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nasty-shy [4]
3 years ago
12

Stat crunchdescribe the sampling distribution of modifyingabove p with caretp. assume the size of the population is 20 comma 000

20,000. nequals=500500​, pequals=0.543
Mathematics
1 answer:
zavuch27 [327]3 years ago
4 0
10860 is what i think the answer is to this
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Birth weights of a sample of newborn babies
Sophie [7]

Answer:  Birth-weight (Kilogram) Number (n) Percentage (%)

4.1 – 4.5 4 1.0%

Total 387 100%

Step-by-step explanation:  In this sample of newborns, the minimum birth-weight was 2.5 Kg, and no babies were low-birth-weight (< 2.5 Kg) according to the WHO criteria. However, the vast majority (51%) of babies weighed between 2.5 Kg and 3.0 Kg at birth

4 0
3 years ago
The area of a circle is 100π in². What is the circumference, in inches? Express your answer in terms of
alekssr [168]

Answer:

20π

Step-by-step explanation:

Area = πr^2

100π = πr^2

solve for r and r = 10

circumference of a circle is 2πr

therefore 2π×10=20π in

4 0
2 years ago
Kate shares a 64-ounce bottle of apple cider with 5 friends. Each person’s serving will be the same number of ounces. Between wh
ruslelena [56]

Answer:

12 and 13

Step-by-step explanation:

The computation of the number of ounces will each person serving be is given below:

Ounces be 64

And, the friends be 5

So for each one it would be

= 64 ÷ 5

= 12.8

That means

12.8 is lies between 12 and 13

8 0
3 years ago
2. 114 out of 150 customers at a
Vilka [71]

Answer: lowkey gonna goes by my math and say 0.76 because I divided 114 by 150 and got 0.76

8 0
3 years ago
You have given an equal sided triangle with side length a. A straight line connects the center
GarryVolchara [31]

Answer:

Where α is an acute angle (first figure)

The area of the shaded triangle = ((√3)·a²/4)·sin(α)·csc(120 - α))

Where α is an obtuse angle (second figure)

The required area of the shaded region = (√3)·a²/4 + (√3)·a²/4)·sin(α)·sec(α + π/6)

Step-by-step explanation:

Where α is an acute angle (first figure)

The given parameters are;

The given triangle = Equilateral Triangle

Let the sides of the equilateral triangle = 2·a

Therefore;

The measure of each interior angles of the given triangle = 60°

Let c represent the side of the shaded triangle opposite ∠α and b represent the side of the shaded triangle opposite ∠60° and c, represent the third side of the shaded triangle, we have;

The sides of the equilateral triangle = 2·a

By sine rule, we have;

c/sin(α) = b/sin(60°) = a/sin(180 - (60 + α)) = a/sin(120 - α))

b = sin(60°) × a/sin(120 - α)) = (√3)/2 × a/sin(120 - α))

The area of the shaded triangle = 1/2 × a × b × sin(α) = 1/2 × a × (√3)/2 × a/sin(120 - α)) × sin(α) = ((√3)·a²/4)·sin(α)·csc(120 - α))

The area of the shaded triangle = ((√3)·a²/4)·sin(α)·csc(120 - α))

Where α is an obtuse angle (second figure)

The required area of the shaded region = The area of the equilateral triangle - The area of the small unshaded triangle, with base side a and interior angles, (180° - α), 60° and ((180 - (180° - α) - 60°) = ) α - 60°

The area of the unshaded triangle is found as follows;

By sine rule, we have;

c/sin(180° - α) = b/sin(60°) = a/sin(α - 60°)

b = sin(60°) × a/sin(α - 60°) = (√3)/2 × a/sin(α - 60°)

The area of the unshaded triangle = 1/2 × a × b × sin(α) = 1/2 × a × (√3)/2 × a/sin(α - 60°) × sin(α) = -((√3)·a²/4)·sin(α)·sec(α + π/6)

The area of the shaded triangle =  -((√3)·a²/4)·sin(α)·sec(α + π/6)

The required area of the shaded region = 1/2×a²·sin(60°)  - (-((√3)·a²/4)·sin(α)·sec(α + π/6))

The required area of the shaded region = (√3)·a²/4 + (√3)·a²/4)·sin(α)·sec(α + π/6)

4 0
3 years ago
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