Answer:
4.72 hours/day
Step-by-step explanation:
Mean time spent watching TV (μ) = 2.8 hours a day
Standard deviation (σ) = 1.5 hours a day
The 90th percentile (upper 10%) of a normal distribution has an equivalent z-score of roughly z = 1.282. The minimum time spent watching TV, X, at the 90th percentile is:
On a typical day, you must watch at least 4.72 hours of TV to be in the upper 10%.
y - 1 = ⁵/₆(x - 4)
y - 1 = ⁵/₆(x) - ⁵/₆(4)
y - 1 = ⁵/₆x - 3¹/₃
+ 1 + 1
y = ⁵/₆x - 2¹/₃
⁻⁵/₆x + y = ⁵/₆x - ⁵/₆x - 2¹/₃
-6(⁻⁵/₆x + y) = -6(-2¹/₃)
-6(⁻⁵/₆x) - 6(y) = 14
5x - 6y = 14
They should get $7.50, because you multiply the adults by two, then the children by three, and add it all together
Answer:
Results can be 
Step-by-step explanation:
-The solution for this question:
is equal to
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-The calculation for
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