Answer: The alternate hypothesis would disprove the null hypothesis and state that there are a significant difference in preferences/proportions between the two villages.
For instance, let's say:
p₁ = proportion of preference from Dobbs Ferryp₂ = proportion of preference from IrvingtonThe null hypothesis would be that p₁ = p₂, while the alternative hypothesis would be that p₁ ≠ p₂.
Instructions: State what additional information is required in order
to know that the triangles in the image below are congruent for the
reason given
Reasory. SAS Postulate
Answer:
HJ = FG
Step-by-step explanation:
SAS means side - (included) angle - side.
we have one angle confirmed (at H and at G).
we have actually one side confirmed (HG), because the graphic shows that this side is shared between the triangles. so, implicitly it is not only congruent but really identical.
so, we need the confirmation of the second side enclosing the confirmed angle.
Mr. Rowley has 16 homework papers and 14 exit tickets to return. Ms. Rivera has 64 homework papers and 60 exit tickets to retum. For each teacher, write a ratio to represent the number of homework papers to number of exit tickets they have to return. Are the ratios equivalent? Explain.
Answer:
Mr. Rowley=16:14
=8:7
Ms.Rivera= 64:60
=16:15
Find the missing side lengths leave your answer as a racials simplest form
Answer:
y=3 and x=3*sqrt(3)
Step-by-step explanation:
sin(30)=y/6
1/2=y/6, y=3. As it's a right angled triangle, 6^2-3^2=x^2, x=3*sqrt(3)
Answer:
x = 3 sqrt(3)
y = 3
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
sin theta = opp / hyp
sin 30 = y /6
6 sin 30 = y
6 (1/2) = y
3 = y
cos theta = adj / hyp
cos 30 = x /6
6 cos 30 =x
6 ( sqrt(3)/2) =x
3 sqrt(3) = x
Many freeways have service (or logo) signs that give information on attractions, camping, lodging, food, and gas services prior to off-ramps. These signs typically do not provide information on distances. An article reported that in one investigation, six sites along interstate highways where service signs are posted were selected. For each site, crash data was obtained for a three-year period before distance information was added to the service signs and for a one-year period afterward. The number of crashes per year before and after the sign changes were as follows.
Before 13 22 65 123 56 63
After 14 21 43 84 75 72
1. The article included the statement "A paired t-test was performed to determine whether there was any change in the mean number of crashes before and after the addition of distance information on the signs." Carry out such a test. (Note: The relevant normal probability plot shows a substantial linear pattern.)
a. State and test the appropriate hypotheses. (Use α = 0.05.)
b. Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)
t = _____
p-value = _____
c. State the conclusion in the problem context.
A. Fail to reject H0. The data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.
B. Reject H0. The data suggests a significant mean difference in the average number of accidents after information was added to road signs.
C. Fail to reject H0. The data suggests a significant mean difference in the average number of accidents after information was added to road signs.
D. Reject H0. The data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.
2. If a seventh site were to be randomly selected among locations bearing service signs, between what values would you predict the difference in the number of crashes to lie? (Use a 95% prediction interval. Round your answers to two decimal places.)
Answer:
Test statistic = 0.63
Pvalue = 0.555
A. Fail to reject H0. The data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.
Step-by-step explanation:
Given :
Before 13 22 65 123 56 63
After_ 14 21 43 84 75 72
To perform a paired t test :
H0 : μd = 0
H1 : μd ≠ 0
We obtain the difference between the two dependent sample readings ;
Difference, d = -1, 1, 22, 39, -19, -9
The mean of difference, Xd = Σd/ n = 33/6 = 5.5
The standard deviation, Sd = 21.296 (calculator).
The test statistic :
T = Xd ÷ (Sd/√n) ; where n = 6
T = 5.5 ÷ (21.296/√6)
T = 5.5 ÷ 8.6940555
T = 0.6326
The Pvalue : Using a Pvalue calculator ;
df = n - 1 = 6 - 1 = 5
Pvalue(0.6326, 5) = 0.5548
Decision region :
Reject H0 ; If Pvalue < α; α = 0.05
Since 0.5548 > 0.05 ; we fail to reject the Null and conclude that the data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.
Find the probability that z lies between 0 and 1.56.
Answer:
P(0 < z < 1.56)=0.4406
Step-by-step explanation:
Write an equivalent fraction for each 1/4 = /10
Answer:
[tex]x = \frac{5}{2}[/tex]
Step-by-step explanation:
Step 1: Find an equivalent fraction
[tex]\frac{1}{4} = \frac{x}{10}[/tex]
[tex]4(x) = 1(10)[/tex]
[tex]4x = 10[/tex]
[tex]x = \frac{10}{4}[/tex]
[tex]x = \frac{5}{2}[/tex]
Answer: [tex]x = \frac{5}{2}[/tex]
f(x) = 1 + 9. Find f '(x) and its domain.
A. f-1 (z) = (x – 9)?: x2 9
B. f-1(x) = (x – 9)?; x20
c. f-1 (2) = x2 – 9;x2 9
D. f-1 (x) = x2 – 9; x2 0
Answer:
B
Step-by-step explanation:
f(x) = sqrt(x) + 9
f(x) - 9=sqrt(x)
f^(-1)(x)=(x-9)^2 and it's domain is greater than 0
What is the reference angle for 293°?
Tyra has recently inherited $5400, which she wants to deposit into an IRA account. She has determined that her two best bets are an account that compounds semi-
annually at an annual rate of 3.1 % (Account 1) and an account that compounds continuously at an annual rate of 4 % (Account 2).
Step 2 of 2: How much would Tyra's balance be from Account 2 over 3.7 years? Round to two decimal places.
The focus here is the use of "Compounding interest rate" and these entails addition of interest to the principal sum of the deposit.
Tyra will definitely prefer the Account 2 over the Account 1 Tyra balance from account 2 over 3.7 years is $6,261.37
The below calculation is to derive maturity value when annual rate of 3.1% is applied.
Principal = $5,400
Annual rate = 3.1% semi-annually for 1 years
A = P(1+r/m)^n*t where n=1, t=2
A = 5,400*(1 + 0.031/2)^1*2
A = 5,400*(1.0155)^2
A = 5,400*1.03124025
A = 5568.69735
A = $5,568.70.
In conclusion, the accrued value she will get after one years for this account is $5,568.70,
- The below calculation is to derive maturity value when the amount compounds continuously at an annual rate of 4%
Principal = $5,400
Annual rate = 4% continuously
A = P.e^rt where n=1
A = 5,400 * e^(0.04*1)
A = 5,400 * 1.04081077419
A = 5620.378180626
A = 5620.378180626
A = $5,620.39.
In conclusion, the accrued value she will get after one years for this account is $5,620.39.
Referring to how much would Tyra's balance be from Account 2 over 3.7 years. It is calculated as follows:
Annual rate = 4% continuously
A = P.e^rt where n=3.7
A = 5,400 * e^(0.04*3.7)
A = 5,400 * e^0.148
A = 5,400 * 1.15951289636
A = 6261.369640344
A = $6,261.37
Therefore, the accrued value she will get after 3.7 years for this account is $6,261.37
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A projectile is fired from a cliff feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of feet per second. The height h of the projectile above the water is given by
where x is the horizontal distance of the projectile from the face of the cliff. Use this information to answer the following.
(a) At what horizontal distance from the face of the cliff is the height of the projectile a maximum?
(Simplify your answer.)
(b) Find the maximum height of the projectile.
(Simplify your answer.)
(c) At what horizontal distance from the face of the cliff will the projectile strike the water?
(d) Using a graphing utility, graph the function h, Which of the following shows the graph of h(x)?
In all graphs, the window is by
A.
A coordinate system has a horizontal axis labeled from 0 to 230 in increments of 20 and a vertical axis labeled from 0 to 260 in increments of 50. From left to right, a curve starts at (0, 180), rises to a maximum at (74, 230), and then falls to (230, 10). All coordinates are approximate.
B.
A coordinate system has a horizontal axis labeled from 0 to 230 in increments of 20 and a vertical axis labeled from 0 to 260 in increments of 50. From left to right, a curve starts at (0, 210), rises to a maximum at (40, 230), and then falls to (176, 0). All coordinates are approximate.
C.
A coordinate system has a horizontal axis labeled from 0 to 230 in increments of 20 and a vertical axis labeled from 0 to 260 in increments of 50. From left to right, a curve starts at (0, 210), rises to a maximum at (56, 240), and then falls to (220, 0). All coordinates are approximate.
D.
