Answer:
Top right option
Step-by-step explanation:
Hallie can use the equation p = 4l + 4w + 4h to determine the sum of the lengths of the edges of a rectangular prism. She begins to solve the equation for h but runs out of time. Her partial work is shown below:
p = 4l + 4w + 4h
= l + w + h
h = –
Which expression should follow the subtraction in Hallie’s equation?
Answer:
h = p - l - w
Step-by-step explanation:
p = 4l + 4w + 4h Divide l, w, and h by 4
p = l + w + h Set the equation equal to h
h = p - l - w
Answer:
A just did it on edge<3
g A psychic was tested for extrasensory perception (ESP). The psychic was presented with cards face down and asked to determine if each of the cards was one of four symbols: a star, cross, circle, or square. Let p represent the probability that the psychic correctly identified the symbols on the cards in a random trial. How large a sample n would you need to estimate p with margin of error 0.01 and 95% confidence?
Answer:
Step-by-step explanation:
Hello!
The objective is to test ESP, for this, a psychic was presented with cards face down and asked to determine if each of the cards was one of four symbols: a star, cross, circle, square.
Be X: number of times the psychic identifies the symbols on the cards correctly is a size n sample.
p the probability that the psychic identified the symbol on the cards correctly
You have to calculate the sample size n to estimate the proportion with a confidence level of 95% and a margin of error of d=0.01
The CI for the population proportion is constructed "sample proportion" ± "margin of error" Symbolically:
p' ± [tex]Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )[/tex]
Where [tex]d= Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )[/tex] is the margin of error. As you can see, the formula contains the sample proportion (it is normally symbolized p-hat, in this explanation I'll continue to symbolize it p'), you have to do the following consideration:
Every time the psychic has to identify a card he can make two choices:
"Success" he identifies the card correctly
"Failure" he does not identify the card correctly
If we assume that each symbol has the same probability of being chosen at random P(star)=P(cross)=P(circle)=P(square)= 1/4= 0.25
Let's say, for example, that the card has the star symbol.
The probability of identifying it correctly will be P(success)= P(star)= 1/4= 0.25
And the probability of not identifying it correctly will be P(failure)= P(cross) + P(circle) + P(square)= 1/4 + 1/4 + 1/4= 3/4= 0.75
So for this experiment, we'll assume the "worst case scenario" and use p'= 1/4 as the estimated probability of the psychic identifying the symbol on the card correctly.
The value of Z will be [tex]Z_{1-\alpha /2}= Z_{0.975}= 1.96[/tex]
Now using the formula you have to clear the sample size:
[tex]d= Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )[/tex]
[tex]\frac{d}{Z_{1-\alpha /2}} = \sqrt{\frac{p'(1-p')}{n} }[/tex]
[tex](\frac{d}{Z_{1-\alpha /2}})^2 =\frac{p'(1-p')}{n}[/tex]
[tex]n*(\frac{d}{Z_{1-\alpha /2}})^2 = p'(1-p')[/tex]
[tex]n = p'(1-p')*(\frac{Z_{1-\alpha /2}}{d})^2[/tex]
[tex]n = (0.25*0.75)*(\frac{1.96}{0.01})^2= 7203[/tex]
To estimate p with a margin of error of 0.01 and a 95% confidence level you have to take a sample of 7203 cards.
I hope this helps!
Answer:
The sample size should be 6157
Step-by-step explanation:
Given that the margin of error (e) = ± 0.01 and the confidence (C) = 95% = 0.95.
Let us assume that the guess p = 0.25 as the value of p.
α = 1 - C = 1 - 0.95 = 0.05
[tex]\frac{\alpha }{2} =\frac{0.05}{2}=0.025[/tex]
The Z score of α/2 is the same as the z score of 0.475 (0.5 - 0.025) which is 1.96. Therefore [tex]Z_\frac{\alpha }{2}=Z_{0.025}=1.96[/tex]
To determine the sample size n, we use the formula:
[tex]Z_{0.025}*\sqrt{\frac{p(1-p)}{n} }\leq e\\Substituting:\\1.96*\sqrt{\frac{0.2(1-0.2)}{n} } \leq 0.01\\\sqrt{\frac{0.2(0.8)}{n} }\leq \frac{1}{196}\\\sqrt{0.16} *196 \leq \sqrt{n}\\78.4\leq \sqrt{n}\\ 6146.56\leq n\\n=6157[/tex]
A small college has 1460 students. What is the approximate probability that more than six students were born on Christmas day? Assume that birthrates are constant throughout the year and that each year has 365 days.
Answer:
The approximate probability that more than six students were born on Christmas day is P=0.105.
Step-by-step explanation:
This can be modeled as a binomial variable, with n=1460 and p=1/365.
The sample size n is the total amount of students and the probability of success p is the probability of each individual of being born on Christmas day.
As the sample size is too large to compute it as a binomial random variable, we approximate it to the normal distribution with the following parameters:
[tex]\mu=n\cdot p=1460\cdot (1/365)=4\\\\\sigma=\sqrt{n\cdot p(1-p)}=\sqrt{1460\cdot(1/365)\cdot(364/365)}=\sqrt{3.989}=1.997[/tex]
We want to calculate the probability that more than 6 students were born on Christmas day. Ww apply the continuity factor and we write the probability as:
[tex]P(X>6.5)[/tex]
We calculate the z-score for X=6.5 and then calculate the probability:
[tex]z=\dfrac{X-\mu}{\sigma}=\dfrac{6.5-4}{1.997}=\dfrac{2.5}{1.997}=1.252\\\\\\P(X>6.5)=P(z>1.252)=0.105[/tex]
Evaluate 1/2 + 1/2 ÷ 18
Answer:
1/18
Step-by-step explanation:
First you would add 1/2 and 1/2 to get 1 then you would divide it by 18 to get 1/18
Answer:
1/18
Step-by-step explanation:
plz mark me brainliest.
The Sunshine Droogs are unhappy as they have not yet been paid for their concert. It was agreed they would be paid eleven thousand, four hundred and fifty three pounds for the concert. What is this amount in numbers?
Answer:
11453
Step-by-step explanation:
A fast food hamburger restaurant uses 3,500 lbs. of hamburger each week. The manager of the restaurant wants to ensure that the meat is always fresh i.e. the meat should be no more than two days old on average when used. How much hamburger should be kept in the refrigerator as inventory
Answer:
The peak inventory will be 2 sales days of hamburguers, which is equivalent to 7,000 lbs. As they are consumed in 2 days, the average inventory is 3,500 lbs.
Step-by-step explanation:
If the meat should be no more than two days old on average when used, the stock of hamburguer in the refrigerator has to be at most the equivalent to 2 day of sales.
The "2 days old" represents the inventory turnover.
If we use all the hamburguers in the refrigerator and refill inmediatly, the average inventory is:
[tex]\bar I=\dfrac{\text{Beginning inventory}+\text{Ending inventory}}{2}\\\\\\\bar I=\dfrac{2*3,500+0}{2}=3,500[/tex]
The peak inventory will be 2 sales days of hamburguers, which is equivalent to 7,000 lbs. As they are consumed in 2 days, the average inventory is 3,500 lbs.
Using the data in the table, use the exponential smoothing method with alpha=0.5 and a February forecast of 500 to forecast
sales for May
Month Demand
January 480
February 520
March 535
April 550
May 590
June 630
Answer:
Step-by-step explanation:
The formula to calculate the forecast could be determine by using the exponential smoothing method :
[tex]Ft = F(t-1) + \alpha [A(t-1) - F(t-1)][/tex]
Where ,Ft is the Forecast for period t
F(t-1) is the Forecast for the period previous to t
A(t-1) is the Actual demand for the period previous to t
[tex]\alpha[/tex] = Smoothing constant
To get the forecast for may and june the above formula with [tex]\alpha =0.5[/tex] and april forecast of 500 will be used
For march
[tex]=500+0.5(520-500)\\\\=500+0.5\times20\\\\=500+10\\\\=510[/tex]
For April
[tex]=510+0.5(535-510)\\\\=510+0.5\times25\\\\=510+12.5\\\\=522.5[/tex]
For May
[tex]=522.5+0.5(550-5225)\\\\=522.5+0.5\times27.5\\\\=522.5+13.75\\\\=536.25[/tex]
So forecast for May = 536.25
Researchers recorded that a certain bacteria population declined from 750,000 to 250 in 48 hours after the administration of medication. At this rate of decay, how many bacteria will there be in 8 hours?
