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Musya8 [376]
3 years ago
5

Mary owns a small business. She spent $4,500 to get the business started. She has noticed that her monthly operating costs have

been consistent. After five months, she has spent a total of $5,250 in start-up costs and monthly operating costs.
In a linear model of this situation, which of the following statements applies?
A.
An additional month of business operation is associated with an additional $500 in costs.
B.
An additional month of business operation is associated with an additional $750 in costs.
C.
An additional month of business operation is associated with an additional $150 in costs.
D.
An additional month of business operation is associated with an additional $50 in costs.
Mathematics
1 answer:
slavikrds [6]3 years ago
5 0
First, let’s find how much she spent on monthly operating costs.

5250-4500=750

Then, because all 5 months have been consistent, we can divide by 5

750/5=150

The answer is C
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Step-by-step explanation:

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The College Board reports that 2% of students who take the SAT each year receive special accommodations because of documented di
Sonbull [250]

Answer:

(a) P(X=1) = 0.3079

(b) P(X≥1) = 0.3965

(c) P(X≥2) = 0.0886

(d) P(X≤1.9) = 0.9114

(e) Expected no. of hours = 3.594 hours

Step-by-step explanation:

We have,

p = 0.02

n = 25

q = 1-p

q = 0.98

We will use the binomial distribution formula to solve this question. The formula is:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

where n = total no. of trials

           x = no. of successful trials

           p = probability of success

           q = probability of failure

Let X be the number of students who received a special accommodation.

(a) P(X=1) = ²⁵C₁ (0.02)¹ (0.98)²⁵⁻¹

          = 25*0.02*0.61578

P(X=1) = 0.3079

(b) P(X≥1) = 1 - P(X<1)

               = 1 - P(X=0)

               = 1 - (²⁵C₀ (0.02)⁰ (0.98)²⁵⁻⁰)

              = 1 - 0.6035

   P(X≥1) = 0.3965

(c)  P(X≥2) = 1 - P(X<2)

                 = 1 - [P(X=0) + P(X=1)]

                 = 1 - (0.6035 + 0.3079)

                 = 1 - 0.9114

     P(X≥2) = 0.0886

(d) The probability that the number among 25 who received a special accommodation is within 2 standard deviations of the expected number of accommodations. This means we need to compute the probability P(X-μ≤2σ). For this we need to calculate the mean and standard deviation of this distribution.

μ = np = (25)*(0.02) = 0.5

σ = \sqrt{npq} = √(25)*(0.02)*(0.98) = √0.49 = 0.7

P(X-μ≤2σ) = P(X - 0.5≤ 2(0.7)) = P(X≤ 1.4 + 0.5) = P(X≤1.9)

P(X≤1.9) = P(X=0) + P(X=1)

             = 0.6035 + 0.3079

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(e) Student who does not receive a special accommodation i.e. X=0 is given 3 hours for the exam whereas an accommodated student P(X>0) is given 4.5 hours. The expected average number of hours given on the exam can be calculated as:

Expected no. of hours = ∑x*P(x)

                                     = 3*P(X=0) + 4.5*P(X>0)

                                     = 3*0.6035 + 4.5(1 - P(X≤0))

                                     = 1.8105 + 4.5(1 - 0.6035)

                                     = 1.8105 + 1.78425

Expected no. of hours = 3.594 hours

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