Here is the correct question.
The manufacturer of a smart watch claims that individuals who pay attention to how many steps they take per day will inadvertently take more steps per day than individuals who pay no attention to how many steps they take per day. To investigate this claim, the manufacturer conducts a study to estimate the difference in the mean number of steps taken by those that pay attention to how many steps they take per day and those that do not. To do so, 40 volunteers are recruited. Half of the volunteers are randomly assigned to receive a smart watch and are taught how to use it to track their steps. The other half of the volunteers are given a wristband to wear, but are not informed that the wristband is tracking their steps. The volunteers are monitored for 30 days. The mean and standard deviation of the number of steps taken per day are computed for each group. Here are the data:
[tex]n[/tex] [tex]\bar x[/tex] [tex]S_x[/tex]
Pay attention 20 10,244 1,580
Do not pay attention 20 8.,648 2,350
Which of the following is a 99% confidence interval for the difference in the mean number of steps taken by all people like these that do and do not pay attention to the number of steps they take per day using df - 19 ?
[tex](A) \ 1596 \ \pm 2.861 \sqrt{ \dfrac{1580^2}{20} + \dfrac{2350^2}{20}} \\ \\ \\ (B) \ 1596 \ \pm 2.861 \sqrt{ \dfrac{1580^2}{20} - \dfrac{2350^2}{20}} \\ \\ \\ (C) \ 1596 \pm 2.576 \sqrt{ \dfrac{1580^2}{20} + \dfrac{2350^2}{20}} \\ \\ \\ (D) 1596 \pm 2.576 ( \dfrac{1580^2}{\sqrt{20}} + \dfrac{2350^2}{\sqrt{20}}) \\ \\ \\ (E) 1596 \pm 2.576 ( \dfrac{1580^2}{\sqrt{20}} - \dfrac{2350^2}{\sqrt{20}})[/tex]
Answer:
[tex]\mathbf{(A) \ 1596 \ \pm 2.861 \sqrt{ \dfrac{1580^2}{20} + \dfrac{2350^2}{20}}}[/tex]
Step-by-step explanation:
Given that :
significance level [tex]\alpha = \mathbf{0.01}[/tex]
From the Given data;
Using Excel with the function : TINV(0.01,19);
Critical value t* = 2.861
The margin of error can now be represented by the illustration:
Margin of error = [tex]t^* \sqrt{ \dfrac {s_1 ^2}{n_1} + \dfrac {s_2 ^2}{n_2}[/tex]
Lower Limit = [tex](\bar x_1 - \bar x_2)- (Margin \ of \ error)[/tex]
Upper Limit = [tex](\bar x_1 - \bar x_2)+ (Margin \ of \ error)[/tex]
Thus; the confidence interval for the difference in the mean number of steps taken by all people like these that do and do not pay attention to the number of steps they take per day using df - 19 is:
[tex]\mathbf{(A) \ 1596 \ \pm 2.861 \sqrt{ \dfrac{1580^2}{20} + \dfrac{2350^2}{20}}}[/tex]
Some college professors make bound lecture notes available to their classes in an effort to improve teaching effectiveness. A study of business student's opinions of lecture notes. Two groups of students were surveyed - 86 students enrolled in a promotional strategy class that required the purchase of lecture notes, and 35 students enrolled in a sales/retailing elective that did not offer lecture notes. At the end of the semester :"Having a copy of the lecture notes was helpful in understanding the material." Responses were measured on a nine-point semantic difference scale, where 1="strongly disagree" and 9=" strongly agree." A summary of the results is reported in the follow:
Classes Buying Lecture Notes Classes Not Buying Lecture Notes
n1=86 n2=35
X1=8.48 X2=7.80
S21=.94 S22=2.99
a. Describe the two populations involved in the comparison.
b. Do the samples provides sufficient evidence to conclude that there is a difference in the mean responses of the two groups of the students? Test using α=.01
c. Construct a 99% confidence interval for (μ1-μ2). Interpret the result.
d. Would a 95% confidence interval for (μ1-μ2) be narrow or wider than the one you found in part c? Why?
Answer:
Step-by-step explanation:
a) The number of students sampled in both populations are large. We can assume that the populations are normally distributed. The populations are also independent.
b) This is a test of 2 independent groups. Let μ1 be the mean responses of students buying lecture notes and μ2 be the mean responses of students not buying lecture notes.
