Three students want to estimate the mean backpack weight of their schoolmates. To do this, they each randomly chose 8 schoolmates and weighed their backpacks. Then as per the given sample data,
(a) The sample means of the backpacks are: 6.375,6.375,6.625
(b) Range of sample means: 0.25
(c)The true statement is: The closer the range of the sample means is to 0, the less confident they can be in their estimate.
For the first sample, mean= 6.375
For the second sample, mean= 6.375
For the third sample, mean= 6.625
Range of sample means=Maximum Mean- Minimum Mean
= 6.625 - 6.375
= 0.25
The students will estimate the average backpack weight of their classmates using sample means, the true statement is:
The closer the range of the sample means is to 0, the more confident they can be in their estimate.
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Answer:
Step-by-step explanation:
<h2 /><h2>
<u>Consider</u></h2>
<h2>
<u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>
So, on substituting all these values, we get
<h2>Hence,</h2>
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
<h2>ADDITIONAL INFORMATION :-</h2>
Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ
(4⁰ = 1 too small)
4¹ = 4
4² = 16
4³ = 64
4⁴ = 256
(4⁵ = 1,024 too big)
Step-by-step explanation:
Let the length of the bridge be x
Since, triangles ADF and BCF are similar, ratio of their corresponding sides are equal.
Therefore,
60 / 144 = 40 / x
X (length of the bridge) = <u>96 m</u>
Answer:
35 mi²
Step-by-step explanation:
Let's subdivide the figure, as shown.
The lower part is a rectangle whose area is (5 mi)(18 mi) = 90 mi².
The upper part is a trapezoid whose area is found by averaging the length and multiplying the result by the width (8 mi - 5 mi), or 3 mi.
Area of trapezoid:
12 mi + 18 mi
------------------------ = 15 mi Width of trapezoid = 3 mi
2
Thus, the area of the trapezoid is (3 mi)(15 mi) = 45 mi²
and the total area of the entire figure is
45 mi² + 90 mi² = 135 mi²