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Luden [163]
2 years ago
5

2x + 5y = 14 4x + 2y = -4 solving with elimination​

Mathematics
2 answers:
melamori03 [73]2 years ago
8 0

Answer:

(- 3, 4 )

Step-by-step explanation:

2x + 5y = 14 → (1)

4x + 2y = - 4 → (2)

multiplying (1) by - 2 and adding to (2) will eliminate the x- term

- 4x - 10y = - 28 → (3)

add (2) and (3) term by term to eliminate x

0 - 8y = - 32

- 8y = - 32 ( divide both sides by - 8 )

y = 4

substitute y = 4 into either of the 2 equations and solve for x

substituting into (1)

2x + 5(4) = 14

2x + 20 = 14 ( subtract 20 from both sides )

2x = - 6 ( divide both sides by 2 )

x = - 3

solution is (- 3, 4 )

m_a_m_a [10]2 years ago
6 0

Answer: x=-\frac{3}{179} =\\-0.016759777\\y=-\frac{142}{179} =\\-0.793296089

Step-by-step explanation: steps using substitution

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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

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(c)  P(X≥2) = 1 - P(X<2)

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

                 = 1 - (0.6035 + 0.3079)

                 = 1 - 0.9114

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(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

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             = 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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