The random variable is μ1 - μ2 = difference in the mean responses of students buying lecture notes and the mean responses of students not buying lecture notes.
We would set up the hypothesis.
The null hypothesis is
H0 : μ1 = μ2 H0 : μ1 - μ2 = 0
The alternative hypothesis is
H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0
This is a two tailed test.
Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is
(x1 - x2)/√(s1²/n1 + s2²/n2)
From the information given,
x1 = 8.48
x2 = 7.8
s1 = 0.94
s2 = 2.99
n1 = 86
n2 = 35
t = (8.48 - 7.8)/√(0.94²/86 + 2.99²/35)
t = 1.32
The formula for determining the degree of freedom is
df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²
df = [0.94²/86 + 2.99²/35]²/[(1/86 - 1)(0.94²/86)² + (1/35 - 1)(2.99²/35)²] = 0.0706/0.00192021883
df = 37
We would determine the probability value from the t test calculator. It becomes
p value = 0.195
c) Since alpha, 0.01 < than the p value, 0.195, then we would fail to reject the null hypothesis. Therefore, at 5% significance level, the samples do not provide sufficient evidence to conclude that there is a difference in the mean responses of the two groups of the students.
d) The formula for determining the confidence interval for the difference of two population means is expressed as
Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)
For a 99% confidence interval, the z score is 1.2.58. This is determined from the normal distribution table.
x1 - x2 = 8.48 - 7.8 = 0.68
z√(s1²/n1 + s2²/n2) = 2.58√(0.94²/86 + 2.99²/35) = 1.33
The confidence interval is
0.68 ± 1.33
The upper boundary for the confidence interval is
0.68 + 1.01 = 2.01
The lower boundary for the confidence interval is
0.68 - 1.33 = - 0.65
We are confident that the difference in population means responses between the students buying lecture notes and the students not buying lecture notes is between - 0.65 and 2.01
d) For a 95% confidence interval, the z score is 1.96.
z√(s1²/n1 + s2²/n2) = 1.96√(0.94²/86 + 2.99²/35) = 1.01
The confidence interval is
0.68 ± 1.01
The upper boundary for the confidence interval is
0.68 + 1.01 = 1.69
The lower boundary for the confidence interval is
0.68 - 1.01 = - 0.33
Therefore, a 95% confidence interval for (μ1-μ2) would be narrower. This is seen in the values in both scenarios.
What’s the correct explanation for this question?
Step-by-step explanation:
=> The volume of a triangular pyramid can be found using the formula V = 1/3AH where A = area of the triangle base, and H = height of the pyramid
=> The volume of a cone can be found by V = 1/3(Ab)(H) where Ab is base area and H is the height of the cone
The difference between both is that is it's base. A cone has a polygonal base while a pyramid has a tetragonal base
Last weekend, Lena worked 7.5 hours on Friday, 9.75 hours on Saturday, and 6.25 hours on Sunday.
She earns £8.60 per hour. How much did she earn in total?
Answer:202.1
Step-by-step explanation:
7.5hrs +9.75hrs+6.25hrs=23.5
8.60 X 23.5 =202.1 pounds
Which expression is equivalent to log Subscript 8 Baseline 4 a (StartFraction b minus 4 Over c Superscript 4 Baseline EndFraction)?
Answer:
[tex]\log_84+\log_8a+\log_8(b-4)-4\log_8c[/tex].
Step-by-step explanation:
The given expression is
[tex]\log_84a\left(\dfrac{b-4}{c^4}\right)[/tex]
Using the properties of logarithm, we get
[tex]\log_84+\log_8a+\log_8\left(\dfrac{b-4}{c^4}\right)[/tex] [tex][\because \log_a mn=\log_a m+\log_a n][/tex]
[tex]\log_84+\log_8a+\log_8(b-4)-\log_8c^4[/tex] [tex][\because \log_a \frac{m}{n}=\log_a m-\log_a n][/tex]
[tex]\log_84+\log_8a+\log_8(b-4)-4\log_8c[/tex] [tex][\because \log_a x^n =n\log_a x][/tex]
Therefore, the required expression is [tex]\log_84+\log_8a+\log_8(b-4)-4\log_8c[/tex].
