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3 years ago

Extension question (provide a full explanation of your method(s):

Mathematics

1 answer:

Volgvan3 years ago

7 0

Answer:

Ann has little chance to win if she is presented with 4 counters.

Ann can always win from a pile of 6 counters.

(both are explained below)

Step-by-step explanation:

If Ann is presented with 4 counters, and

1. if she takes out 3, she will lose since the opponent will pull out 1 and the last one.

2. if she takes 2 her opponent will take out 1 and she can't pull out the last 1 since her opponents last move was to pull out 1 counter so she will lose.

3. If she takes out 1 and her opponent takes out 3 in the next move she loses.

but if instead of 3 her opponent takes out 2 and in the last move Ann takes out the last 1 then she will win.

So, If Ann is presented with 4 counters she has little chance to win provided in the move just before, her opponent didn't move 1 counter.

Now,

if there is 6 counters to Ann, and

1., if Ben's previous move was 1 then Ann can win if she takes out 3 or 2.

If she takes out 3 Ben can take out 1 or 2 and in the last move she will take out 2 or 1 (respectively) and winning the game.

If she takes out 2 Ben can take out 1 or 3 and in the last move Ann wins by pulling out 3 or 1 respectively.

2. if Ben's previous move was 2 then Ann can win if she takes out 1 or 3.

If she takes out 1 Ben can take out 2 or 3 and in the last move she will take out 3 or 2(respectively) and winning the game.

If she takes out 3 Ben can take out 1 or 2 and in the last move Ann wins by pulling out 2 or 1 respectively.

2. if Ben's previous move was 3 then Ann can win if she takes out 1 or 2.

If she takes out 1 Ben can take out 2 or 3 and in the last move she will take out 3 or 2(respectively) and winning the game.

If she takes out 2 Ben can take out 1 or 3 and in the last move Ann wins by pulling out 3 or 1 respectively.

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Answer:

C and E

Step-by-step explanation:

We are given that

$x_1=0.2,y_1=0.3$

$\frac{y}{x}=\frac{0.3}{0.2}=\frac{3}{2}$

$k=\frac{y}{x}=\frac{3}{2}$

A. $x_2=1.2,y_2=2.3$

$\frac{y_2}{x_2}=\frac{2.3}{1.2}=\frac{23}{12}\neq=\frac{3}{2}$

Hence, it is not in proportional relationship with (0.2,0.3)

B.(2.7,4.3)

$x_3=2.7,y_3=4.3$

$\frac{y_3}{x_3}=\frac{4.3}{2.7}=\frac{43}{27}\neq\frac{3}{2}$

Hence, it is not in proportional relationship with (0.2,0.3).

C.(3.2,4.8)

$x_4=3.2,y_4=4.8$

$\frac{y_4}{x_4}=\frac{4.8}{3.2}=\frac{3}{2}$

Hence, the ordered pair (3.2,4.8) are in a proportional relationship with (0.2,0.3).

D.(3.5,5.3)

$\frac{y_5}{x_5}=\frac{5.3}{3.5}=\frac{53}{35}\neq \frac{3}{2}$

Hence, the ordered pair (3.5,5.3) are not in a proportional relationship with (0.2,0.3).

E.(5.2,7.8)

$\frac{y_6}{x_6}=\frac{7.8}{5.2}=\frac{3}{2}$

Hence, the ordered pair (5.2,7.8) are in a proportional relationship with (0.2,0.3).

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Step-by-step explanation:

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20. Based on student records, 25% of the students at a large high school have a

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Answer:

(E) 0.71

Step-by-step explanation:

Let's call A the event that a student has GPA of 3.5 or better, A' the event that a student has GPA lower than 3.5, B the event that a student is enrolled in at least one AP class and B' the event that a student is not taking any AP class.

So, the probability that the student has a GPA lower than 3.5 and is not taking any AP classes is calculated as:

P(A'∩B') = 1 - P(A∪B)

it means that the students that have a GPA lower than 3.5 and are not taking any AP classes are the complement of the students that have a GPA of 3.5 of better or are enrolled in at least one AP class.

Therefore, P(A∪B) is equal to:

P(A∪B) = P(A) + P(B) - P(A∩B)

Where the probability P(A) that a student has GPA of 3.5 or better is 0.25, the probability P(B) that a student is enrolled in at least one AP class is 0.16 and the probability P(A∩B) that a student has a GPA of 3.5 or better and is enrolled in at least one AP class is 0.12

So, P(A∪B) is equal to:

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P(A∪B) = 0.29

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P(A'∩B') = 1 - P(A∪B)

P(A'∩B') = 1 - 0.29

P(A'∩B') = 0.71

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