Answer:
a) Option D is correct.
H0: μ = 71
Ha: μ > 71
b) Option F is correct
z > 1.28
c) z = 2.85
d) Option C is correct.
Reject H0.There is sufficient evidence at the α = 0.10 level of significance to conclude that the true mean heart rate during laughter exceeds 71 beats per minute.
Step-by-step explanation:
a) For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
This question aims to test the the true mean heart rate during laughter exceeds 71 beats per minute.
Hence, the null hypothesis is that there isn't sufficient evidence to say that the true mean heart rate during laughter exceeds 71 beats per minute. That is, the true mean doesn't exceed 71 beats per minute.
And the alternative hypothesis is that there is sufficient evidence to say that the true mean heart rate during laughter exceeds 71 beats per minute.
Mathematically,
The null hypothesis is represented as
H₀: μ = 71
The alternative hypothesis is represented as
Hₐ: μ > 71
b) Using z-distribution, the rejection area is obtained from the confidence level at which the test is going to be performed. Since the hypothesis test tests only in one direction,
Significance level = (100% - confidence level)/2
0.10 = 10% = (100% - confidence level)/2
20% = 100% - (confidence level)
Confidence level = 100% - 20% = 80%
Critical value for 80% confidence level = 1.28
And since we are testing if the true mean heart rate during laughter exceeds 71 beats per minute, the rejection area would be
z > 1.28
c) The test statistic is given as
z = (x - μ)/σₓ
x = sample mean = 73.4
μ = 71
σₓ = standard error = (σ/√n)
σ = 8
n = Sample size = 90
σₓ = (8/√90) = 0.8433
z = (73.4 - 71) ÷ 0.8433
z = 2.846 = 2.85
d) Since the z-test statistic obtained, 2.85, is firmly in the rejection area, z > 1.28, we reject the null hypothesis, accept the alternative hypothesis and say that there is sufficient evidence at the α = 0.10 level of significance to conclude that the true mean heart rate during laughter exceeds 71 beats per minute.
Hope this Helps!!!
segment AB is dilated from the origin to create segment A prime B prime at A' (0, 6) and B' (6, 9). What scale factor was segment AB dilated by?
1/2
2
3
4
Answer:
the answer is 3
Step-by-step explanation:
i took the test
Find the area of a triangle that has the base of 5 inches and a height of 3 3/4 inches
Answer:
9.375 in^2
Step-by-step explanation:
Consider the set of sequences of seven letters chosen from W and L. We may think of these sequences as representing the outcomes of a match of seven games, where W means the first team wins the game and L means the second team wins the game. The match is won by the first team to win four games (thus, some games may never get played, but we need to include their hypothetical outcomes in the points in order that we have a probability space of equally likely points).A. What is the probability that a team will win the match, given that it has won the first game?B. What is the probability that a team will win the match, given that it has won the first two games? C. What is the probability that a team will win the match, given that it has won two out of the first three games?
Answer:
a) Probability that a team will win the match given that it has won the first game = 0.66
b) Probability that a team will win the match given that it has won the first two games= 0.81
c) Probability that a team will win the match, given that it has won two out of the first three games = 0.69
Step-by-step explanation:
There are a total of seven games to be played. Therefore, W and L consists of 2⁷ equi-probable sample points
a) Since one game has already been won by the team, there are 2⁶ = 64 sample points left. If the team wins three or more matches, it has won.
Number of ways of winning the three or more matches left = [tex]6C3 + 6C4 + 6C5 + 6C6[/tex]
= 20 + 15 + 6 + 1 = 42
P( a team will win the match given that it has won the first game) = 42/64 = 0.66
b) Since two games have already been won by the team, there are 2⁵ = 32 sample points left. If the team wins two or more matches, it has won.
Number of ways of winning the three or more matches left = [tex]5C2 + 5C3 + 5C4 + 5C5[/tex] = 10 + 10 + 5 +1 = 26
P( a team will win the match given that it has won the first two games) = 26/32 = 0.81
c) Probability that a team will win the match, given that it has won two out of the first three games
They have played 3 games out of 7, this means that there are 4 more games to play. The sample points remain 2⁴ = 16
They have won 2 games already, it means they have two or more games to win.