A coordinate system has a horizontal axis labeled from 0 to 230 in increments of 20 and a vertical axis labeled from 0 to 260 in increments of 50. From left to right, a curve starts at (0, 240), rises to a maximum at (28, 245), and then falls to (194, 0). All coordinates are approximate.
(e) When the height of the projectile is 100 feet above the water, how far is it from the cliff?
Answer:
$170 Feet
Step-by-step explanation:
It is very long process
Which of the following words is generally used to describe what managers do as opposed to what leaders do b) Organize c) Inspire O d) Innovate
Answer:
Innovate
Step-by-step explanation:
Straight forward
Find the missing side lengths leave your answer as a racials simplest form
Answer:
x = 20
y = 10
it's a 30-60-90 triangle
When A = 200, solve the equation x2 - 40x + A=0 using the quadratic formula. Show all your working and give your answers correct to 2 decimal places.
Answer:
Solution given:
equation is:
x²-40x+A=0
when A=200
equation becomes
x²-40x+200=0
Comparing above equation with ax²+bx+c=0 we get
a=1
b=-40
c=200
By using quadratic equation formulax=[tex]\displaystyle \frac{-b±\sqrt{b²-4ac}}{2a}[/tex]
substituting value
x=[tex]\displaystyle \frac{-*-40±\sqrt{(-40)²-4*1*200}}{2*1}[/tex]
x=[tex]\displaystyle \frac{40±\sqrt{800}}{2}[/tex]
x=[tex]\displaystyle \frac{40±20\sqrt{2}}{2}[/tex]
taking positive
x=[tex]\displaystyle \frac{40+20\sqrt{2}}{2}[/tex]
x=34.14
taking negative
x=[tex]\displaystyle \frac{40-20\sqrt{2}}{2}[/tex]
x=5.86
x=34.14 or 5.86Researchers studied symptom distress and palliative care designation among a sample of 710 hospitalized patients. Controlling for age, they used a t-test to compare average distress from nausea scores in men and women. Lower scores indicated less distress from nausea. They report men had an average score of 1.02 and woman had an average score of 1.79. Which statement is correct?
(2pts)
Select Men had significantly less distress from nausea. as your answer
Men had significantly less distress from nausea.
Select Men had half as much distress from nausea as woman but we can not determine if this is a significant difference. as your answer
Men had half as much distress from nausea as woman but we can not determine if this is a significant difference.
Select Men had less distress from nausea on average than women but we can not determine if this is a significant difference. as your answer
Men had less distress from nausea on average than women but we can not determine if this is a significant difference.
Select There is a positive correlation between distress from nausea and gender. as your answer
There is a positive correlation between distress from nausea and gender.
Answer:
A. Men had less distress from nausea on average than women but we can not determine if this is a significant difference.
Step-by-step explanation:
Working based on the information given, the mean values of each group with with men having an average score of 1.02 and women have an average of 1.79 this reveals that distressing nausea on average is higher in women than in men . However, to test if there is a significant difference would be challenging as the information given isn't enough to make proceed with the test as the standard deviations of the two groups aren't given and no accompanying sample data is given.
Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean region).
Let A = the event that a country is in Asia.
Let E = the event that a country is in Europe.
Let F = the event that a country is in Africa.
Let N = the event that a country is in North America. Let O = the event that a country is in Oceania.
Let S = the event that a country is in South America.
18. What is the probability of drawing a red card in a standard deck of 52 cards?
19. What is the probability of drawing a club in a standard deck of 52 cards?
The probabilities we found in this exercise are.
0.2268 = 22.68% probability that a country is in Asia.0.2423 = 24.23% probability that a country is in Europe.0.2784 = 27.84% probability that a country is in Africa.0.1186 = 11.86% probability that a country is in North America.0.0722 = 7.22% probability that a country is in Oceania.0.0619 = 6.19% probability that a country is in South America.0.5 = 50% probability of drawing a red card in a standard deck of 52 cards.0.25 = 25% probability of drawing a club in a standard deck of 52 cards.In this exercise, probability concepts are used.
A probability is the number of desired outcomes divided by the number of total outcomes.
Total number of countries:
23 + 12 + 47 + 44 + 54 + 14 = 194
Let A = the event that a country is in Asia.
44 of the 194 countries are in Asia, thus:
[tex]P(A) = \frac{44}{194} = 0.2268[/tex]
0.2268 = 22.68% probability that a country is in Asia.
Let E = the event that a country is in Europe.
