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soldi70 [24.7K]

2 years ago

11

Dre' wants to add 1 foot of grass all around his doghouse for a dog run. If the doghouse is 2 feet by 5 feet, what would the are

a of the grassy region be

2 answers:

IrinaK [193]2 years ago

5 0

Answer:

10 feet

Step-by-step explanation:

you would just have to do 5*2 which equals 10

nexus9112 [7]2 years ago

4 0

10 ft

all you would need to do is 5x2

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Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 5.08 and angle θA = 25.6 ∘ measured in the sense from the +x

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Answer:

(A⃗ ×B⃗ )⋅C⃗ = - 76.415

Step-by-step explanation:

First we need to calculate (A⃗ ×B⃗ ) :

(A⃗ ×B⃗ ) = A.B.sin (α).n

Where A is the magnitude of A⃗

Where B is the magnitude of B⃗

Where α is the angle between A⃗ and B⃗ = 63.9 - 25.6 = 38.3

Finally n is the vector orthogonal to A⃗ and B⃗

n magnitude is 1 and his direction is given by the right hand-rule

so n = ( 0 , 0 , 1 )

(A⃗ ×B⃗ ) = A.B.sin (α).n = 5.08 . 3.94 . sin (38.3) . (0 , 0 , 1 ) = (0,0,12.4)

C⃗ can be written as C.(0,0,-1) because of his +z - direction

C.(0,0,-1) = 6.16.(0,0,-1) = (0,0,-6.16)

(A⃗ ×B⃗ )⋅C⃗ = (0,0,12.4).(0,0,-6.16) = -76.41480787 = -76.415

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An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of

ANEK [815]

Answer:

(a) 0.152

(b) 0.789

(c) 0.906

Step-by-step explanation:

Let's denote the events as follows:

<em>F</em> = The word free occurs in an email

<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, $P(F|S)=0.60$
The probability of the word free occurring in a valid message is, $P(F|V)=0.04$
The probability of spam messages is,

$P(S)=0.20$

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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2 years ago