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Answer:
The probability that it will still be working after one week is
Step-by-step explanation:
Given :
Total number of bulbs = 25
Number of bulbs which are good condition and will function for at least 30 days = 5
Number of bulbs which are partially defective and will fail in their second day of use = 10
Number of bulbs which are totally defective and will not light up = 10
To find : What is the probability that it will still be working after one week?
Solution :
First condition is a randomly chosen bulb initially lights,
i.e. Either it is in good condition and partially defective.
Second condition is it will still be working after one week,
i.e. Bulbs which are good condition and will function for at least 30 days
So, favorable outcome is 5
The probability that it will still be working after one week is given by,
The function is
f(x) = (1/3)x² + 10x + 8
Write the function in standard form for a parabola.
f(x) = (1/3)[x² + 30x] + 8
= (1/3)[ (x+15)²- 225] + 8
= (1/3)(x+15)² -75 + 8
f(x) = (1/3)(x+15)² - 67
This is a parabola with vertex at (-15, -67).
The axis of symmetry is x = -15
The curve opens upward because the coefficient of x² is positive.
As x -> - ∞, f -> +∞.
As x -> +∞, f -> +∞
The domain is all real values of x (see the graph below).
Answer: The domain is (-∞, ∞)
f(x) = (1/3)x² + 10x + 8
Write the function in standard form for a parabola.
f(x) = (1/3)[x² + 30x] + 8
= (1/3)[ (x+15)²- 225] + 8
= (1/3)(x+15)² -75 + 8
f(x) = (1/3)(x+15)² - 67
This is a parabola with vertex at (-15, -67).
The axis of symmetry is x = -15
The curve opens upward because the coefficient of x² is positive.
As x -> - ∞, f -> +∞.
As x -> +∞, f -> +∞
The domain is all real values of x (see the graph below).
Answer: The domain is (-∞, ∞)