiplied by a constant factor, the median is also multiplied by that factor. However, since the median is the middle value of an ordered list of data, and we are not changing the order or adding or removing any data points, the median in terms of yards and feet is the same numerical value, even though the units are different:
New Median = [tex]Original Median Ã[/tex]— Conversion Factor
= 7 yards (since the median is the same numerical value)
For the variance, when each data point is multiplied by a constant, the variance is multiplied by the square of that constant:
New Variance = [tex]Original Variance × (Conversion Factor)^2[/tex]
= [tex]16 yards^2 × (3 feet/yard)^2[/tex]
= [tex]16 yards^2 × 9 (feet^2/yard^2)[/tex]
= [tex]144 feet^2[/tex]
Finally, the standard deviation is the square root of the variance, so to find the new standard deviation, we take the square root of the new variance:
New Standard Deviation = [tex]√New Variance[/tex]
= [tex]√144 feet^2[/tex]
= 12 feet
The height of Jake's window is 5x - 3 inches and the width is 3x + 2 inches. What is the perimeter of Jake's window ?
Answer: (16x - 2)inches
Step-by-step explanation:
The shape of Jake's window is a rectangle. It means that the two opposite sides are equal. The perimeter of the rectangular window is the distance round it.
The perimeter of a rectangle is expressed as 2(length + width)
From the dimensions given
Length = height of the window
= 5x - 3 inches
Width = 3x + 2 inches
Perimeter of the window =
2(5x - 3 + 3x + 2) = 10x - 6 + 6x + 4
= 10x + 6x + 4 - 6
Perimeter of the window =
(16x - 2)inches
Answer:The perimeter of window can be calculated by the following formula;
Step-by-step explanation: