Tessa’s employee benefits include family health care coverage. She contributes 18% of the cost. Tessa gets paid biweekly and $10
1 answer:
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Given that 
, then 
The slope of a tangent line in the polar coordinate is given by:
Thus, we have:
Part A:
For horizontal tangent lines, m = 0.
Thus, we have:
Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:
</span><span>θ = 0
</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488
Part B:
For vertical tangent lines,
Thus, we have:
</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:
</span>θ = <span>4.91718592528713</span>
The slope of a tangent line in the polar coordinate is given by:
Thus, we have:
Part A:
For horizontal tangent lines, m = 0.
Thus, we have:
Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:
</span><span>θ = 0
</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488
Part B:
For vertical tangent lines,
Thus, we have:
</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:
</span>θ = <span>4.91718592528713</span>
Answer:
a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.
b) 0.996 is the probability that more than half of the vehicles carry just one person.
Step-by-step explanation:
We are given the following information:
A) Binomial distribution
We treat vehicle on road with one passenger as a success.
P(success) = 64% = 0.64
Then the number of vehicles follows a binomial distribution, where
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 10
We have to evaluate:
0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.
B) By normal approximation
Sample size, n = 92
p = 0.64
We have to evaluate the probability that more than 47 cars carry just one person.
After continuity correction, we will evaluate
Calculation the value from standard normal z table, we have,
0.996 is the probability that more than half out of 92 vehicles carry just one person.