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3 years ago

2. Draw a table to show the sample space of a toss of two dice. Denote the event of a sum of 7 or 11 on the table.

Mathematics

1 answer:

frozen [14]3 years ago

4 0

The sample space of a toss of two dice is:

$\end{bmatrix} (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) \\ (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) \\ (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) \\ (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) \\ (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) \\ (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) \end{bmatrix}$

The sample space denoting the sum of 7 or 11 is:

$\begin{bmatrix} (1,6) (6,1) (3,4) (4,3) (5,2) (2,5) (5,6) (6,5) \end{bmatrix}$

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3 years ago

Sue either travels by bus or walks when she visits the shops. The probability that she catches the bus from the shops is 0.7. Sh

Alina [70]

Corrected Question

Sue travels by bus or walks when she visits the shops. The probability that she catches the bus to the shops is 0.4. The probability she catches the bus from the shops is 0.7. Show the probability that Sue walks at one way is 0.72

Answer:

$P(A \cup B) =0.72$ (Proved)

Step-by-step explanation:

Sue travels by bus or walks when she visits the shops.

Let the event that she catches the bus to the shop=A

Let the event that she catches the bus from the shop=B

P(A)=0.4

P(B)=0.7

Both A and B are independent events.

Therefore,Probability that she catches the bus to and from the shop:

P(A∩B) = 0.4 X 0.7= 0.28

Probability Sue walks at least one way $P(A \cup B) = 1 - P(A \cap B)$

$= 1 - 0.28\\= 0.72$

Hence, the probability that Sue walks at least one way is 0.72.

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3 years ago

Write down the quadratic equation whose roots are $x = -7$ and $x = 1,$ and the coefficient of $x^2$ is 1. Enter your answer in

pav-90 [236]

<h2>Steps:</h2>

So firstly, since we know that the coefficient of x² is 1, this means that this is our base equation:

y = x² + bx + c

Now, since we know that the roots are -7 and 1, set y = 0 and set x = -7 and 1 and simplify:

$0=(-7)^2+b(-7)+c\\0=49-7b+c\\-49=-7b+c\\\\0=1^2+b(1)+c\\0=1+b+c\\-1=b+c\\\\-49=-7b+c\\-1=b+c$

Now with this, we can set up a system of equations to solve for b and c. For this, I will be using the elimination method. For this, subtract the 2 equations:

$\begin{alignedat}{2}-49&=-7b+c\\-(-1&=b+c)\\-48&=-8b\end{alignedat}$

Now that the c variable has been eliminated we can solve for b. For this, divide both sides by -8 and your first part of your answer is b = 6.

Now that we know the value of b, plug it into either equation to solve for c:

$-49=-7(6)+c\\-49=-42+c\\-7=c\\\\-1=6+c\\-7=c$

<h2>Answer:</h2>

<u>Putting it together, your final answer is x² + 6x - 7 = 0.</u>

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3 years ago