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3 years ago

WILL GIVE BRAINLIEST FOR BEST ANSWER DONT ANSWER IF YOU DONT KNOW AND NO VIRUS LINKS OR I WILL REPORT YOU!!!

Mathematics

2 answers:

Digiron [165]3 years ago

5 0

56 is the answer :)

VLD [36.1K]3 years ago

4 0

Answer:

5/6

Step-by-step explanation:

Number of options not including 2 = 5

Number of total options = 6

(Will add the other questions if asked)

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What is the length of the line segment that is graphed from (+2, +5) to (+2, +8)?

Natasha_Volkova [10]

The length of the line segment is 3.

Hope this helps!

6 0

3 years ago

5. After paying a tax of 8 palsa per rupee a man saves Rs. 18,400 per month Then his monthly income is 2 Points) 20,000

Hunter-Best [27]

Answer:

92% of Income is Rs 18400

Step-by-step explanation:

I = 18400/.92 = Rs 20000

8 0

2 years ago

2+2..................

Nana76 [90]

Answer:

4..............21...........

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

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

Dakota has \dfrac4{10} \text { kg} 10 4 kgstart fraction, 4, divided by, 10, end fraction, start text, space, k, g, end text o

Savatey [412]

Answer:

There are 2.22 grams of clay are in each pot.

Step-by-step explanation:

Given:

Dakota has 4/10 kg of clay. He divides the clay to make 888 equal-sized pots.

Now, to find the number of grams of clay are in each pot.

Dakota has clay = $\frac{4}{10} \ kg.$

So, by using conversion factor we convert it into grams:

$\frac{4}{10} \times 1000\\\\=\frac{4000}{10} \\\\=400\ grams.$

Quantity of clay he has = 400 grams.

Number of equal-sized pots to make = 888.

Now, to get the quantity of grams of clay are in each pot we divide number of equal-sized pots to make by quantity of clay he has:

$\frac{888}{400}$

$=2.22\ grams.$

Therefore, there are 2.22 grams of clay are in each pot.

3 0

3 years ago

*WILL GIVE BRAINLIEST FOR BEST ANSWER DONT ANSWER IF YOU DONT KNOW AND NO VIRUS LINKS OR I WILL REPORT YOU!!!*

WILL GIVE BRAINLIEST FOR BEST ANSWER DONT ANSWER IF YOU DONT KNOW AND NO VIRUS LINKS OR I WILL REPORT YOU!!!