Jessica is selling books during the summer to earn money for college. She

Divide:5.52 x 1031.2 x 105Write your answer in scientific notation. For example, you'd enter 2.3*10^4 for 2.3 x 104

IrinaK [193]

The question can be solved as follows:

$\frac{5.52\times10^3}{1.2\times10^5}=\frac{5.52}{1.2}\times10^{-2}=4.6\times10^{-2}$

Hence the value of the given expression in scientific notation is expressed as:

$\frac{5.52\times10^3}{1.2\times10^5}=4.6\times10^{-2}$

8 0

1 year ago

Please help me with my math homework??

qaws [65]

-2 1/2 / 6 would be -5/12, just turn - 2 1/2 into a decimal and divide by 6 :)

4 0

3 years ago

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:

F = The word free occurs in an email

S = The email is spam

V = 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.

4 0

3 years ago

Plzzzzzz help I don’t understand I know what to do I need to use PEMDAS but my answer is still wrong

Effectus [21]

simplify exponent: 20*(9-4)/50

simplify parentheses: 20*5/50

multiply: 100/50

divide: 2

6 0

3 years ago

Please help. Don’t understand this math problem!

oksian1 [2.3K]

Answer:

{ 4 }

Step-by-step explanation:

First you need to solve the equation. Then you need to write the solutions as a set.

__

I like to use a graphing calculator to solve equations involving roots. That helps avoid extraneous solutions. Here, the graph of the left side intersects the graph of the right side at x=4. There is exactly one solution, so the solution set is ...

{ 4 }

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Solving this algebraically, you would isolate the radical, then "undo" it by squaring both sides of the equation. Then solve the resulting quadratic. That will have 2 solutions, only one of which will work in the original equation.

√(5x +16) -2 = 3x -8 . . . . given

√(5x +16) = 3x -6 . . . . . . add 2

5x +16 = (3x -6)^2 . . . . . square both sides

5x +16 = 9x^2 -36x +36 . . . . expand the square

9x^2 -41x +20 = 0 . . . . . . subtract (5x+16) to put into standard form

Factors of 9·20 = 180 that have a sum of 41 are 5 and 36, so we can rewrite this equation as ...

9x^2 -36x -5x +20 = 0

9x(x -4) -5(x -4) = 0 . . . . . . factor by pairs

(9x -5)(x -4) = 0

The values of x that make this equation true are the values that make the factors be zero: 5/9 and 4. Trying these in the equation, we find ...

√(5(5/9) +16) -2 = 13/3 -2 = 2 1/3 . . . . value of the left side for x = 5/9

3(5/9) -8 = -6 1/3 . . . . . . value of the right side for x=5/9; not the same

x = 5/9 is NOT A SOLUTION

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√(5(4)+16) -2 = √36 -2 = 4 . . . . value of the left side for x = 4

3(4) -8 = 12 -8 = 4 . . . . . . value of the right side for x=4. These are the same, so ...

x = 4 is a solution

The set of solutions is then ...

{ 4 } . . . . . solution set

5 0

3 years ago