47 out of 194 countries are in Europe, thus:
[tex]P(E) = \frac{47}{194} = 0.2423[/tex]
0.2423 = 24.23% probability that a country is in Europe.
Let F = the event that a country is in Africa.
54 out of 194 countries are in Africa, thus:
[tex]P(F) = \frac{54}{194} = 0.2784[/tex]
0.2784 = 27.84% probability that a country is in Africa.
Let N = the event that a country is in North America.
23 out of 194 countries are in North America, thus:
[tex]P(N) = \frac{23}{194} = 0.1186[/tex]
0.1186 = 11.86% probability that a country is in North America.
Let O = the event that a country is in Oceania.
14 out of 194 countries are in Oceania, thus:
[tex]P(O) = \frac{14}{194} = 0.0722[/tex]
0.0722 = 7.22% probability that a country is in Oceania.
Let S = the event that a country is in South America.
12 out of 194 countries are in South America, thus:
[tex]P(S) = \frac{12}{194} = 0.0619[/tex]
0.0619 = 6.19% probability that a country is in South America.
18. What is the probability of drawing a red card in a standard deck of 52 cards?
In a standard deck of 52 cards, 26 are red, and thus:
[tex]p = \frac{26}{52} = 0.5[/tex]
0.5 = 50% probability of drawing a red card in a standard deck of 52 cards.
19. What is the probability of drawing a club in a standard deck of 52 cards?
In a standard deck of 52 cards, 13 are clubs, and thus:
[tex]p = \frac{13}{52} = 0.25[/tex]
0.25 = 25% probability of drawing a club in a standard deck of 52 cards.
For more about probabilities, you can check https://brainly.com/question/24104122
A local grocery store receives strawberries from suppliers in Florida and California. Currently there are 18 strawberry containers on the shelf and 11 of them are from Florida. A shopper selects three containers to purchase. What is the probability that exactly one of the containers is from the Florida supplier
Using the hypergeometric distribution, it is found that there is a 0.2831 = 28.31% probability that exactly one of the containers is from the Florida supplier.
The containers are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
x is the number of successes. N is the size of the population. n is the size of the sample. k is the total number of desired outcomes.In this problem:
There are 18 containers, hence [tex]N = 18[/tex]11 of those are in Florida, hence [tex]k = 11[/tex].A sample of 3 containers is taken, hence [tex]n = 3[/tex]The probability that exactly one of the containers is from the Florida supplier is P(X = 1), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 1) = h(1,18,3,11) = \frac{C_{11,1}C_{7,2}}{C_{18,3}} = 0.2831[/tex]
0.2831 = 28.31% probability that exactly one of the containers is from the Florida supplier.
A similar problem is given at https://brainly.com/question/24826394
Y=square root of x compare to y= - square root of x how they differ and why
Answer:
Simply because x=y2 doesn't imply that y=
√
x
.
A hose is left running for 240 minutes to 2 significant figures. The amount of water coming out of the hose each minute is 2.1 litres to 2 significant figures. Calculate the lower and upper bounds of the total amount of water that comes out of the hose.
Answer:
Hello,
Step-by-step explanation:
Let say t the time the hose is left running
235 ≤ t < 245 (in min)
Let say d the amount of water coming out of the hose each minute
2.05 ≤ d < 2.15 (why d : débit in french)
235*2.05 ≤ t*d < 245*2.15
481.75 ≤ t*d < 526.75 (litres)
Answer:
Lower bound: [tex]495\; \rm L[/tex] (inclusive.)
Upper bound: [tex]505\; \rm L[/tex] (exclusive.)
Step-by-step explanation:
The amount of water from the hose is the product of time and the rate at which water comes out.
When multiplying two numbers, the product would have as many significant figures as the less accurate factor.
In this example, both factors are accurate to two significant figures. Hence, the product would also be accurate to two significant figures. That is:
[tex]240 \times 2.1 = 5.0 \times 10^{2}\; \rm L[/tex] ([tex]500\; \rm L[/tex] with only two significant figures.)
Let [tex]x[/tex] denote the amount of water in liters. For [tex]x\![/tex] to round to [tex]5.0 \times 10^{2}\; \rm L[/tex] only two significant figures are kept, [tex]495 \le x < 505[/tex]. That gives a bound on the quantity of water from the hose.
Weekly wages at a certain factory are
normally distributed with a mean of
$400 and a standard deviation of $50.
