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3 years ago

Help me with this problem please a and b

Mathematics

1 answer:

nadya68 [22]3 years ago

5 0

(a) Compare your quadratic for h to the general quadratic ax² +bx +c. Perhaps you can see that ...
a = -16
b = 128
You use these numbers in the given formula to find the time when the ball is highest.
t = -b/(2a) = -128/(2(-16)) = 4 . . . . . . the time at which the ball is highest

(b) Evaluate the quadratic to find the height at t=4.
h = -16(4)² +128(4) +21
h = -256 +512 +21
h = 277
The maximum height of the ball is 277 ft.

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Answer:

x<-2

Step-by-step explanation:

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3 years ago

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3 years ago

Susan is taking Western Civilization this semester on a pass/fail basis. The department teaching the course has a history of pas

Volgvan

Answer:

a. P(n) = 0.85 * (0.15)^(n-1)

b. P(n=1) = 0.85

c. P(n= 2) = 0.1275

d. P(n≥3) = 0.0225

e. Expected number of attempts is 1.176

Step-by-step explanation:

Given

p = success = 85% = 0.85

q = failure = 1 - q = 1 - 0.85 = 0.15

The results of passing/failing takes a Bernoulli distribution

Since, there are independent trials

The number of trials until the first successful event occurs is given by

P(n = k) = p . (1 - p)^(k-1)

P(n = k) = p.q^(k-1)

This is so because it is a Bernoulli distribution and it is modeled by a geometric distribution.

Substitute 0.85 for p

P(n) = 0.85 * (0.15)^(n-1)

Given

n = 1

Using P(n=1) = 0.85 * (0.15)^(n-1)

P(1) = 0.85 * 0.15^(1-1)

P(1) = 0.85 * 0.15°

P(1) = 0.85 * 1

P(1) = 0.85

Therefore, the probability that Susan passes on the first try is 0.85.

n = 2

Using P(n=2) = 0.85 * (0.15)^(2-1)

P(2) = 0.85 * 0.15^(2-1)

P(2) = 0.85 * 0.15¹

P(2) = 0.85 * 0.15

P(2) = 0.1275

Therefore, the probability that Susan passes on the first try is 0.1275

We'll make use of the probability of Susan passing the course after an infinite number of trials is 1.

i.e.

P(n=1) + P(n=2) + P(n=3) + P(n=4) + ......... = 1 --- This is then simplified to

P(n=1) + P(n=2) + P(n≥3) = 1

P(n≥3) = 1 - P(n=1) - P(n=2)

P(n≥3) = 1 - 0.85 - 0.1275

P(n≥3) = 0.0225

Therefore, the probability that Susan needs at least 3 attempts to pass is 0.0225

In (a) above, we explained that the distribution is modeled by an exponential distribution.

The Expected Value for this is inverse of p, where p = 0.85

So, E(n) = 1/p

E(n) = 1/0.85

E(n) = 1.176470588235294

E(n) = 1.176 --- Approximated

Hence the Expected number of attempts is 1.176

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Answer:

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3 years ago