The circle on the right has center O. Its radius is 3 yd., and the central angle a measures 130°. What is the area of the shaded
region? Give the exact answer in terms of π, and be sure to include the correct unit in your answer.
1 answer:
Answer:
area of the sector = 3.25π yard²
Step-by-step explanation:
The radius of the circle is 3 yards . The central angle is 130° let us say it is the sector angle of the circle. The angle is 130°. If the shaded area of the circle is the sector area of the circle the area of the sector can be computed below.
area of a sector = ∅/360 × πr²
where
∅ = center angle
r = radius
area of the sector = 130/360 × π × 9
area of the sector = 1170π/360
area of the sector = 3.25π yard²
If the shaded area is segment. The shaded area can be solved with the formula.
Area of segment = area of sector - area of the triangle
Area of segment = ∅/360 × πr² - 1/2 sin∅ r²
The picture demonstrate the area of sector and the segment of a circle with illustration on how to compute the area of the triangle
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Answer:
Probability that the sample average is at most 3.00 = 0.98030
Probability that the sample average is between 2.65 and 3.00 = 0.4803
Step-by-step explanation:
We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.
Also, a random sample of 25 specimens is selected.
Let X bar = Sample average sediment density
The z score probability distribution for sample average is given by;
Z = ~ N(0,1)
where, = population mean = 2.65
= standard deviation = 0.85
n = sample size = 25
(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)
P(X bar <= 3) = P( <=
) = P(Z <= 2.06) = 0.98030
(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)
P(X bar < 3) = P( <
) = P(Z < 2.06) = 0.98030
P(X bar <= 2.65) = P( <=
) = P(Z <= 0) = 0.5
Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .