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FromTheMoon [43]
3 years ago
15

Select the correct answer. A baseball is hit from an initial height of 3 feet and reaches a maximum height of 403 feet. Which fu

nction could be used to model this situation, where is the height, in feet, after t seconds?
Mathematics
1 answer:
Oliga [24]3 years ago
7 0

Answer:

ΔS = 1/2gt²

Step-by-step explanation:

Given

Initial height of the ball = 3 feet

Maximum height of the  ball = 403 feet

To get the function that best suits the situation after t seconds, we will use the equation of motion;

ΔS = ut + 1/2gt²

ΔS is the change in height

u is the initial velocity of the body = 0m/s

g is the acceleration due to gravity = 9.81m/s²

t is the time taken for the ball to reach its maximum height.

ΔS = 403-3

ΔS = 400ft

Substituting the given values into the formula;

ΔS = ut + 1/2gt²

400 = 0 + 1/2(9.81)t²

400 = 4.905t²

t² = 400/4.905

t² = 81.55

t = √81.55

t = 9.03secs

The situation that best model the situation is ΔS = ut + 1/2gt²

ΔS = 0 + 1/2gt²

ΔS = 1/2gt²

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(c) 0.906

Step-by-step explanation:

Let's denote the events as follows:

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<em>S</em> = The email is spam

<em>V</em> = The email is valid.

The information provided to us are:

  • The probability of the word free occurring in a spam message is,             P(F|S)=0.60
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        P(S)=0.20

First let's compute the probability of valid messages:

P (V) = 1 - P(S)\\=1-0.20\\=0.80

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})

The probability that a message contains the word free is:

P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

The probability that a message is spam provided that it contains free is:

P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

The probability that a message is valid provided that it does not contain free is:

P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566

The probability that a message is valid provided that it does not contain free is approximately 0.906.

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