Answer:
B on edge
Step-by-step explanation:
ASAP! GIVING BRAINLIEST! Please read the question THEN answer CORRECTLY! NO guessing. I say no guessing because people usually guess on my questions.
Answer:
B. F(x) = 3(x - 2)² - 2
Step-by-step explanation:
→The function F(x) narrowed, meaning the absolute value being multiplied to the function is greater than 1.
→The function F(x) flipped over the x-axis, this means that the number being multiplied has to be a negative.
→The function F(x) shifted to the left 2 units, this means there needs to be a 2 being added.
→The function F(x) shifted downwards 2 units, meaning there needs to be a 2 being subtracted from the whole function.
This gives us the correct answer of "B. F(x) = 3(x - 2)² - 2."
What’s the correct answer for this question?
Answer:
C
Step-by-step explanation:
A cylinder is formed when rotating the 3-D figure around y-axis
Any help would be great
Answer:
30%
Step-by-step explanation:
fat ÷ total
15 ÷ 50
.3
30%
Answer:
30%
Step-by-step explanation:
To find the percent from fat, take the calories from fat and divide by the total
15/50
.3
Multiply by 100%
30%
Please help me explain your answer only answer if you are sure
Answer:
The answer of top prism is 262
and down prism is 478
The upper figure is triangular prism.
so, we use bh+2ls+lb formula
B=5
h=3
s=4
l=19
Now,
surface area of triangular prism = bh+2ls+lb
= 5×3+2×19×4+19×5
= 262
The down figure is rectangular prism.
so, we use 2lw+2lh+2hw
l=5
h=6
w=19
Now,
The area of rectangular prism = 2lw+2lh+2hw
= 2×5×19+2×19×6+2×5×6
= 478
divide the following polynomials ( 9 x 4 + 3 x 3 y − 5 x 2 y 2 + x y 3 ) ÷ ( 3 x 2 + 2 x 2 y − x y 2 )
Answer:
2(-2y+9)/3+y
Step-by-step explanation:
Translate to a system of equations: Twice a number plus three times a second number is negative one. The first number plus four times the second number is two.
Answer:
work is shown and pictured
The length of a field is twice it's breadth. If the length is 30cm. Calculate the perimeter of the field.
Answer:
b=30/2=15
peri= 90
Step-by-step explanation:
Solve the problem.
If a boat uses 25 gallons of gas to go 73 miles, how many miles
can the boat travel on 75 gallons of gas?
24 mi
438 mi
219 mi
239 mi
Answer:
For this problem we can use the following proportional rule:
[tex] \frac{73 mi}{25 gal}= \frac{x}{75 gal}[/tex]
Where x represent the number of miles that we can travel with 75 gallons. For this case we can use this proportional rule since by definition [tex] D=vt[/tex]. If we solve for x we got:
[tex] x =75 gal (\frac{73 mi}{25 gal}) =219mi[/tex]
And the best answer would be:
219 mi
Step-by-step explanation:
For this problem we can use the following proportional rule:
[tex] \frac{73 mi}{25 gal}= \frac{x}{75 gal}[/tex]
Where x represent the number of miles that we can travel with 75 gallons. For this case we can use this proportional rule since by definition [tex] D=vt[/tex]. If we solve for x we got:
[tex] x =75 gal (\frac{73 mi}{25 gal}) =219mi[/tex]
And the best answer would be:
219 mi
Line segment ON is perpendicular to line segment ML
What is the length of chord ML?
0
20 units
24 units
26 units
30 units
13
P
8
M
N
Mark this and return
Answer:
The correct answer is B (24 units)
Step-by-step explanation:
HELP PLEASE!!
NEED ANSWER ASAP!!!
A farmer in China discovers a mammal
hide that contains 54% of its original
Find age of the mammal hide to the nearest year.
amount of C-14
N=N0e^-kt
N = Noe
No = inital amount of C-14 (at time t = 0)
N = amount of C-14 at time t
k = 0.0001
t = time, in years
Answer:
6163.2 years
Step-by-step explanation:
A_t=A_0e^{-kt}
Where
A_t=Amount of C 14 after “t” year
A_0= Initial Amount
t= No. of years
k=constant
In our problem we are given that A_t is 54% that is if A_0=1 , A_t=0.54
Also , k=0.0001
We have to find t=?
