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Genrish500 [490]
2 years ago
9

Describe the behavior of the function p around its vertical asymptote at x = -9

Mathematics
1 answer:
nydimaria [60]2 years ago
5 0

Answer:

Step-by-step explanation:

The graph of P approaches -∞from the left and +∞ from the right of the asymptote

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779x^2/ 3 is the answer.
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A softball pitcher has a 0.507 probability of throwing a strike for each pitch. If the softball pitcher throws 15 pitches, what
Elena L [17]

Answer:

The probability that the pitcher throws exactly 8 strikes out of 15 pitches is approximately 0.199

Step-by-step explanation:

The given probability that the pitcher throws a strike, p = 0.507

The number of pitches thrown by the pitcher = 15 pitches

The probability that the pitcher does not throw a strike, q = 1 - P

∴  q = 1 - 0.507 = 0.493

By binomial theorem, we have;

P(X = r) = \dbinom{n}{r}p^{r} \cdot q^{n-r}=  \dbinom{n}{r}p^{r} \cdot \left (1-p  \right )^{n-r}

When X = r =  8, and n = 15, we get;

The probability that the pitcher throws exactly 8 strikes out of 15 pitches, P(8), is given as follows

P(8) = ₁₅C₈ × 0.507⁸ × (1 - 0.507)⁽¹⁵ ⁻ ⁸⁾ = 6,435 × 0.507⁸ × 0.493⁷ ≈ 0.199

8 0
3 years ago
What is the upper bound of the function f(x)=4x4−2x3+x−5?
inessss [21]

Answer:

(no global maxima found)

Step-by-step explanation:

Find and classify the global extrema of the following function:

f(x) = 4 x^4 - 2 x^3 + x - 5

Hint: | Global extrema of f(x) can occur only at the critical points or the endpoints of the domain.

Find the critical points of f(x):

Compute the critical points of 4 x^4 - 2 x^3 + x - 5

Hint: | To find critical points, find where f'(x) is zero or where f'(x) does not exist. First, find the derivative of 4 x^4 - 2 x^3 + x - 5.

To find all critical points, first compute f'(x):

d/( dx)(4 x^4 - 2 x^3 + x - 5) = 16 x^3 - 6 x^2 + 1:

f'(x) = 16 x^3 - 6 x^2 + 1

Hint: | Find where f'(x) is zero by solving 16 x^3 - 6 x^2 + 1 = 0.

Solving 16 x^3 - 6 x^2 + 1 = 0 yields x≈-0.303504:

x = -0.303504

Hint: | Find where f'(x) = 16 x^3 - 6 x^2 + 1 does not exist.

f'(x) exists everywhere:

16 x^3 - 6 x^2 + 1 exists everywhere

Hint: | Collect results.

The only critical point of 4 x^4 - 2 x^3 + x - 5 is at x = -0.303504:

x = -0.303504

Hint: | Determine the endpoints of the domain of f(x).

The domain of 4 x^4 - 2 x^3 + x - 5 is R:

The endpoints of R are x = -∞ and ∞

Hint: | Evaluate f(x) at the critical points and at the endpoints of the domain, taking limits if necessary.

Evaluate 4 x^4 - 2 x^3 + x - 5 at x = -∞, -0.303504 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

-0.303504 | -5.21365

∞ | ∞

Hint: | Determine the largest and smallest values that f achieves at these points.

The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:

The open endpoints of the domain are marked in gray

x | f(x) | extrema type

-∞ | ∞ | global max

-0.303504 | -5.21365 | global min

∞ | ∞ | global max

Hint: | Finally, remove the endpoints of the domain where f(x) is not defined.

Remove the points x = -∞ and ∞ from the table

These cannot be global extrema, as the value of f(x) here is never achieved:

x | f(x) | extrema type

-0.303504 | -5.21365 | global min

Hint: | Summarize the results.

f(x) = 4 x^4 - 2 x^3 + x - 5 has one global minimum:

Answer: f(x) has a global minimum at x = -0.303504

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