Answer:
There will be 66 bacteria in 8 hours.
Step-by-step explanation:
The number of bacteria after t hours is given by the following formula.
[tex]P(t) = P(0)(1-r)^{t}[/tex]
In which P(0) is the initual number of bacteria and r is the decay rate.
Researchers recorded that a certain bacteria population declined from 750,000 to 250 in 48 hours after the administration of medication.
This means that [tex]P(0) = 750000, P(48) = 250[/tex]
We use this to find r. So
[tex]P(t) = P(0)(1-r)^{t}[/tex]
[tex]250 = 750000(1-r)^{48}[/tex]
[tex](1-r)^{48} = \frac{250}{750000}[/tex]
[tex]\sqrt[48]{(1-r)^{48}} = \sqrt[48]{\frac{250}{750000}}[/tex]
[tex]1-r = 0.84637[/tex]
So
[tex]P(t) = 750000(0.84637)^{t}[/tex]
How many bacteria will there be in 8 hours?
8 hours from now, in this context, is 8 + 48 = 56 hours. So this is P(56).
[tex]P(56) = 750000(0.84637)^{56} = 65.83[/tex]
Rounding to the nearest number
There will be 66 bacteria in 8 hours.
Answer:
197,488
Step-by-step explanation:
This problem requires two main steps. First, we must find the unknown rate, k. Then, we use that value of k to help us find the unknown number of bacteria.
Identify the variables in the formula.
AA0ktA=250=750,000=?=48hours=A0ekt
Substitute the values in the formula.
250=750,000ek⋅48
Solve for k. Divide each side by 750,000.
13,000=e48k
Take the natural log of each side.
ln13,000=lne48k
Use the power property.
ln13,000=48klne
Simplify.
ln13,000=48k
Divide each side by 48.
ln13,00048=k
Approximate the answer.
k≈−0.167
We use this rate of growth to predict the number of bacteria there will be in 8 hours.
AA0ktA=?=750,000=ln13,00048=8hours=A0ekt
Substitute in the values.
A=750,000eln13,00048⋅8
Evaluate.
A≈197,488.16
At this rate of decay, researchers can expect 197,488 bacteria.
"The chance that a person selected at random has blue eyes is 16%. Two people are chosen at random (and are independent of each other). Find the probability at least one of them does not have blue eyes. Round your answer to 4 decimal places."
Answer:
[tex]P(X=0)=(2C0)(0.84)^0 (1-0.84)^{2-0}=0.0256[/tex]
And replacing we got:
[tex] P(X \geq 1)=1 -P(X<1) = 1-P(X=0)=1-0.0256=0.9744[/tex]
Step-by-step explanation:
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=2, p=1-0.16=0.84)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And we can find this probability:
[tex] P(X \geq 1)[/tex]
And we can solve this probability like this:
[tex] P(X \geq 1)=1 -P(X<1) = 1-P(X=0)[/tex]
And if we use the probability mass function we got:
[tex]P(X=0)=(2C0)(0.84)^0 (1-0.84)^{2-0}=0.0256[/tex]
And replacing we got:
[tex] P(X \geq 1)=1 -P(X<1) = 1-P(X=0)=1-0.0256=0.9744[/tex]
Write an equation of a line that passes through (-6, 1), parallel to y = 2x – 6.
Answer:
y = -1/2x - 2
Step-by-step explanation:
If it's parallel, that means that the slope is the opposite of the one in the given equation, meaning that 2 would be flipped and turned negative into -1/2.
Then, fill in the x and y values to get the y-intercept.
1 = -1/2(-6) + b
1 = 3 + b
-2 = b
So your answer is y = -1/2x - 2
is 7.68 bigger than 7.680
Answer:
literally 7.68=7.680
Please answer this correctly
Answer:
[tex]h=\sqrt{1.44}\\h = 1.2[/tex]
Step-by-step explanation:
Base of the triangle on the left = 0.5
Use pythagorean theorem
[tex]a^{2} + b^{2} = c^{2}[/tex]
Substitute
[tex]0.5^{2} + b^{2} = 1.3^{2}[/tex]
[tex]b^{2} = 1.3^2 - 0.5^2[/tex]
[tex]b^2 = 1.44[/tex]
[tex]b = \sqrt{1.44} \\[/tex]
[tex]b = 1.2[/tex]
in this case b is the height
so
[tex]h=\sqrt{1.44}\\h = 1.2[/tex]
Factories fully 4ab + 8ac
Answer:
Hello!
I believe your answer is:
4a(b+2c)
Step-by-step explanation:
I hope this worked for you! Good luck!
In a particular region, for families with a combined income of $75,000 or more, 15% of these families have no children, 35% of the families have one child, 45% have two children, and 5% have three children. Use this information to construct the probability distribution for X, where x represents the number of children per family for this income group. Arrange x in increasing order and write the probabilities P(x) as decimals
Answer:
The probability distribution for x:"number of children per family for this income group" is:
[tex]\text{P(x=0)}=0.15\\\\\text{P(x=1)}=0.35\\\\\text{P(x=2)}=0.45\\\\\text{P(x=3)}=0.05\\\\[/tex]
Step-by-step explanation:
With the information given we have the relative frequencies of each category.
We know:
[tex]\text{P(x=0)}=0.15\\\\\text{P(x=1)}=0.35\\\\\text{P(x=2)}=0.45\\\\\text{P(x=3)}=0.05\\\\[/tex]
Please answer this correctly
Answer:
Number of people
6
5
5
6
3
1
Step-by-step explanation:
All you had to do was the count how much numbers there were on the list.
Like there were 6 0s.
Answer:
Hope this helps
Step-by-step explanation:
6 people did 0 sit ups
5 people did 1 sit ups
5 People did 2 sit ups
6 people did 3 sit ups
3 people did 4 sit ups
1 person did 5 sit ups
On average, a major earthquake (Richter scale 6.0 or above) occurs three times a decade in a certain California county. Find the probability that at least one major earthquake will occur within the next decade. A. .1992 B. .7408 C. .9502 D. .1494
What’s the correct answer for this question?
Answer:
C.
Step-by-step explanation:
Volume of cylinder = πr²h
= (3.14)(4)(0.75)
= 9.4
Since she'll fill it half so
Amount of water to be filled = 4.7
FIND P(NOT 6) WHEN YOU ROLL A STANDARD NUMBER CUBE THEN DESCRIBE THE LIKELIHOOD OF THE EVENT WRITE IMPOSSIBLE ,UNLIKELY , EQUALLY LIKELY , LIKLEY OR CERAIN
Answer: LIKLEY
Step-by-step explanation:
Formula : Probability [tex]=\dfrac{\text{Number of favorable outcomes}}{\text{Total outcomes}}[/tex]
A standard cube has six numbers on it (1,2,3,4,5 and 6).
P( NOT 6) =[tex]\dfrac{\text{Numbers that are not 6}}{\text{Total numbers}}[/tex]
[tex]=\dfrac{5}{6}=0.8333[/tex]
We know that when the probability of any event lies between 0.5 and 1then the event is said to be likely to happen.
Since , P(not 6)=0.8333 which lies between 0 and 0.5.
That means, it is likely to happen.
Note :
When probability of having A = 0 , we call A as uncertain event.
When probability of having A = 1 , we call A as certain event.
When probability of having A = 0.5 , we call A as equally likely event.
When probability of having A lies between 0 and 0.5 , we call A as unlikely event.
When probability of having A lies between 0.5 and 1 , we call A as likely event.
Find the amount to which $2,500 will grow if interest of 6.75% is compounded quarterly for 10
years.
Find the amount to which $2,500 will grow if interest of 6.75% is compounded daily for 10
years.
Answer:
Part a
For this case n = 4. If we use the future value formula we got:
[tex] A= 2500 (1+ \frac{0.0675}{4})^{4*10}= 4882.506[/tex]
Part b
For this case n = 365. If we use the future value formula we got:
[tex] A= 2500 (1+ \frac{0.0675}{365})^{365*10}= 4909.776[/tex]
Step-by-step explanation:
We can use the future vaue formula for compound interest given by:
[tex] A= P(1+ \frac{r}{n})^{nt}[/tex]
Where P represent the present value, r=0.0675 , n is the number of times that the interest is compounded in a year and t the number of years.