Find the probability that a worker
selected at random makes between
$350 and $400.
Answer:
b or a
Step-by-step explanation:
Assuming that a person going to community college can't afford to go to a four-year college is an example of a) a generalization. b) discrimination. O c) a stereotype. O d) tolerance.
Answer:
a) generalization
Step-by-step explanation:
The statement is an example of a generalization. This is because the statement is assumming that all individuals who go to community college are poor. Therefore, this is why they cannot go to a four-year college, and instead go to a community college which is far cheaper. This assumption is being applied to all individuals who attend community college, without any further or more-specific information about each individual, therefore generalizing the entire situation.
A jet flew 2660 miles in 4.75 hours. What is the rate of speed in miles per hour? (The proportion would be 2660 : 4.75 ::X:1 Set the proportion in fractional form and proceed to find x.)
Answer:
X = 560
Step-by-step explanation:
Speed = distance / time
Distance = 2660 miles
Time taken = 4.75 hours
Writing the equation in terms of proportion :
Rate or speed = Distance : time
Speed = 2660 : 4.75
Reducing to lowest term ;
Divide both sides by 4.75
Hence, we have ;
Speed = 2660/4.75 : 4.75/4.75
Speed = 560 : 1
Comparing with the proportion in the question, X = 560
Suppose a certain study reported that 27.7% of high school students smoke.
Random samples are selected from high school that has 632 students.
(i) If a random sample of 60 students is selected, what is the probability that
fewer than 19 of the students smoke?
(ii) If a random sample of 75 students is selected, what is the probability that
more than 17 of the students smoke?
The correct answer of the question is "0.7062" and "0.835". The further solution is provided below.
Given:
Probability of student smoke,
P = 27.7%
= 0.277
Number of students (n) = 632
[tex]q = 1-p[/tex]
[tex]=1-0.277[/tex]
[tex]=0.723[/tex]
(i)
Here,
Number of students (n) = 60
then,
⇒ [tex]n_P=60\times 0.277[/tex]
[tex]=16.62[/tex]
⇒ [tex]n_q=60\times 0.723[/tex]
[tex]=43.38[/tex]
We can see that [tex]n_P > 10[/tex] and [tex]n_q>10[/tex] so the normal approximation condition are met.
Now,
[tex]\mu = n_P= 16.62[/tex]
[tex]\sigma = \sqrt{n_{Pq}}[/tex]
[tex]= \sqrt{60\times 0.277\times 0.723}[/tex]
[tex]=3.9664[/tex]
Now,
⇒ [tex]P(X<19) = P(X<18.5)[/tex]
[tex]=P(Z_{18.5})[/tex]
The Z-score is:
= [tex]\frac{18.5-16.62}{3.4664}[/tex]
= [tex]0.5423[/tex]
hence,
The probability will be:
⇒ [tex]P(Z_{18.5}) = 0.7062[/tex]
or,
⇒ [tex]P(Z<19) = 0.7062[/tex]
(ii)
Here,
Number of students (n) = 75
[tex]\mu = n_P = 75\times 0.277[/tex]
[tex]=20.775[/tex]
[tex]\sigma = \sqrt{n_{Pq}}[/tex]
[tex]=\sqrt{75\times 0.277\times 0.723}[/tex]
[tex]=3.8756[/tex]
Now,
⇒ [tex]P(X>17) = P(X> 17.5)[/tex]
[tex]=1-P(X \leq 17.5)[/tex]
[tex]=1-P(Z_{17.5})[/tex]
The Z-score is:
= [tex]\frac{17.5-20.775}{3.8756}[/tex]
= [tex]-0.9740[/tex]
then, [tex]P(Z_{17.5}) = 0.165[/tex]
hence,
The probability will be:
⇒ [tex]P(X>17) = 1-0.165[/tex]
[tex]=0.835[/tex]
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How are the functions y = x and y = x+ 5 related? How are their graphs related?
a. Each output for y = x + 5 is 5 less than the corresponding output for y = x.
The graph of y = x+ 5 is the graph of y = x translated down 5 units.
b. Each output for y = x+ 5 is 5 more than the corresponding output for y = x.
The graph of y = x+ 5 is the graph of y = x translated up 5 units.
Each output for y = x+5 is 5 more than the corresponding output for y = x.
The graph of y = x+5 is the graph of y = x translated down 5 units.
d. Each output for y = x + 5 is 5 less than the corresponding output for y=x.