Let us substitute these values in the formula
0.54=1* e^{-0.0001t}
Taking log on both sides to the base 10 we get
log 0.54=log e^{-0.0001t}
-0.267606 = -0.0001t*log e
-0.267606 = -0.0001t*0.4342
t=\frac{-0.267606}{-0.0001*0.4342}
t=6163.20
t=6163.20 years
PLEASE MARK BRAINLY
Consider the following dice game, as played at a certain gambling casino: players 1 and 2 roll a pair of dice in turn. the bank then rolls the dice to determine the outcome according to the following rule: player i,i=1,2, wins if his roll is strictly
Ii={1 if i wins, 0 otherwise}
and show that I1 and I2 are positively correlated. Explain why this result was to be expected.
Answer:
they are positively correlated.
Step-by-step explanation:
We can calculate the individal expectations first. FIrst player will win if that player's roll is greater than the bank's roll. There are (6 possible rolls of player 1 * 6 possible rolls of bank =) 36 total possible rolls, out of which player 1 will win in 15 cases.
[tex]\therefore E(I_i) = 1\cdot \frac{15}{36} + 0 \cdot \frac{21}{36} = \frac{5}{12} \approx 0.4167[/tex]
For the joint expectation, there are (6 possible rolls of player 1 * 6 possible rolls of player 2 * 6 possible rolls of bank =) 216 total possible rolls.
Cases where both players win: Expectation = $2.
If bank rolls 1, both players will win in 5*5 = 25 cases. P1 is one of {2,3,4,5,6}, P2 is one of {2,3,4,5,6}
If bank rolls 2, both players will win in 4*4 = 16 cases.
If bank rolls 3, both players will win in 3*3 = 9 cases.
If bank rolls 4, both players will win in 2*2 = 4 cases.
If bank rolls 5, both players will win in 1*1 = 1 cases.
If bank rolls 6, both players will win in 0*0 = 0 cases.
Total cases = 25+16+9+4+1+0 = 55 cases.
Cases where player 1 wins $1 and player 2 loses: Expectation = $1.
If bank rolls 1, player 1 will win and player 2 will lose in 5*1 = 5 cases. P1 is one of {2,3,4,5,6}, P2 is {1}
If bank rolls 2, player 1 will win and player 2 will lose in 4*2 = 8 cases.
If bank rolls 3, player 1 will win and player 2 will lose in 3*3 = 9 cases.
If bank rolls 4, player 1 will win and player 2 will lose in 2*4 = 8 cases.
If bank rolls 5, player 1 will win and player 2 will lose in 1*5 = 5 cases.
If bank rolls 6, player 1 will win and player 2 will lose in 0*6 = 0 cases.
Total cases = 5+8+9+8+5+0 = 35
Cases where player 2 wins $1 and player 1 loses: Expectation = $1.
This is the same as above with player 1 and 2 exchanged.
Total cases = 35
Cases where both players lose: Expectation = $0.
If bank rolls 1, both players will lose in 1*1 = 1 cases. P1 is {1}, P2 is {1}
If bank rolls 2, both players will lose in 2*2 = 4 cases.
If bank rolls 3, both players will lose in 3*3 = 9 cases.
If bank rolls 4, both players will lose in 4*4 = 16 cases.
If bank rolls 5, both players will lose in 5*5 = 25 cases.
If bank rolls 6, both players will lose in 6*6 = 36 cases.
Total cases = 1+4+9+16+25+36 = 91 cases.
Total of all cases (we expect this to be 216 as mentioned above) = 55+35+35+91=216
So, joint expectation is:
[tex]E(I_1I_2) = \frac{2\cdot 55 +1\cdot 35+1\cdot 35+0\cdot 91}{216} = \frac{180}{216}= \frac{5}{6} \approx 0.8333[/tex]
So, the covariance is given by:
[tex]\texttt{Cov}(I_1I_2) =E(I_1I_2) -E(I_1)\cdot E(I_2)= \frac{5}{6}-\frac{5}{12}\cdot\frac{5}{12}=\frac{95}{144} \approx 0.6597[/tex]
As this is greater than 0 and closer to 1, they are positively correlated.
