Change 5 5/16 into a decimal and show work
1 answer:
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Answer:
1 - (π/6) - (√3/4)
Step-by-step explanation:
First, calculate the area of ABCD. That's just
Then, we are told that arcs BD and AC are circular, meaning they are portions of a circle. Therefore, let's redraw this image a little. Look below for the image.
Essentially, what I've done is created an equilateral triangle and two sectors. You may be wondering why the triangle is equilateral. Well, we firstly know it's isosceles because the two original arcs are circular and congruent. Their overlap creates two congruent lines that can be extend from point M to the vertices A and D. These two lines also happen to be radii of the quarter circles because the two sectors have AM AND AD as radii. AM and AD = 1, so AMD is an equilateral triangle.
Draw altitude OD. This is a right angle. Since equilateral triangles are also isosceles triangles, and isosceles triangles have the property of sharing the median to base and altitude to base, we can say that OD = AD =
. MD = 1, so this triangle is a 30-60-90 triangle. Why? This is because of the 30º angle converse theorem. The shorter leg is half of the hypotenuse MD, so <OMD is a 30º angle, making <ODM a 60º angle and inevitably <MDC a 30º angle (because <D is a right angle) and <ABM as well because the two sectors are congruent.
Because m<ABM = m<MDC = 30º, these two sectors are 30/360 of the individual circles or 1/12 of the original circles, which have areas of π (because 1^2*π = π). Each is 1/12π, combined they are 1/6π.
The area of the equilateral triangle is √3/8. This is because MO can be calculated, using the 30-60-90 ratios, to be √3/2. (1 * √3/2)/2 = (√3/2)/2 = √3/4; this uses the area of a triangle. Now, time to calculate the shaded area.
The area of the square is 1, the area of the sectors π/6, and the equilateral triangle √3/4. Subtract the equilateral triangle and sectors' areas from the area of the square to be left with the remaining shaded part. 1-(π/6)-(√3/4) is your final answer. To put that a little more cleanly below in the image.
Hope this helps, have a great day.
Complete question:
Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby this mention is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect so the probability of a girl is 0.5. Assume that the group consists of 36 couples.
A) Find the mean and standard deviation for the number of girls in groups of 36 births.
B) Use the range rule of thumb to find the values separating results that are significantly low and significantly high.
C) Is the result of 33 girls significantly high? A result of 33 girls would suggest the method is effective or is not effective?
Answer:
a) mean = 18
Standard deviation =3
b) low range = 12
High range = 24
c) The result of 33 girls is significantly high. Yes, the method is effective.
Step-by-step explanation:
Given:
p = 0.5
n = 36
a) The mean is the product of n and p
Mean u = np
u = 36 * 0.5 = 18
The standard deviation is the square root of the product of n and p&q.
S.d ó =
b) To find the range rule of thumb:
• For low range
Low range = u - 2ó
= 18 - (2 * 3)
= 12
• High range
= u + 2ó
= 18 + (2*3)
= 24
c) The result is significantly high, because 33 is greater than 24 girls.
A result of 33 girls would prove the method as effective.