Part a
For this case n = 4. If we use the future value formula we got:
[tex] A= 2500 (1+ \frac{0.0675}{4})^{4*10}= 4882.506[/tex]
Part b
For this case n = 365. If we use the future value formula we got:
[tex] A= 2500 (1+ \frac{0.0675}{365})^{365*10}= 4909.776[/tex]
An appliance repairman charges $25 plus $40 per hour for house calls. Write the rule as an equation that relates hours worked x and his fee y.
To get the total fee, you need to multiply the hourly rate by number of hours worked and add that to the flat fee of $25.
The equation would be y = 40x + 25
For a particular diamond mine, 78% of the diamonds fail to qualify as "gemstone grade". A random sample of 106 diamonds is analyzed. Find the mean μ.
Answer:
Mean of the binomial distribution μ = 82.68
Step-by-step explanation:
Explanation:-
Given sample size 'n' = 106 diamonds
The probability that the diamonds fail to qualify as "gemstone grade
p = 78% =0.78
We will use binomial distribution
Mean of the binomial distribution
μ = n p
μ = 106 × 0.78
μ = 82.68
conclusion:-
Mean of the binomial distribution μ = 82.68
Q5. Calculate the median value of this data set. 24 -8 -17 32 -1 -28
Answer:
The median value in this set is -4.5
Step-by-step explanation:
Reorder the numbers from least to greatest
-28,-17,-8,-1,24,32
Then, since there is 6 digits in this data set there is no defined median value. In the numbers 1 to 8 there are 8 different numbers, the middle of 1 to 8 is 4.5. Then since were using the numbers -8,-1 the middle is -4.5
Which set of ordered pairs does NOT represent a function ?
Answer:
The answer is C.
Step-by-step explanation:
For a function, we do a vertical line test. If there is more than one point in one single x-position, it is not a function. Example, the ordered pairs (1, 1) and (1, 2) do NOT describe a function because there are more than one point on x=1.
A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought in to the time repairs are completed. A random sample of 12 repair records showed the following repair times (in days): 5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.
H0: \mu \leq 5 days versus H1: \mu > 5 days. At \alpha = .05, choose the right option.
a) Reject H0 if tcalc < 1.7960
b) Reject H0 if tcalc >1.7960
Answer:
The degrees of freedom first given by:
[tex]df=n-1=12-1=11[/tex]
Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:
[tex] t_{\alpha}= 1.796[/tex]
And for this case the rejection region would be:
b) Reject H0 if tcalc >1.7960
Step-by-step explanation:
Information given
5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.
System of hypothesis
We want to test if the true mean is higher than 5, the system of hypothesis are :
Null hypothesis:[tex]\mu \leq 5[/tex]
Alternative hypothesis:[tex]\mu > 5[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
The degrees of freedom first given by:
[tex]df=n-1=12-1=11[/tex]
Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:
[tex] t_{\alpha}= 1.796[/tex]
And for this case the rejection region would be:
b) Reject H0 if tcalc >1.7960
A city has just added 100 new female recruits to its police force. The city will provide a pension to each new hire who remains with the force until retirement. In addition, if the new hire is married at the time of her retirement, a second pension will be provided for her husband. A consulting actuary makes the following assumptions: (i) Each new recruit has a 0.4 probability of remaining with the police force until retirement. (ii) Given that a new recruit reaches retirement with the police force, the probability that she is not married at the time of retirement is 0.25. (iii) The events of different new hires reaching retirement and the events of different new hires being married at retirement are all mutually independent events. Calculate the probability that the city will provide at most 90 pensions to the 100 new hires and their husbands. (A) 0.60 (B) 0.67 (C) 0.75 (D) 0.93 (E) 0.99
Answer:
E) 0.99
Step-by-step explanation:
100 recruits x 0.4 chance of retiring as police officer = 40 officers
probability of being married at time of retirement = (1 - 0.25) x 40 = 30 officers
each new recruit will result in either 0, 1 or 2 new pensions
0 pensions when the recruit leaves the police force (0.6 prob.)1 pension when the recruit stays until retirement but doesn't marry (0.1 prob.)2 pensions when the recruit stays until retirement and marries (0.3 prob.)mean = µ = E(Xi) = (0 x 0.6) + (1 x 0.1) + (2 x 0.3) = 0.7
σ² = (0² x 0.6) + (1² x 0.1) + (2² x 0.3) - µ² = 0 + 0.1 + 1.2 - 0.49 = 0.81
in order for the total number of pensions (X) that the city has to provide:
the normal distribution of the pension funds = 100 new recruits x 0.7 = 70 pension funds
the standard deviation = σ = √100 x √σ² = √100 x √0.81 = 10 x 0.9 = 9
P(X ≤ 90) = P [(X - 70)/9] ≤ [(90 - 70)/9] = P [(X - 70)/9] ≤ 2.22
z value for 2.22 = 0.9868 ≈ 0.99
1/216^-2/3 + 1/256^-3/4 + 1/243^-1/5
Answer:
103
Step-by-step explanation:
[tex]\dfrac{1}{216}^{-2/3}+\dfrac{1}{256}^{-3/4}+\dfrac{1}{243}^{-1/5}= \\\\\\\sqrt[3]{216^2}+\sqrt[4]{256^3}+\sqrt[5]{243}=\\\\\\6^2+4^3+3=\\\\\\36+64+3=\\\\\\103[/tex]
Hope this helps!