The reason why this result is expected is because the same bank roll is being used for both players. So, it is very likely that both players will win if the bank roll is 1 or even 2. Also, it is very likely that both players will lose if the bank roll is 6, 5, or even 4. This shows positive correlation between the events.
help asap, will get branliest !!
Answer:
D
Step-by-step explanation:
Lines EF and GH are already parallel. Translating them 2 units to the side without changing how far apart they are vertically means they won't intersect and will remain the same distance apart.
Answer:
D
Step-by-step explanation:
They are parallel lines
Solve the equation.
3(x + 1)-1=3x+2
Answer:
0=0
Step-by-step explanation:Let's solve your equation step-by-step.
3(x+1)−1=3x+2
Step 1: Simplify both sides of the equation.
3(x+1)−1=3x+2
(3)(x)+(3)(1)+−1=3x+2(Distribute)
3x+3+−1=3x+2
(3x)+(3+−1)=3x+2(Combine Like Terms)
3x+2=3x+2
3x+2=3x+2
Step 2: Subtract 3x from both sides.
3x+2−3x=3x+2−3x
2=2
Step 3: Subtract 2 from both sides.
2−2=2−2
0=0
mp
Bob has 54 more five-dollar bills than ten-dollar bills. The number of five-dollar bills he has
is 7 times that of ten-dollar bills. How many dollars does Bob have in all?
Answer:5000 sum
Step-by-step explanation:
which is pattern 12,24,36,48
Answer:
multiples of 12
Step-by-step explanation: when looking at the GCF, the answer is 12
In 2018, the number of students at The Villages High School was 975 and is increasing at a rate of 2.5% per year. Write and use an exponential growth function to project the populating in 2025. Round to the nearest whole number. Help plzzz
Answer:
[tex]A(t)=975(1.025)^t[/tex]
In 2025,the number of students at the villages high school=1159
Step-by-step explanation:
We are given that in 2018
Number of students at the villages high school=975
Increasing rate,r=2.5%=0.025
We have to write and use of exponential growth function to project the populating in 2025.
[tex]A_0=975,t=0[/tex]
According to question
Number of students at the villages high School is given by
[tex]A(t)=A_0(1+r)^t[/tex]
Substitute the values
[tex]A(t)=975(1+0.025)^t=975(1.025)^t[/tex]
t=7
Substitute the value
Then, the number of students at the villages high school in 2025
[tex]A(7)=975(1.025)^7=1158.96\approx 1159[/tex]
Answer:
1,159 students
Step-by-step explanation:
the exponential growth rate formula:
A = P ( 1 + r)ⁿ
A = amount after growth = ??P = current/original amount = 975 studentsr = yearly growth rate = 2.5% or 0.025n = number of years = 2025 - 2018 = 7Pop. 2025 = 975 (1 + 0.025)⁷
Pop. 2025 = 975 x 1.025⁷ = 1,158.97 ≈ 1,159 students
Consider the following set of sample data.
18 26 30 42 50 52 52 76 78 84
For the given data, the mean is_______, the median is________, and the mode is_______.
Suppose the value 76 in the data is mistakenly recorded as 55 instead of 76. For the sample with this error, the mean is_________, the median is______, and the mode is_______. The mean_____, the median_______, and the mode______. Suppose the value 76 in the original sample is inadvertently removed from the sample. For the sample with this value removed, the mean is_______, the median is_______, and the mode is________. The mean_________, the median_______, and the mode________.
Answer:
For the given data, the mean is 50.8, the median is 51, and the mode is 52.
For the sample with this error, the mean is 48.7, the median is 51, and the mode is 52.
For the sample with this value removed, the mean is 43.2, the median is 50, and the mode is 52.
Step-by-step explanation:
We are given the following set of sample data below;
18, 26, 30, 42, 50, 52, 52, 76, 78, 84.
The formula for calculating mean is given by;
Mean = [tex]\frac{\text{Sum of all data values}}{\text{Total number of observations}}[/tex]
= [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 76+ 78+ 84}{10}[/tex]
= [tex]\frac{508}{10}[/tex] = 50.8
For calculating median, we have to observe that the number of observations (n) in our data is even or odd, i.e;
If n is odd, then the formula for calculating median is given by;Median = [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]
If n is even, then the formula for calculating median is given by;Median = [tex]\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}[/tex]
Now, here in our data the number of observations is even, i.e. n = 10.
So, Median = [tex]\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}[/tex]
= [tex]\frac{(\frac{10}{2})^{th}\text{ obs.} +(\frac{10}{2}+1)^{th}\text{ obs.} }{2}[/tex]
= [tex]\frac{5^{th}\text{ obs.} +6^{th}\text{ obs.} }{2}[/tex]
= [tex]\frac{50+52 }{2}[/tex] = 51
A Mode is a value that appears the maximum number of times in our data.
In our data, the value 52 is appera]ing maximum number of times, i.e. 2 times which means that mode of our data is 52.
Now, suppose the value 76 in the data is mistakenly recorded as 55 instead of 76. For the sample with this error,
Mean will be changed as value has been changed.New Mean = [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 55+ 78+ 84}{10}[/tex]
= [tex]\frac{487}{10}[/tex] = 48.7
There will be no change in median because there is no change in the 5th and 6th observation of the data.Also, there will be no change in mode as stiil 52 appears maximum number of times in our data.Now, suppose the value 76 in the original sample is inadvertently removed from the sample. For the sample with this value removed,
Mean will be changed as value has been removed from data.New Mean = [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 78+ 84}{9}[/tex]
= [tex]\frac{432}{10}[/tex] = 43.2
Median will also get changed because the number of observation is now odd, i.e. n = 9So, Median = [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]
= [tex](\frac{9+1}{2})^{th} \text{ obs.}[/tex]
= [tex]5^{th} \text{ obs.}[/tex] = 50
Also, there will be no change in mode as stiil 52 appears maximum number of times in our data.A hot dog has about 1/4 the amount of protein as 3 ounces of hamburger. Together, they have about 25 grams of protein. How many grams of protein are in a 3 oz hamburger?
Answer:
(1) protein in hot dog = ¼ * protein in 3 ounces of hamburger
(2) protein in hot dog + protein in 3 ounces of hamburger = 25
So we need to re-arrange (1) and (2) to solve for the protein in 3 ounces of hamburger!
(re-arrange (1)): 4 * protein in hot dog = protein in 3 ounces of hamburger
(re-arrange (2)): protein in hot dog = 25 - protein in 3 ounces of hamburger
(plugging re-arranged (2) into re-arranged (1)):
4 * (25 - protein in 3 ounces of hamburger) = protein in 3 ounces of hamburger ( multiplying )
100- 4 protein in 3 ounces of hamburger = protein in 3 ounces of hamburger
solving for the protein in 3 ounces of hamburger:
5 * protein in 3 ounces of hamburger = 100
protein in 3 ounces of hamburger = 20 gram
normally distributed with an unknown population mean and a population standard deviation of 4.5 points. A random sample of 45 scores is taken and gives a sample mean of 84. Find a 90% confidence interval
Answer:
= ( 82.90, 85.10) points
Therefore at 90% confidence interval (a,b)= ( 82.90, 85.10) points
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
Given that;
Mean x = 84
Standard deviation r = 4.5
Number of samples n = 45
Confidence interval = 90%
z(at 90% confidence) = 1.645
Substituting the values we have;
84+/-1.645(4.5/√45)
84+/-1.645(0.670820393249)
84+/-1.10
= ( 82.90, 85.10) points
Therefore at 90% confidence interval (a,b)= ( 82.90, 85.10) points
What is the simplified value of the exponential expression 27 1/3?
1/3
1/9
3
9
Answer:3
Step-by-step explanation:
Answer:
C.3
Step-by-step explanation:
The following data represent the number of flash drives sold per day at a localcomputer shop and their prices.Price Units Sold34 336 432 635 530 938 240 1a. Develop the estimated regression equation that could be used to predict thequantity sold given the price. Interpret the slope.b. Did the estimated regression equation provide a good fit? Explain.c. Compute the sample correlation coefficient between the price and the number offlash drives sold. Use a= 0.01 to test the relationship between price and units sold.d. How many units can be sold per day if the price of flash drive is set to $28.
Answer:
a)3145 x 0.01 = 31.45 3145- 31.45 = 3113.55
Compute the sample correlation 3113.55 -? we find the least square pressing at least 15x on the calculator then minus this from 3113.55 to find a better fit and minimum regression.
