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lesya [120]
3 years ago
12

The graphs of functions f(x) and g(x) = f(x) + k are shown below:

Mathematics
2 answers:
allochka39001 [22]3 years ago
8 0
The correct answer for k would be 2
Nataly [62]3 years ago
7 0
Here's the info for f(x):  We are going to find the slope of the line and then write the equation for the line using one of the given points.  The coordinate points we are given are (0, 0) and (2, 4).  Using the slope formula:
m= \frac{y_{2} - y_{1} }{ x_{2}- x_{1}  }
gives us a slope equation of:
m= \frac{4-0}{2-0} and the slope is 2.  Using the point (0, 0) to write the equation of the line for f(x) looks like this in the slope-intercept form of the equation:
y- y_{1} =m(x- x_{1}) where m is the sloppe of 2 that we found and y_{1}  and  x_{1}  are the coordinates of one of the points.  It doesn't matter which one you choose; you will get the same answer whether you use (0, 0) or (2, 4): y-0=2(x-0)   Distributing that 2 into the parenthesis and simplifying gives you the equation of y = 2x, or in our function notation, f(x) = 2x.  Since f(x) is the first part of g(x), so far for g(x) we have that g(x) = 2x + k.  Now we will do the same thing for g(x) that we did for f(x) as far as writing its equation down; we don't need to find the slope cuz the slope of g(x) is the function f(x).  The equation for g(x), using the point (0, 2) (again, you could have used either point; I just picked (0, 2) cuz the other one has a decimal in it!): y - 2 = 2(x - 0).  Distributing that 2 into the parenthesis gives you this: y - 2 = 2x - 0; y = 2x + 2.  So 2 is your k value!

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The formula for E(x) would be SUM[ X(n)P(n)] from n = 1 to n = infinity until we get the prime number

In the given problem, a 10-sided die with 1-10 numbers is rolled until we get a prime number

Prime numbers between 1-10 = 2,3,5,7 (4 prime numbers)

Non - prime numbers = 1,4,6,8,9,10( 6 non-prime number)

Let X be the number of times the die is rolled

The probability of getting a prime =

P(Prime) = \frac{4}{10}

Now, the value of

E(X)=∑x.P(x) [ x-> {1, infinity}]

= P(1)X(1) +P(2)X(2)+P(3)X(3)+P(4)X(4)........+ P(n)X(n)

= 1.\frac{4}{10} + 2. \frac{6}{10}.\frac{4}{10} + 3.\frac{6}{10}.\frac{6}{10}\frac{4}{10} + 4. \frac{6}{10}.\frac{6}{10}.\frac{6}{10}.\frac{4}{10}.......

Hence, this is the value of E(x)

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