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horrorfan [7]
3 years ago
15

39 percent of all customers who enter a store will make a purchase. Suppose that 6 customers enter the store and that these cust

omers make independent purchase decisions. (1) Use the binomial formula to calculate the probability that exactly five customers make a purchase.
Mathematics
1 answer:
aleksklad [387]3 years ago
3 0

Answer:the probability that exactly five customers make a purchase is 0.033

Step-by-step explanation:

The formula for binomial distribution is expressed as

P(x = r) = nCr × q^(n - r) × p^r

Where

p = probability of success

q= probability of failure

n = number of sample

From the information given,

39 percent of all customers who enter a store will make a purchase. It means that the probability of success, p = 39/100 = 0.39

The probability of failure q, would be 1 - p = 1 - 0.39 = 0.61

n = 6

The probability that exactly five customers make a purchase is expressed as

P(x = 5) = 6C5 × 0.61^(6 - 5) × 0.39^5

P(x = 5) = 6 × 0.61× 0.39^5 = 0.033

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  See attached for graphs

  g(x) -- domain: -∞ < x < ∞; range: 0 < y < ∞

  g^-1(x) -- domain: 0 < x < ∞; range: -∞ < y < ∞

Step-by-step explanation:

g(x) is an exponential decay function. Its base is 1/3, so each increase of 1 unit in x will multiply the y-value by a factor of 1/3. The graph will rapidly approach its horizontal asymptote of y=0 as x gets large. The y-intercept is (0, 1). Just as y gets smaller as x increases, so it gets larger as x decreases. Each decrease of x by 1 unit causes the y-value to be multiplied by 3.

__

The graph of g^-1(x) is the graph of g(x) reflected across the line y=x. That is, each coordinate pair (x, y) on the graph of g(x) becomes a point (y, x) on the graph of the inverse function. In order to graph g^-1(x), you don't need to write down the function, you only need to know the relationship between the graphs.

Just as x- and y- are interchanged on the graph, so the domain, range, and intercepts are interchanged. g^-1(x) will have a vertical asymptote of x=0, and an x-intercept of (1, 0). The domain of g^-1(x) is the range of g(x): 0 < x < ∞; and the range of g^-1(x) is the domain of g(x): -∞ < y < ∞.

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The attached graph shows g(x) in red and g^-1(x) in blue. As you can see, we created the graph simply by interchanging x and y. The line y=x is shown for reference, so you can see that each curve is a reflection of the other across that line.

_____

<em>Additional comment</em>

The explicit expression for g^-1(x) can be found by solving for y:

  x = g(y)

  x=\left(\dfrac{1}{3}\right)^y=\dfrac{1}{3^y}=3^{-y}\\\\ \log(x)=-y\cdot\log(3)\qquad\text{take logarithms}\\\\y=-\dfrac{\log{x}}{\log{3}}=-\log_3{x}\qquad\text{use the change of base relation}\\\\\boxed{g^{-1}(x)=-\log_3{x}}

If you're familiar with the log function, you know it has an x-intercept of 1 and a vertical asymptote at x=0. The base of the log function is simply a vertical scale factor. The minus sign reflects it across the x-axis.

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