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Aleksandr [31]
2 years ago
8

Write an equation showing that the distance traveled on the first day plus the distance traveled on the second day is equal to 4

25 miles.
Mathematics
2 answers:
Alexandra [31]2 years ago
8 0
It's simple you did not asked to solve this problem, but the equation for this is x+y = 425  X is the distance traveled on the first day, and Y is the distance traveled on the second day
Radda [10]2 years ago
4 0
X+y=425. This is as simple it can be, unless theres more information. x is the distance traveled on the first day and y is the distance traveled on the second day. 


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The California State University (CSU) system consists of 23 campuses, from San Diego State in the south to Humboldt State near t
miv72 [106K]

Answer:

For this case since since they not want to differentiate between the campuses, so is good to use a simple random sampling or SRS, that is a procedure in order to select a sample of size n from a population of size N known, and each element of the population have the sample probability of being selected p=\frac{1}{N}

So then the administrator can select a random sample of n students from all the campuses from the CSU University and obtain the distance from hometwon to campuses for the selected sample and then calculate the average and use this to inference.

Other possibility is use a systematic random sampling for example they can define a random number k and they can select 1 student each k individuals and then create a random sample of size n and calculate the average for this in order to do inference.

The convenience sampling is not too useful since is not a probabilistic method.

Step-by-step explanation:

For this case we can clasify this study as an enumerative study and inferential since they want to identify the average distance between the hometowns of students and their campuses.

For this case the sampling frame represent all the 23 campuses, and is known for the researcher,

For this case since since they not want to differentiate between the campuses, so is good to use a simple random sampling or SRS, that is a procedure in order to select a sample of size n from a population of size N known, and each element of the population have the sample probability of being selected p=\frac{1}{N}

So then the administrator can select a random sample of n students from all the campuses from the CSU University and obtain the distance from hometwon to campuses for the selected sample and then calculate the average and use this to inference.

Other possibility is use a systematic random sampling for example they can define a random number k and they can select 1 student each k individuals and then create a random sample of size n and calculate the average for this in order to do inference.

The convenience sampling is not too useful since is not a probabilistic method.

6 0
2 years ago
this month drew worked six hours less than twice the number of hours H he worked last month. what expression represents the numb
Klio2033 [76]

Answer:

2x-6

Step-by-step explanation:

I think, I might be wrong

5 0
2 years ago
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Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 45
vovikov84 [41]

Answer:

a. The 95% confidence interval is 1,022.94559 < μ < 1,003.0544

b. There is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. i. The 95% confidence interval for the change in average test score is; -18.955390 < μ₁ - μ₂ < 6.955390

ii. There are no statistical significant evidence that the prep course helped

d. i. The 95% confidence interval for the change in average test scores is  3.47467 < μ₁ - μ₂ < 14.52533

ii. There is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

Step-by-step explanation:

The mean of the standardized test = 1,000

The number of students test to which the test is administered = 453 students

The mean score of the sample of students, \bar{x} = 1013

The standard deviation of the sample, s = 108

a. The 95% confidence interval is given as follows;

CI=\bar{x}\pm z\dfrac{s}{\sqrt{n}}

At 95% confidence level, z = 1.96, therefore, we have;

CI=1013\pm 1.96 \times \dfrac{108}{\sqrt{453}}

Therefore, we have;

1,022.94559 < μ < 1,003.0544

b. From the 95% confidence interval of the mean, there is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. The parameters of the students taking the test are;

The number of students, n = 503

The number of hours preparation the students are given, t = 3 hours

The average test score of the student, \bar{x} = 1019

The number of test scores of the student, s = 95

At 95% confidence level, z = 1.96, therefore, we have;

The confidence interval, C.I., for the difference in mean is given as follows;

C.I. = \left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm z_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore, we have;

C.I. = \left (1013- 1019  \right )\pm 1.96 \times \sqrt{\dfrac{108^{2}}{453}+\dfrac{95^{2}}{503}}

Which gives;

-18.955390 < μ₁ - μ₂ < 6.955390

ii. Given that one of the limit is negative while the other is positive, there are no statistical significant evidence that the prep course helped

d. The given parameters are;

The number of students taking the test = The original 453 students

The average change in the test scores, \bar{x}_{1}- \bar{x}_{2} = 9 points

The standard deviation of the change, Δs = 60 points

Therefore, we have;

C.I. = \bar{x}_{1}- \bar{x}_{2} + 1.96 × Δs/√n

∴ C.I. = 9 ± 1.96 × 60/√(453)

i. The 95% confidence interval, C.I. = 3.47467 < μ₁ - μ₂ < 14.52533

ii. Given that both values, the minimum and the maximum limit are positive, therefore, there is no zero (0) within the confidence interval of the difference in of the means of the results therefore, there is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

5 0
2 years ago
The image of ΔABC after a reflection across Line E G is ΔA'B'C'. 2 triangles are shown. A line of reflection is between the 2 tr
nirvana33 [79]

Answer:  

It's A - F is the midpoint of AA' because Line E G bisects AA'

Step-by-step explanation: :D

6 0
2 years ago
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What inverse operation is used to solve the equation 2x=10
Jlenok [28]
It’s c multiply by 2 on both sides
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