Number of ways of winning the three or more matches left = [tex]4C2 + 4C3 + 4C4[/tex] = 6 + 4 + 1 = 11
Probability that a team will win the match, given that it has won two out of the first three games = 11/16
Probability that a team will win the match, given that it has won two out of the first three games = 0.69
All math teachers are smart. Ms. Smith is your math teacher, so she is smart. What type of reasoning is this? inductive or deductive
Answer:
I believe it is Inductive Reasoning.
Step-by-step explanation:
Inductive Reasoning is a type of logical thinking that involves forming generalizations based on specific incidents you've experienced, observations you've made, or facts you know to be true or false.
Deductive Reasoning is a basic form of valid reasoning.
The figure shows a square floor plan with a smaller square area that will accommodate a combination fountain and pool.The floor with the fountain pool area removed has an area of 33 Square meters and a perimeter of 36 meters. Find the dimensions of the floor and the dimensions of the square that will accommodate the fountain and pool.
Answer:
(x, y) = (7, 4) meters
Step-by-step explanation:
The area of the floor without the removal is x^2, so with the smaller square removed, it is x^2 -y^2.
The perimeter of the floor is the sum of all side lengths, so is 4x +2y.
The given dimensions tell us ...
x^2 -y^2 = 33
4x +2y = 36
From the latter equation, we can write an expression for y:
y = 18 -2x
Substituting this into the first equation gives ...
x^2 -(18 -2x)^2 = 33
x^2 -(324 -72x +4x^2) = 33
3x^2 -72x + 357 = 0 . . . . write in standard form
3(x -7)(x -17) = 0 . . . . . factor
Solutions to this equation are x=7 and x=17. However, for y > 0, we must have x < 9.
y = 18 -2(7) = 4
The floor dimension x is 7 meters; the inset dimension y is 4 meters.
An Undergraduate Study Committee of 6 members at a major university is to be formed from a pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is the probability that precisely half of the members will be women?
Answer:
5/33649= approx 0.00015
Step-by-step explanation:
Total number of outcomes are C24 6= 24!/(24-6)!/6!=19*20*21*22*23*24/(2*3*4*5*6)= 19*14*22*23
Half of the Committee =3 persons. That mens that number of the women in Commettee=3. 3 women from 6 can be elected C6 3 ways ( outputs)=
6!/3!/3!=4*5*6*/2/3=20
So the probability that 3 members of the commettee are women is
P(women=3)= 20/(19*14*22*23)=5/(77*19*23)=5/33649=approx 0.00015
The probability that precisely half of the members will be women is;
P(3 women) = 0.1213
This question will be solved by hypergeometric distribution which has the formula;
P(x) = [S_C_s × (N - S)_C_(n - s)]/(NC_n)
where;
S is success from population
s is success from sample
N is population size
n is sample size
We are give;
s = 3 women (which is precisely half of the members selected)
S = 6 women
N = 24 men and women
n = 6 people selected
Thus;
P(3 women) = (⁶C₃ * ⁽¹⁸⁾C₍₃₎)/(²⁴C₆)
P(3 women) = (20 * 816)/134596
P(3 women) = 0.1213
Read more at; https://brainly.com/question/5733654
Consider random samples of size 900 from a population with proportion 0.75 . Find the standard error of the distribution of sample proportions. Round your answer for the standard error to three decimal places. standard error
Answer:
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] n=900[/tex] represent the sample size selected
[tex]p = 0.75[/tex] represent the population proportion
We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:
[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]
And the standard error is given;
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]
Please answer this correctly I want genius expert or ace people to answer this correctly as soon as possible as my work is due today
Answer:
25%
Step-by-step explanation:
The last percentile always contains 25% of the observations.
how many nickels equal $18.45? (show your work)
Answer:
369
Step-by-step explanation:
One nickel = 0.05
0.05x=18.45
x=369
Brainliest to whoever gets this correct Which of the following is equal to the rational expression when x ≠ -3? x^2-9/x+3
Answer:
see below
Step-by-step explanation:
We presume you want to simplify ...
[tex]\dfrac{x^2-9}{x+3}=\dfrac{(x-3)(x+3)}{x+3}=\boxed{x-3}[/tex]
__
The numerator is the difference of squares, so is factored accordingly. One of those factors cancels the denominator.
