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4 years ago

12

Allison has five times as many baseball cards as football cards and all she has 120 baseball and football cards how many basebal

l cards does Allison have

1 answer:

Shtirlitz [24]4 years ago

7 0

The answer is 600 because you have to multiply 120 by 5 and the sum is 600

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What is the area of the figure?

Scrat [10]

Answer:

15 sq. cm.

Step-by-step explanation:

Each of the full squares are 1 so each of the half squares are .5

There are 12 full squares and 6 half squares. If you add those half squares you will get 15 sq. cm.

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3 years ago

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Please help me. Hurry, linear equations.

Lisa [10]

Answer:

Let's simplify step-by-step.

3−3/4x−1/4x−7

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3 years ago

6. In the diagram, find the value of y that makes b paralo to,

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Answer:

Step-by-step explanation:

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3 years ago

Find the equation of the exponential function represented by the table below:

marshall27 [118]

Answer:

$y=0.2(0.5)^x$ is the answer.

Step-by-step explanation:

Let the equation of the exponential function is,

$y=a(b)^x$

Two points from the given table which lie on the graph are (0, 0.2) and (1, 0.1)

For a point (0, 0.2),

0.2 = $a(b)^0$

a = 0.2

For another point (1, 0.1),

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Therefore, exponential function defined by the given table is,

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4 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:

<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.

4 0

4 years ago