Answer:
b
Step-by-step explanation:
the graph gets translated 5 units above its parent graph of y = x
what is the least common factor for 9 8 7
Answer:504
This is the answer
504
Find the degree of each polynomial and indicate whether the
polynomial is a monomial, binomial, trinomial, or none of these.
Answer:
1. Degree = 1, monomial
2. Degree = 2, monomial
3. degree = 2, trinomial
4. Degree = 2, binomial
5. Degree = 2, binomial
Step-by-step explanation:
A psychologist conducted a survey of the attitude towards the sustainability of American energy consumption with 250 randomly selected individuals several years ago. The psychologist believes that these attitudes have changed over time. To test this he randomly selects 250 individuals and asks them the same questions. Can the psychologist confirm his theory that the attitudes have changed from the first survey to the second survey?
Attitude 1st Survey 2nd Survey
Optimistic 7% 6%
Slightly Optimistic 9% 6%
Slightly Pessimistic 31% 37%
Pessimistic 53% 51%
Step 4 of 10: Find the expected value for the number of respondents who are optimistic. Round your answer to two decimal places.
Answer:
Yes. the Psychologist can confirm his theory that the attitudes have changed over time, based on the first and second surveys.
The expected value for the number of respondents who are optimistic is:
= 16.25
Step-by-step explanation:
Attitude 1st Survey 2nd Survey
Optimistic 7% 6%
Slightly Optimistic 9% 6%
Slightly Pessimistic 31% 37%
Pessimistic 53% 51%
Expected value of optimistic respondents:
Attitude
Optimistic Expected Value
1st Survey 8.75 (250 * 7% * 50%)
2nd Survey 7.50 (250 * 6% * 50%)
Total EV 16.25
During a test period, an experimental group of 10 vehicles using an 85 percent ethanol-gasoline mixture showed mean CO2 emissions of 667 pounds per 1000 miles, with a standard deviation of 20 pounds. A control group of 14 vehicles using regular gasoline showed mean CO2 emissions of 679 pounds per 1000 miles with a standard deviation of 15 pounds. At α = 0.05, in a left-tailed test (assuming equal variances) the test statistic is:______.
A. 1.321.
B. -2.508.
C. -2.074.
D. -1.717.
Answer:
-1.683
Step-by-step explanation:
Given :
Group 1 :
x1 = 667 ; n1 = 10 ; s1 = 20
Group 2 :
x2 = 679 ; n2 = 14 ; s2 = 15
The test statistic assuming equal variance :
x1 - x2 / √[Sp² * (1/n1 + 1/n2)]
sp² = [(n1 - 1)*s1² + (n2 - 1)*s2²] ÷ (n1 + n2 - 2)
Sp² = [(10 - 1)*20² + (14 - 1)*15²] = 296.59
Test statistic =
(667 - 679)/ √[296.59 * (1/10 + 1/14)]
-12 / 7.1304978
Test statistic = - 1.682
HELP PLEASE HELP HELP
Answer:
The first one is the only one that makes sense.
Step-by-step explanation:
hope it helps!
Determine whether the following polygons are similar. If yes, type 'yes' in the Similar box and type in the similarity statement and scale factor. If no, type 'None' in the blanks. For the scale factor, please enter a fraction. Use the forward dash (i.e. /) to create a fraction (e.g. 1/2 is the same as 12
1
2
).
Given:
The figures of two polygons.
To find:
Whether the polygons are similar and then find the scale factor (if similar).
Solution:
From the given figures it is clear that both polygons are rectangles and their all interior angles are right angles.
The ratio of their longer sides:
[tex]\dfrac{32}{26}=\dfrac{16}{13}[/tex]
The ratio of their shorter sides:
[tex]\dfrac{18}{12}=\dfrac{3}{2}[/tex]
Since the ratio of their corresponding sides are not equal, therefore the two polygons are not similar.
Therefore the required solutions are:
Similar : No
Similarity statement : None
Scale factor : None
Roulette is a casino game that involves spinning a ball on a wheel that is marked numbered squares that are red, black, or green. Half of the numbers 1 - 36 are colored red and half are black and the numbers 0 and 00 are green. Each number occurs only once on the wheel. What is the probability of landing on a green space
Answer:
1/19
Step-by-step explanation:
There are a total of 36+2 = 38 spaces
2 are green
P(green) = green / total
= 2/38
=1/19
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