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Agata [3.3K]
3 years ago
5

24 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Nostrana [21]3 years ago
6 0
Red:green

5:3
5+3=8
8 units
4qt=8 units
divide by 8
0.5qt=1unit
5 unit=red
5*0.5=2.5qt, red=2.5qt
3unit=lime
0.5*3=1.5qt
1.5qt, 2/3 of that is yellow, 1/3 is blue, 1.5 times 2/3=1, 1.5 times 1/3=0.5
result:
2.5qt red
1qt yellow
0.5qt blue

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True or False? This table represents a function. y 3 12 6 1 -7 -18 -5 -12 8 | 27​
Alex73 [517]

Answer:

true

Step-by-step explanation:

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Create your own quadratic function in standard form with a Y intercept is 9
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Answer:

\boxed{\sf \ \ x^2+9 \ \ }

Step-by-step explanation:

Hello,

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3 years ago
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

P(X≤1.9) = 0.9114

(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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