Answer:
I think this is the answer I hope I have helped you
Step-by-step explanation:
a
b
Your school is 20 blocks east and 12 blocks south of your house. The mall is 10 blocks north and 7 blocks west of your house. Yo
u plan on going to the mall right after school. Find the distance between your school and the mall assuming there is a road directly connecting the school and the mall. One block is 0.1 mile. Round your answer to the nearest tenth of a mile.
Your school is 20 blocks east and 12 blocks south of your house. The mall is 10 blocks north and 7 blocks west of your house. You plan on going to the mall right after school. Find the distance between your school and the mall assuming there is a road directly connecting the school and the mall. One block is 0.1 mile. Round your answer to the nearest tenth of a mile.
Your school is 20 blocks east and 12 blocks south of your house. The mall is 10 blocks north and 7 blocks west of your house. You plan on going to the mall right after school. Find the distance between your school and the mall assuming there is a road directly connecting the school and the mall. One block is 0.1 mile. Round your answer to the nearest tenth of a mile.
1 answer:
Answer:
3.5 miles
Step-by-step explanation:
The straight line distance from the school to the mall is found using the Pythagorean theorem. That distance is the hypotenuse of a triangle that is 27 blocks in the east-west direction and 22 blocks in the north-south direction.
d = √(27^2 +22^2) = √1213 ≈ 34.83 . . . blocks
Since each block is 0.1 miles, the distance in miles is ...
distance to mall = (0.1 miles/block)(34.83 blocks) = 3.483 miles
Rounded to tenths, the distance from school to mall is 3.5 miles.
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Answer: no, the confidence interval for the standard deviation σ cannot be expressed as 15.7 
4.7
There are three ways in which you can possibly express a confidence interval:
1) inequality
The two extremities of the interval will be each on one side of the "less then" symbol (the smallest on the left, the biggest on the right) and the symbol for the standard deviation (σ) will be in the middle:
11.0 < σ < 20.4
This is the first interval given in the question and it means <span>that the standard deviation can vary between 11.0 and 20.4
2) interval
</span>The two extremities will be inside a couple of round parenthesis, separated by a comma, always <span>the smallest on the left and the biggest on the right:
(11.0, 20.4)
This is the second interval given in the question.
3) point estimate </span><span> margin of error</span>
This is the most common way to write a confidence interval because it shows straightforwardly some important information.
However, this way can be used only for the confidence interval of the mean or of the popuation, not for he confidence interval of the variance or of the standard deviation.
This is due to the fact that in order to calculate the confidence interval of the standard variation (and similarly of the variance), you need to apply the formula:
which involves a χ² distribution, which is not a symmetric function. For this reason, the confidence interval we obtain is not symmetric around the point estimate and the third option cannot be used to express it.
There are three ways in which you can possibly express a confidence interval:
1) inequality
The two extremities of the interval will be each on one side of the "less then" symbol (the smallest on the left, the biggest on the right) and the symbol for the standard deviation (σ) will be in the middle:
11.0 < σ < 20.4
This is the first interval given in the question and it means <span>that the standard deviation can vary between 11.0 and 20.4
2) interval
</span>The two extremities will be inside a couple of round parenthesis, separated by a comma, always <span>the smallest on the left and the biggest on the right:
(11.0, 20.4)
This is the second interval given in the question.
3) point estimate </span><span>
This is the most common way to write a confidence interval because it shows straightforwardly some important information.
However, this way can be used only for the confidence interval of the mean or of the popuation, not for he confidence interval of the variance or of the standard deviation.
This is due to the fact that in order to calculate the confidence interval of the standard variation (and similarly of the variance), you need to apply the formula:
which involves a χ² distribution, which is not a symmetric function. For this reason, the confidence interval we obtain is not symmetric around the point estimate and the third option cannot be used to express it.