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Nady [450]
2 years ago
9

(Please help worth 50 points) Part A

Mathematics
1 answer:
Marina86 [1]2 years ago
6 0

x_{123} \frac{x}{y} \sqrt{x} \left \{ {{y=2} \atop {x=2}} \right. x^{2} \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \leq \leq x_{123} \frac{x}{y}

   

                                                .                                     .                    

  •                                          

                                         

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Answer:

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Step-by-step explanation:

We see the relationship is closer to linear in the interval 30 to 40

<u>Approximate points:</u>

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<u>The slope of the line:</u>

  • m = (54 - 34)/(34- 30) = 20/4 = 5

<u>The line is:</u>

  • y - 34 = 5(x - 30)
  • y = 5x - 116

<u>At x = 40, the value of y would be:</u>

  • y = 5*40 - 116 = 84

The closet option is B

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To graduate this semester, you must pass Statistics 314; and you estimate that you have an 85% chance of passing. You need to pa
Nitella [24]

Answer:

Probability of graduating this semester is 0.7344

Step-by-step explanation:

Given the data in the question;

let A represent passing STAT-314

B represent passing at least in MATH-272 or MATH-444

M1 represent passing in MATH-272

M2 represent passing in MATH-444

C represent  passing GERMAN-32

now

P( A ) = 0.85, P( C ) = 90, P( M1 ) = P( M2 ) = 0.8

P( B ) = P( pass at least one of either MATH-272 or MATH-444 ) = P( pass in MATH-272 but not MATH-444 ) + ( pass in MATH-444 but not in MATH 272) + P( pass both )

P( B ) =  P( M1 ) × ( 1 - P( M2 ) ) + ( 1 - P( M1 ) ) × P( M2 ) + P( M1 ) × P( M2 )

we substitute

⇒ 0.8×0.2 + 0.2×0.8 + 0.8×0.8 = 0.16 + 0.16 + 0.64 = 0.96

∴ the probability of graduating this semester will be;

⇒ P( A ) × P( B ) × P( C )

we substitute

⇒ 0.85 × 0.96 × 90

⇒ 0.7344

Probability of graduating this semester is 0.7344

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