Those are vertical angles,so they are equal
150=5m
m=150/5
m=30
A + B + C = 518 ft^3
A = 1/3B, or 3A = B
B = C
A + B + C = 518 ft^3
1/3B + B + B = 518
7/3B = 518
B = 518×3/7 = 222 ft^3
A = 1/3B = 1/3×222 = 74 ft^3
C = B = 222 ft^3
Answer:
1.
Step-by-step explanation:
1. sin 241 = sin(61 + π) = -sin61 = -t^(1/2)
2. sin^2 + cos^2 =1, cos 61= (1-t)^(1/2)
3. [2sin^2 (x)sin(x) +2cos^2 (x)sin(x)]/2cos(x) = 2sin(x) / 2cos(x) = tan(x)
4. tan (x), x do not equal to (π/2 +or- kπ)
x do not equal to 90,270。
To find the z-score for a weight of 196 oz., use

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.