What is the slope of the line with the two
points A(-4, 8) and B(-9, 12)?
Answer:
slope = -4/5
Step-by-step explanation:
A line passes two points (x1, y1) and (x2, y2).
The slope of this line can be calculate by the formula:
s = (y2 - y1)/(x2 - x1)
=>The line that passes A(-4, 8) and B(-9, 12) has the slope:
s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5
Hope this helps!
A 2011 survey, by the Bureau of Labor Statistics, reported that 91% of Americans have paid leave. In January 2012, a random survey of 1000 workers showed that 89% had paid leave. The resulting p-value is .0271; thus, the null hypothesis is rejected. It is concluded that there has been a decrease in the proportion of people, who have paid leave from 2011 to January 2012. What type of error is possible in this situation?
Answer:
Is possible to make a Type I error, where we reject a true null hypothesis.
Step-by-step explanation:
We have a hypothesis test of a proportion. The claim is that the proportion of paid leave has significantly decrease from 2011 to january 2012. The P-value for this test is 0.0271 and the nunll hypothesis is rejected.
As the conclusion is to reject the null hypothesis, the only error that we may have comitted is rejecting a true null hypothesis.
The null hypothesis would have stated that there is no significant decrease in the proportion of paid leave.
This is a Type I error, where we reject a true null hypothesis.
The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.5 ppm and standard deviation 1.4 ppm. 18 randomly selected large cities are studied. Round all answers to two decimal places.
A. xBar~ N( ____) (____)
B. For the 18 cities, find the probability that the average amount of pollutants is more than 9 ppm.
C. What is the probability that one randomly selected city's waterway will have more than 9 ppm pollutants?
D. Find the IQR for the average of 18 cities.Q1 =
Q3 =
IQR:
2. X ~ N(30,10). Suppose that you form random samples with sample size 4 from this distribution. Let xBar be the random variable of averages. Let ΣX be the random variable of sums. Round all answers to two decimal places.
A. xBar~ N(___) (____)
B. P(xBar<30) =
C. Find the 95th percentile for the xBar distribution.
D. P(xBar > 36)=
E. Q3 for the xBar distribution =
Answer:
1)
A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552
C) P(X> 9)= 0.36317
D) IQR= 0.4422
2)
A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)
B) P( [tex]\frac{}{X}[/tex]<30)= 0.50
C) P₉₅= 32.60
D) P( [tex]\frac{}{X}[/tex]>36)= 0
E) Q₃: 31.0586
Step-by-step explanation:
Hello!
1)
The variable of interest is
X: pollutants found in waterways near a large city. (ppm)
This variable has a normal distribution:
X~N(μ;σ²)
μ= 8.5 ppm
σ= 1.4 ppm
A sample of 18 large cities were studied.
A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
The population mean is the same as the mean of the variable
μ= 8.5 ppm
The standard deviation is
σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108
So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B)
P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)
To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.
Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]
Then using the Z table you'll find the probability of
P(Z≤1.51)= 0.93448
Then
1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552
C)
In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:
P(X> 9)= 1 - P(X ≤ 9)
Z= (X-μ)/δ= (9-8.5)/1.44
Z= 0.347= 0.35
P(Z≤0.35)= 0.63683
Then
P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317
D)
The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:
Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25
Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:
P(Z≤z₁)= 0.25
Using the table you have to identify the value of Z that accumulates 0.25 of probability:
z₁= -0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)
z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)
[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5= 8.2789 ppm
The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
Using the table you have to identify the value of Z that accumulates 0.75 of probability:
z₃= 0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5= 8.7211 ppm
IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422
2)
A)
X ~ N(30,10)
For n=4
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
Population mean μ= 30
Population variance σ²/n= 10/4= 2.5
Population standard deviation σ/√n= √2.5= 1.58
[tex]\frac{}{X}[/tex] ~ N(30;2.5)
B)
P( [tex]\frac{}{X}[/tex]<30)
First you have to standardize the value and then look for the probability:
Z= ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0
P(Z<0)= 0.50
Then
P( [tex]\frac{}{X}[/tex]<30)= 0.50
Which is no surprise since 30 y the value of the mean of the distribution.
