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artcher [175]
2 years ago
11

What equation is graphed in this figure?

Mathematics
1 answer:
Ksivusya [100]2 years ago
7 0

to get the equation of any straight line, we simply need two points off of it, let's use those two points in the picture below.

(\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{0}}}\implies \cfrac{3}{1}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_2=m(x-x_2) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_2}{5}=\stackrel{m}{3}(x-\stackrel{x_2}{1})

keeping in mind that for the point-slope form, either point will do, in this case we used the second one, but the first one would have worked just the same.

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Find a cubic polynomial whose zeros are 1/2,1 and -3.
Alexxx [7]

Answer:

2x³+3x²-8x+3

Step-by-step explanation:

So we know x=½,x=1 and x=-3

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5 0
2 years ago
A coin is tossed four times What is the probability of getting four heads? (1 mark) a) b) What is the probability of getting exa
BlackZzzverrR [31]
<h2>Answer:</h2>

1. A coin is tossed four times. What is the probability of getting four heads?

Each toss has a 1/2 chance of getting a head.

So the chance of getting all four heads can be calculated as :

1/2\times1/2\times1/2\times1/2=1/16

2. A coin is tossed four times. What is the probability of getting two heads?

Each toss can have 2 results, so 4 flips will have 2^{4} or 16 results.

Getting two heads means getting two tails also. So, we can get the number of times two heads come = \frac{4!}{2!2!} = 6

We can write the groups like - {HHTT,HTHT,HTTH,THHT,THTH,TTHH}

So, the required probability is : 6/16 or 3/8.

3. Not getting two heads means getting 3 tails and 1 head or all tails.

Probability of having all tails = 1/2\times1/2\times1/2\times1/2=1/16

Probability of one head(1st trial) and three tails = 1/16

Probability of one head (2nd trial) and three tails (the 1st, 3rd and 4th trials) = 1/16

Probability of one head (3rd trial) and three tails (the 1st, 2nd and 4th trials) = 1/16

Probability of one head (4th trial) and three tails (the 1st, 2nd and 3rd trials) = 1/16

So, the total probability showing only one or none head and at least three tails = 1/16+1/16+1/16+1/16+1/16=5/16

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3 years ago
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