We add the differences of units then divide by distribution as seen below.
b) unsure.
c) = (see below) just test each number shown unit sold per day / price then x can show the differences in each number from day 1 to day 2.
d) = 16 sold.
Step-by-step explanation:
a) We count the units up and deduct from it from the equation p is recognized as units sold. R1 is cost R2 is total days.
b) The line of best fit is described by the equation ŷ = bX + a, where b is the slope of the line and a is the intercept (i.e., the value of Y when X = 0).
c) r 2= decimal ; the regression equation has accounted for percentage of the total sum of squares. You cna do this one.
d) = 16 sold at $28 each. - Why ? We using 7 day data and prove a how many units can be sold p/d if the price of flash drive is set to $28 each per unit.
Day 1 = 34 / 28 = 1 = 1.21428571429 = 1 no difference day prior.
Day 2 = 336 / 28 = 12 = 12 = difference day prior is 11
Day 3 = 432 / 28 = 15 = 15.4285714286 = 15 difference day prior is 3
Day 4 = 635 / 28 = 23 = 22.6785714286 = 23 difference day prior is 8
Day 5 = 530 / 28 = 19 = 18.9285714286 = 19 difference day prior is minus - 4
Day 6 = 938 / 28 = 34 = 33.5 = 34 difference day prior is 15
Day 7 = 240 / 28 = 9 = 8.57142857143 = 9 difference day prior is minus -25
Total days 7 = Total revenue / price = average units sold
Average units sold total = 1+ 12+15 +23 +19+34+9 = 113 rounded.
Average units sold total = 1.21428571429 + 12 + 15.4285714286
+ 22.6785714286
+18.9285714286
+ 33.5
+ 8.57142857143 = 112.321428572 units sold weekly when priced at $28
To answer D we divide this by 7 to show;
112.321428572/ 7 = 16.0459183674
Daily units sold = 16
the sum of the three numbers in 2003,two of the numbers are 814 and 519 what is the third number
Answer:
idk dont ask me
Step-byi-step explanation:
Answer:
a+b+c=2003
a+b=814
2003-819=189
Step-by-step explanation:
Mist (airborne droplets or aerosols) is generated when metal-removing fluids are used in machining operations to cool and lubricate the tool and work-piece. Mist generation is a concern to OSHA, which has recently lowered substantially the workplace standard. The article "Variables Affecting Mist Generation from Metal Removal Fluids" (Lubrication Engr., 2002: 10-17) gave the accompanying data on x = fluid flow velocity for a 5% soluble oil (cm/sec) and y = the extent of mist droplets having diameters smaller than some value:
x: 89 177 189 354 362 442 965
y: .40 .60 .48 .66 .61 .69 .99
a. Make a scatterplot of the data. By R.
b. What is the point estimate of the beta coefficient? (By R.) Interpret it.
c. What is s_e? (By R) Interpret it.
d. Estimate the true average change in mist associated with a 1 cm/sec increase in velocity, and do so in a way that conveys information about precision and reliability.
e. Suppose the fluid velocity is 250 cm/sec. Find the mean of the corresponding y in a way that conveys information about precision and reliability. Use 95% confidence level. Interpret the resulting interval. By hand, as in part d.
f. Suppose the fluid velocity for a specific fluid is 250 cm/sec. Predict the y for that specific fluid in a way that conveys information about precision and reliability. Use 95% prediction level. Interpret the resulting interval. By hand, as in part d.
