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snow_lady [41]
3 years ago
6

a ball is thrown into the air with a velocity of 40 ft/s, its height in feet seconds later is given by y=40t-16t^2. a) Find the

average velocity for the time period beginning when and lasting (i) 0.5 second (ii) 0.1 second (iii) 0.05 second (iv) 0.01 second (b) Estimate the instantaneous velocity when t=2
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
5 0
V(t) = dy/dt = 40 - 32t
ai). average velocity = (v(0.5) - v(0))/(0.5 - 0) = ((40 - 32(0.5)) - (40 - 32(0)))/0.5 = (40 - 16 - 40)/0.5 = -16/0.5 = -32m / s

ii). average velocity = (v(0.1) - v(0))/(0.1 - 0) = ((40 - 32(0.1)) - (40 - 32(0)))/0.1 = (40 - 3.2 - 40)/0.1 = -3.2/0.1 = -32m / s

iii). average velocity = (v(0.05) - v(0))/(0.05 - 0) = ((40 - 32(0.05)) - (40 - 32(0)))/0.05 = (40 - 1.6 - 40)/0.05 = -1.6/0.05 = -32m / s
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6 0
3 years ago
I toss an unfair coin 12 times. This coin is 65% likely to show up heads. Calculate the probability of the following.
mars1129 [50]

Answer:

a. 0.0368

b. 0.99992131

c. 0.2039

d. 0.0048

e. 0.6533

Step-by-step explanation:

Let the probability of obtaining a head be p = 65% = 13/20 = 0.65. The probability of not obtaining a head is q = 1 - p = 1 -13/20 = 7/20 = 0.35

Since this is a binomial probability, we use a binomial probability.

a. The probability of obtaining 11 heads is ¹²C₁₁p¹¹q¹ = 12 × (0.65)¹¹(0.35) = 0.0368

b. Probability of 2 or more heads P(x ≥ 2) is

P(x ≥ 2) = 1 - P(x ≤ 1)

Now P(x ≤ 1) = P(0) + P(1)

= ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹

= (0.65)⁰(0.35)¹² + 12(0.65)¹(0.35)¹¹

= 0.000003379 + 0.00007531

= 0.0007869

P(x ≥ 2) = 1 - P(x ≤ 1)

= 1 - 0.00007869

= 0.99992131

c. The probability of obtaining 7 heads is ¹²C₇p⁷q⁵ = 792(0.65)⁷(0.35)⁵ = 0.2039

d. The probability of obtaining 7 heads is ¹²C₉q⁹p³ = 220(0.65)³(0.35)⁹ = 0.0048

e. Probability of 8 heads or less P(x ≤ 8) = ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹ + ¹²C₂p²q¹⁰ + ¹²C₃p³q⁹ + ¹²C₄p⁴q⁸ + ¹²C₅p⁵q⁷ + ¹²C₆p⁶q⁶ + ¹²C₇p⁷q⁵ + ¹²C₈p⁸q⁴

= = ¹²C₀(0.65)⁰(0.35)¹² + ¹²C₁(0.65)¹(0.35)¹¹ + ¹²C₂(0.65)²(0.35)¹⁰ + ¹²C₃(0.65)³(0.35)⁹ + ¹²C₄(0.65)⁴(0.35)⁸ + ¹²C₅(0.65)⁵(0.35)⁷ + ¹²C₆(0.65)⁶(0.35)⁶ + ¹²C₇(0.65)⁷(0.35)⁵ + ¹²C₈(0.65)⁸(0.35)⁴

= 0.000003379 + 0.00007531 + 0.0007692 + 0.004762 + 0.01990 + 0.05912 + 0.1281 + 0.2039 + 0.2367

= 0.6533

3 0
3 years ago
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denis23 [38]
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