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3 years ago

Mean median and mode of 12,14,17,19,9,12,7,19,8

Mathematics

2 answers:

Vanyuwa [196]3 years ago

7 0

Median; 12
mean; 13
mode; 12, 19

Artyom0805 [142]3 years ago

6 0

Mean is 13
median is 9
mode is 12 and19

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Hector is buying books at a book store. He buys 2 used books and 1 new book for $26. The new book costs $18. Each used book cost

sergejj [24]

Answer:

4 dollars

Step-by-step explanation:

Mark me as brainliest

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3 years ago

Read 2 more answers

Determine the number of possible solutions for a triangle with B=37 degrees, a=32, b=27

vladimir1956 [14]

Answer:

Two possible solutions

Step-by-step explanation:

we know that

Applying the law of sines

$\frac{a}{sin(A)}=\frac{b}{Sin(B)}=\frac{c}{Sin(C)}$

we have

$a=32\ units$

$b=27\ units$

$B=37\°$

step 1

Find the measure of angle A

$\frac{a}{sin(A)}=\frac{b}{Sin(B)}$

substitute the values

$\frac{32}{sin(A)}=\frac{27}{Sin(37\°)}$

$sin(A)=(32)Sin(37\°)/27=0.71326$

$A=arcsin(0.71326)=45.5\°$

The measure of angle A could have two measures

the first measure-------> $A=45.5\°$

the second measure -----> $A=180\°-45.5\°=134.5\°$

step 2

Find the first measure of angle C

Remember that the sum of the internal angles of a triangle must be equal to $180\°$

$A+B+C=180\°$

substitute the values

$A=45.5\°$

$B=37\°$

$45.5\°+37\°+C=180\°$

$C=180\°-(45.5\°+37\°)=97.5\°$

step 3

Find the first length of side c

$\frac{a}{sin(A)}=\frac{c}{Sin(C)}$

substitute the values

$\frac{32}{sin(37\°)}=\frac{c}{Sin(97.5\°)}$

$c=Sin(97.5\°)\frac{32}{sin(37\°)}=52.7\ units$

therefore

the measures for the first solution of the triangle are

$A=45.5\°$ , $a=32\ units$

$B=37\°$ , $b=27\ units$

$C=97.5\°$ , $b=52.7\ units$

step 4

Find the second measure of angle C with the second measure of angle A

Remember that the sum of the internal angles of a triangle must be equal to $180\°$

$A+B+C=180\°$

substitute the values

$A=134.5\°$

$B=37\°$

$134.5\°+37\°+C=180\°$

$C=180\°-(134.5\°+37\°)=8.5\°$

step 5

Find the second length of side c

$\frac{a}{sin(A)}=\frac{c}{Sin(C)}$

substitute the values

$\frac{32}{sin(37\°)}=\frac{c}{Sin(8.5\°)}$

$c=Sin(8.5\°)\frac{32}{sin(37\°)}=7.9\ units$

therefore

the measures for the second solution of the triangle are

$A=45.5\°$ , $a=32\ units$

$B=37\°$ , $b=27\ units$

$C=8.5\°$ , $b=7.9\ units$

6 0

3 years ago

What's the answer? I need it please help me !!

Softa [21]

I believe this is J with a radius of 6

the formula for circumference is 2(pi)r so i just took the 12(pi) and divided it by 2, which left me with 6(pi), meaning that 6 would be the "r" in the situation

6 0

3 years ago

Hey can you please help me posted picture of question

Olenka [21]

Hello!

The variable "n" represents an exponent.

An exponent is used to multiply a number (b) by itself "n" times. For example, if the expression were to read (b²) it would represent the following:

b × b

The same could be said for (b³):

b × b × b

Even ( $b^{10}$ ):

b × b × b × b × b × b × b × b × b × b

I hope this helps!

6 0

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

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

4 years ago