the time taken by a student to the university has been shown to be normally distributed with mean of 16 minutes and standard deviation of 2.1 minutes. He walks in once a day during term time, 180 days per year, and leaves home 20 minutes before his first lecture. a. Find the probability that he is late for his first lecture. b. Find the number of days per year he is likely to be late for his first lecture.
Answer:
a) 2.84% probability that he is late for his first lecture.
b) 5.112 days
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 16, \sigma = 2.1[/tex]
a. Find the probability that he is late for his first lecture.
This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20 - 16}{2.1}[/tex]
[tex]Z = 1.905[/tex]
[tex]Z = 1.905[/tex] has a pvalue of 0.9716
1 - 0.9716 = 0.0284
2.84% probability that he is late for his first lecture.
b. Find the number of days per year he is likely to be late for his first lecture.
Each day, 2.84% probability that he is late for his first lecture.
Out of 180
0.0284*180 = 5.112 days
5.44 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?
Answer:
The null hypothesis is that all the different teaching methods have the same average test scores.
H0: μ1 = μ2 = μ3 = μ4 = μ5
The alternative hypothesis is that at least one of the teaching methods have a different mean.
Ha: at least one mean is different. (μ1 ≠ μi)
Step-by-step explanation:
The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.
For the case above, let μ represent the average test scores for the teaching methods:
The null hypothesis is that all the different teaching methods have the same average test scores.
H0: μ1 = μ2 = μ3 = μ4 = μ5
The alternative hypothesis is that at least one of the teaching methods have a different mean.
Ha: at least one mean is different. (μ1 ≠ μi)
What’s the correct answer for this question?
Answer:
107 meters
Step-by-step explanation:
Central angle = 123°
In radians
123° = 123π/180
123° = 2.147 radians
Putting in formula
S = r∅
S = (50)(2.147)
S = 107 meters
Liquid suspension contains 125 MG of medication aide for every 300 ML solution this is Spenton is being infused into a patient at the rate of 100 ML per hour if the infusion started at 6 AM and the patient needs 500 MG of the medication a at what time will you need to stop the infusion
Answer:
6 PM
Step-by-step explanation:
125 mg --- 300 mL
500 mg --- x mL
x = 500*300/125 = 1200 mL solution contains 500 mg
rate = 100 mL/h
1200 mL* 1h/100 mL = 12 h
6AM + 12 h = 6 PM
You need to stop infusion at 6 PM
It is found that You need to stop infusion at 6 PM.
What is the unitary method?The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.
Given that Liquid suspension contains 125 MG of medication aide for every 300 ML solution this is Spenton is being infused into a patient at the rate of 100 ML per hour if the infusion started at 6 AM and the patient needs 500 MG of the medication.
125 mg = 300 mL
500 mg = x mL
x = 500*300/125
x = 1200 mL
Here solution contains 500 mg
The rate = 100 mL/h
1200 mL* 1h/100 mL = 12 h
6AM + 12 h = 6 PM
Therefore, You need to stop infusion at 6 PM.
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