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sweet-ann [11.9K]
3 years ago
6

Gerald graphs the function f(x) = (x – 3)2 – 1. Which statements are true about the graph? Check all that apply. The domain is {

x| x ≥ 3}. The range is {y| y ≥ –1}. The function decreases over the interval (–∞, 3). The function increases over the interval (–1, ∞). The axis of symmetry is x = –1. The vertex is (3, –1).
Mathematics
2 answers:
katrin2010 [14]3 years ago
8 0
Hey! Statement 1 is false because the domain of f is all real numbers since the function is quadratic.

Statement 2 is true.

Statement 3 is true because when we graph f, it is a parabola opening upwards with vertex (3, -1). Since it is opening upward, then the value of x from -∞ to 3 is decreasing while increasing from 3 to ∞.

Statement 4 is false because we just stated above that f is decreasing from -∞ to 3. Hence, f is also decreasing from -1 to 3. Hence, f is not increasing from -1 to ∞.

Statement 5 is false because the axis of symmetry is x = 3.

Statement 6 is true.
Hope this helps! :)

elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

2,3,6 are true on edg

Step-by-step explanation:

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A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective
Inessa05 [86]

Answer:

(a) P(X \leq 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....

where, n = number of samples taken = 150

            r = number of success

           p = probability of success which in our question is % of bulbs that

                  are defective, i.e. 10%

<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>

So, Let X = No. of defective bulbs in a box

<u>Mean of X</u>, \mu = n \times p = 150 \times 0.10 = 15

<u>Standard deviation of X</u>, \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.10 \times (1-0.10)} = 3.7

So, X ~ N(\mu = 15, \sigma^{2} = 3.7^{2})

Now, the z score probability distribution is given by;

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(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X \leq 20) = P(X < 20.5)  ---- using continuity correction

    P(X < 20.5) = P( \frac{X-\mu}{\sigma} < \frac{20.5-15}{3.7} ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = n \times p = 150 \times 0.10 = 15.

                                           

5 0
3 years ago
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