At a local fitness center, gym members pay $3 per workout with an annual membership fee of $140. Non-members pay $10 per workout
1 answer:
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Answer:
1. <u>Cost per customer</u>: 10 + x
<u>Average number of customers</u>: 16 - 2x
3. $10, $11, $12 and $13
Step-by-step explanation:
<u>Given information</u>:
- $10 = cost of buffet per customer
- 16 customers choose the buffet per hour
- Every $1 increase in the cost of the buffet = loss of 2 customers per hour
- $130 = minimum revenue needed per hour
Let x = the number of $1 increases in the cost of the buffet
<u>Part 1</u>
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<u>Cost per customer</u>: 10 + x
<u>Average number of customers</u>: 16 - 2x
<u>Part 2</u>
The cost per customer multiplied by the number of customers needs to be <u>at least</u> $130. Therefore, we can use the expressions found in part 1 to write the <u>inequality</u>:
<u>Part 3</u>
To determine the possible buffet prices that Noah could charge and still maintain the restaurant owner's revenue requirements, solve the inequality:
Find the roots by equating to zero:
Therefore, the roots are x = 3 and x = -5.
<u>Test the roots</u> by choosing a value between the roots and substituting it into the original inequality:
As 144 ≥ 130, the <u>solution</u> to the inequality is <u>between the roots</u>:
-5 ≤ x ≤ 3
To find the range of possible buffet prices Noah could charge and still maintain a minimum revenue of $130, substitute x = 0 and x = 3 into the expression for "cost per customer.
[Please note that we cannot use the negative values of the possible values of x since the question only tells us information about the change in average customers per hour considering an <em>increase </em>in cost. It does not confirm that if the cost is reduced (less than $10) the number of customers <em>increases </em>per hour.]
<u>Cost per customer</u>:
Therefore, the possible buffet prices Noah could charge are:
$10, $11, $12 and $13.
Answer:
(a) E(Y) = 4400
sd (Y) =225
(b) P(Y ≤ 4500) = 0.67003
(c) P (X₁ > X₂) = 0.31744
Step-by-step explanation:
(a) Here we have
Y = 2·X₁ + 3·X₂
Therefore E(Y) = 2·E(X₁) + 3·E(X₂) = 2000 + 2400 = 4400
sd(Y) is given by
Variance Y = (sd (Y))² = 2²·(sd (X₁))² + 3²·(sd (X₂))²
= 4·8100 + 9·2025 = 50625
sd (Y) = √50625 = 225
(b) The probability that the revenue does not exceed 4500 is given by
P(Y ≤ 4500) = P(z ≤0.444)
z =
z = = 0.444
Therefore from the normal distribution table, we have
P = 0.67003
(c) The probability that the P(X₁ > X₂)
Since the gas station sells 2 portions of X₁ to 3 portions of X₂
Therefore, the probability that the gas station sells more of X₁ is given by
₅C₀ × 2/5⁰×3/5⁵ = 0.07776
₅C₁ × 2/5¹×3/5⁴= 0.2592
₅C₂ × 2/5²×3/5³ = 0.3456
P (X₁ > X₂) = 1 - 0.68256 = 0.31744