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Tanzania [10]
3 years ago
5

Julie designs gardens for a living. For a year she kept track of the amount of time she spent designing the garden and whether o

r not the customer returned. She recorded the time and probability that a customer returned.
To have the greatest probability of a returning customer, what amount of time should Julie spend per customer?

TIME DESIGNING PROBABILITY OF RETURN
1-2 Hours 5/8
2-3 Hours 5/9
3-4 Hours 7/8
4-5 Hours 3/4
5-6 Hours 1/2
6-7 Hours 3/7
7-8 Hours 2/9

A. 1-2 Hours
B. 2-3 Hours
C. 3-4 Hours
D. 4-5 Hours

Mathematics
2 answers:
lubasha [3.4K]3 years ago
6 0

Answer:C)3-4 hours

Step-by-step explanation:

JulijaS [17]3 years ago
4 0

Answer:

<em>To have the greatest probability of a returning customer, Julie should 3-4 Hours per customer </em>

Step-by-step explanation:

==> To compare the Time Designing Probability of Return, we need to convert the Probabilities into Percentage form by multiplying them with 100.

<em>1-2 Hour = (5÷8) × 100 = 62.5%</em>

<em>2-3 Hours = (5÷9) × 100 = 55.55%</em>

<em>3-4 Hours = (7÷8) × 100 = 87.5%</em>

<em>4-5 Hours = (3÷4) × 100 = 75%</em>

<em>5-6 Hours = (1÷2) × 100 = 50%</em>

<em>6-7 Hours = (3÷7) × 100 = 42.85%</em>

<em>7-8 Hours = (2÷9) × 100 = 22.22%</em>

<em>To have the greatest probability of a returning customer, Julie should 3-4 Hours per customer </em>

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Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x
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{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

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\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}

4.  The equation of the hyperbola is

\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

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