C)
P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95
P( Z≤ z₀)= 0.95
z₀= 1.645
Now you have to reverse the standardization:
z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)
z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)
[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60
P₉₅= 32.60
D)
P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0
E)
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
z₃= 0.67
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586
Q₃: 31.0586
Find the area of the smaller sector.
A
6 in
030°
Area = [? ]in?
B
Round your answer to the nearest hundredth.
Answer:
9.42 in²
Step-by-step explanation:
The area of whole circle S=pi*R² , where pi is appr. 3.14, R= 6 in
S= 3.14*6² =113.04 in²
The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²
The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.
To find the area of a sector, you can use the formula:
Area of sector = (θ/360) × π × r²
where θ is the central angle in degrees, r is the radius of the sector.
The central angle is 30 degrees and the radius is 6 inches.
Plugging these values into the formula:
Area of sector = (30/360) × π × 6²
= (1/12) × π × 36
= (1/12) × 3.14159 × 36
= 9.42478 square inches
To learn more on Area of sector click:
https://brainly.com/question/29055300
#SPJ2
Please answer this question I give brainliest thank you! Number 8
Answer:
The third options
Step-by-step explanation:
Counting we can see that 10 students went to two or less states, and 10 went to three or more
Which of the following is the solution to 9|x-1|=-45
Answer:
No solutions.
Step-by-step explanation:
9|x-1|=-45
Divide 9 into both sides.
|x-1| = -45/9
|x-1| = -5
Absolute value cannot be less than 0.
Answer:
No solution
Step-by-step explanation:
=> 9|x-1| = -45
Dividing both sides by 9
=> |x-1| = -5
Since, this is less than zero, hence the equation has no solutions
Number of multiples of 7 between 200 and 1000
Answer:
114
Step-by-step explanation:
Answer:
144Step-by-step explanation:
Please answer this correctly
Answer:
yes
Step-by-step explanation:
not every person is going to have the same opinion, so it is yes.
// have a great day //
Answer:
Yes, because if Pedro asked them the question "what do you think of public transportation?" the majority would probably say that they like it or something along those lines. This is biased because there may be other city inhabitants who don't think very highly of public transportation. Basically, what I'm trying to say is that not everyone will have the same opinion.
When $\frac{1}{1111}$ is expressed as a decimal, what is the sum of the first 40 digits after the decimal point?
Answer:
90
Step-by-step explanation:
1/1111= 0. (0009) cycles of 0009 after decimal point (one 9 per 4 digits)
Number of digits 9:
40/4= 1010*9= 90Answer:
90
Step-by-step explanation:
Terry has a number cube that is numbered from 1 to 6. She rolls the cube 50 times. Which equation can be used to predict the number of times that she will roll a number that is greater than 4? P (number greater than 4) = StartFraction 1 over 6 EndFraction (50) P (number greater than 4) = StartFraction 2 over 6 EndFraction (50) P (number greater than 4) = StartFraction 3 over 6 EndFraction (50) P (number greater than 4) = StartFraction 4 over 6 EndFraction (50)
Answer:
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
Find the Laplace transform F(s)=L{f(t)} of the function f(t)=sin2(wt), defined on the interval t≥0. F(s)=L{sin2(wt)}= help (formulas) Hint: Use a double-angle trigonometric identity. For what values of s does the Laplace transform exist? help (inequalities)
The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .
The Laplace transform exist when s > 0 .
Here, the given function is f(t) = sin²(wt) .
The Laplace transform of the the function f(t),
F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }
F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }
F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }
F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]
Next,
The above Laplace transform exist if s > 0 .
Know more about Laplace transform,
https://brainly.com/question/31481915
#SPJ4
3z/10 - 4 = -6
someone help?
Answer:
[tex]z=-\frac{20}{3}[/tex]
Step-by-step explanation:
[tex]\frac{3z}{10}-4=-6\\\\\frac{3z}{10}-4+4=-6+4\\\\\frac{3z}{10}=-2\\\\\frac{10\cdot \:3z}{10}=10\left(-2\right)\\\\3z=-20\\\\\frac{3z}{3}=\frac{-20}{3}\\\\z=-\frac{20}{3}[/tex]
Best Regards!
which of the following is equivalent to this?
a: b over a divided by d over c
b: a over b divided by d over c
c: b over a divided by d over c
d: b over a divided by c over d
please help me!
Answer:
b: a over b divided by do over c
Step-by-step explanation:
You can solve this by plugging in numbers for each variable.
For example: a=1, b=4, c=1, d=2
1/4 ÷ 1/2 = 0.125
If you plug in the numbers for all the equations listed, only 1/4 ÷ 2/1 = 0.125.
HELP ASAP!!!The first picture is what each variables equal too
Answer:
Just replace the variables with the number
d5
c4 (uh oh)
a2
b-3
f-7
d-c = 5 - 4 = 1
1/3 - 4(ab+f)
2 x -3 = -6
-6 + -7 = -13
-13 x 4 = -52
1/3 - -52 = 1/3 + 52 =
52 1/3
Hope this helps
Step-by-step explanation:
5/a - 4/b as a single fraction
Answer:
I'm not completely sure what you mean by a, "single fraction," but I'm pretty sure the answer you are looking for is [tex]\frac{5-4}{a-b}[/tex]
Step-by-step explanation:
Need help with this . The picture is enclosed
Answer: (fоg)(24)=5
Step-by-step explanation:
(fоg)(24) is f of g of 24. This means you plug in g(24) into f(x).
[tex]g(24)=\sqrt{24-8}[/tex]
[tex]g(24)=\sqrt{16}[/tex]
[tex]g(24)=4[/tex]
Now that we know g(24), we can plug it into f(x).
f(4)=2(4)-3
f(4)=8-3
f(4)=5
Pet Place sells pet food and supplies including a popular bailed hay for horses. When the stock of this hay drops to 20 bails, a replenishment order is placed. The store manager is concerned that sales are being lost due to stock outs while waiting for a replenishment order. It has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails. The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 bails
Answer:
The probability that demand during lead-time will exceed 20 bails is 0.2033.
Step-by-step explanation:
We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.
Let X = demand during the lead-time
So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu=[/tex] population mean demand = 15 bails
[tex]\sigma[/tex] = standard deviation = 6 bails
Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)
P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)
= 1 - 0.7967 = 0.2033
What is the result of −18⋅16 2/3? Enter the result as an improper fraction and as a mixed number.
Answer:
-30000/100
300 0/1
Step-by-step explanation:
We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.
16 2/3 = 16.67
We do the multiplication:
−18⋅16 2/3 = -300
We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:
-30000/100
How mixed number would it be:
300 0/1
ali's typing rate between 8:00 am and noon is 48 words per minute . after lunch a lunch break, Ali's typing rate between 1:00 pm and 4:00 pm is 2,040 words per hour . what is Ali's average typing rate per minute for the whole time she works?
Answer:
41 word/min
Step-by-step explanation:
Before noon Ali works:
4 hours= 4*60 min= 240 minShe types:
240*48= 11520 wordsAfter lunch she works:
4 hoursShe types:
4*2040= 8160 wordsTotal Ali works= 4+4= 8 hours= 480 min
Total Ali types= 11520+8160= 19680 words
Average typing rate= 19680 words/480 min= 41 word/min
Teaching descriptive statistics. A study compared five different methods for teaching descriptive statistics. The five methods were traditional lecture and discussion, programmed textbook instruction, programmed text with lectures, computer instruction, and computer instruction with lectures. 45 students were randomly assigned, 9 to each method. After completing the course, students took a 1-hour exam.
a. What are the hypotheses for evaluating if the average test scores are different for the different teaching methods?
b. What are the degrees of freedom associated with the F-test for evaluating these hypotheses?
c. Suppose the p-value for this test is 0.0168. What is the conclusion?
Answer:
Step-by-step explanation:
a. The hypotheses are:
Null hypothesis: the average test scores are the same for the different teaching methods.
Alternative hypothesis: the average test scores are different for the different teaching methods.
b. To determine the degree of freedom for the F test: we must find two sources of variation such that we have two variances. The two sources of variation are: Factor (between groups) and the error (within groups) and add this up. Or use (N - 1). N is number in sample
c. With a p value of of 0.0168 and using a standard significance level of 0.05, we will reject the null hypothesis as 0.0168 is less than 0.05 and conclude that the average test scores are different for the different teaching methods.