Answer:
Step-by-step explanation:
a) image attached
b) Lets do the analysis in R , the complete R snippet is as follows
x<- c(89,177,189,354,362,442,965)
y<- c(.4,.6,.48,.66,.61,.69,.99)
# scatterplot
plot(x,y, col="red",pch=16)
# model
fit <- lm(y~x)
summary(fit)
#equation is
#y = 0.4041 + 0.0006211*X
# beta coeffiecients are
fit$coefficients
coef(summary(fit))[, "Std. Error"]
# confidence interval of slope
confint(fit, 'x', level=0.95)
The results are
> summary(fit)
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6 7
-0.05940 0.08595 -0.04151 0.03602 -0.01895 0.01136 -0.01346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.041e-01 3.459e-02 11.684 8.07e-05 ***
x 6.211e-04 7.579e-05 8.195 0.00044 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.05405 on 5 degrees of freedom
Multiple R-squared: 0.9307, Adjusted R-squared: 0.9168 # model is able to capture 93% of the variation of the data
F-statistic: 67.15 on 1 and 5 DF, p-value: 0.0004403 , p value is less than 0.05 , hence model as a whole is significant
> fit$coefficients
(Intercept) x
0.4041237853 0.0006210758
> coef(summary(fit))[, "Std. Error"]
(Intercept) x
3.458905e-02 7.579156e-05
> confint(fit, 'x', level=0.95)
2.5 % 97.5 %
x 0.0004262474 0.0008159042
c)
> x=c(89,177,189,354,362,442,965)
> y=c(0.40,0.60,0.48,0.66,0.61,0.69,0.99)
>
> ### linear model
> model=lm(y~x)
> summary(model)
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6 7
-0.05940 0.08595 -0.04151 0.03602 -0.01895 0.01136 -0.01346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.041e-01 3.459e-02 11.684 8.07e-05 ***
x 6.211e-04 7.579e-05 8.195 0.00044 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.05405 on 5 degrees of freedom
Multiple R-squared: 0.9307, Adjusted R-squared: 0.9168
F-statistic: 67.15 on 1 and 5 DF, p-value: 0.0004403
s_e is the Residual standard error from the model and its estimated value is 0.05405. s_e is the standard deviation of the model.
d) 95% confidence interval
> confint(model, confidence=0.95)
2.5 % 97.5 %
(Intercept) 0.3152097913 0.4930377793
x 0.0004262474 0.0008159042
Comment: The estimated confidence interval of slope of x does not include zero. Hence, x has the significant effect on y at 0.05 level of significance.
e)
> predict(model, newdata=data.frame(x=250), interval="confidence", level=0.95)
fit lwr upr
1 0.5593927 0.5020485 0.616737
f)
> predict.lm(model, newdata=data.frame(x=250), interval="prediction", level=0.95)
fit lwr upr
1 0.5593927 0.4090954 0.7096901
g Question 6 1 pts A 3x3 matrix with real entries can have (select ALL that apply) Group of answer choices three eigenvalues, all of them real. three eigenvalues, all of them complex. two real eigenvalues and one complex eigenvalue. one real eigenvalue and two complex eigenvalues. only two eigenvalues, both of them real. only two eigenvalues, both of them complex. only one eigenvalue -- a real one. only one eigenvalue -- a complex one.
Answer:
(A)three eigenvalues, all of them real.
(D)one real eigenvalue and two complex eigenvalues.
(G)only one eigenvalue -- a real one.
Step-by-step explanation:
Given an [tex]n \times n[/tex] matrix, the characteristic polynomial of the matrix is the degree n polynomial in one variable λ:
[tex]p(\lambda) = det(\lambda I- A)[/tex]
If such [tex]n \times n[/tex] matrix A has real entries, its complex eigenvalues will always occur in complex conjugate pairs.
Therefore, for a [tex]3 \times 3[/tex] matrix with real entries, the following are possible:
(A)three eigenvalues, all of them real.
(D)one real eigenvalue and two complex eigenvalues.
(G)only one eigenvalue -- a real one.
A [tex]3 \times 3[/tex] matrix with real entries cannot have the following:
(B)three eigenvalues, all of them complex.
(C)two real eigenvalues and one complex eigenvalue.
(E)only two eigenvalues, both of them real.
(F)only two eigenvalues, both of them complex.
(H)only one eigenvalue -- a complex one.
Four men are to divide K500 equally among them. When the money was given, 20% was taken away.
How much each did the four men receive?
Answer: 20% of 500= 100
So 500-100 = 400
4x100= 400
Step-by-step explanation:
please help! ill give 24 points just tryna finish before the last day
Answer:
(1,3)
Step-by-step explanation:
Note that the solution for a graphed system of equations is just the point where the two lines intersect.
A point is (x coordinate, y coordinate).
This said, we can find the point where it intersects then see which value it is above for the x axis.
It is directly above 1.
So the x coordinate is 1.
Now, let's look at what coordinate it is next to on the y axis.
It would be 3.
So the y coordinate is 3.
Therefore, the solution to the system of equations graphed below is (1,3)
