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xz_007 [3.2K]
2 years ago
10

A teacher has 24 students in her class. She wants to divide them into 4 equal groups. Write an expression to represent the numbe

r of students in each group.
Mathematics
1 answer:
Karo-lina-s [1.5K]2 years ago
6 0

let's set up an equation

4x=24

solve normally:

x=6

so the expression is 4x=24 and she can have 6 kids in each group

You might be interested in
<img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B12x-7%3D0%20%5C%5C" id="TexFormula1" title="2x^{2} +12x-7=0 \\" alt="2x^{2}
m_a_m_a [10]

Answer:

-2. 46

- 9.54

Step-by-step explanation:

  • 2x²+12x-7=0
  • 2x²+12x+18= 25
  • 2*(x²+6x+9)=25
  • 2*(x+6)²= 5²
  • x+6= 5√2/2 ⇒ x= 5√2/2 -6 ≈ -2. 46
  • x+6= -5√2/2 ⇒ x= -5√2/2 -6 ≈ - 9.54
4 0
2 years ago
Circle has rad of 1.5m what angle subtended at center of circle by arc of 1m
Mumz [18]
120 degrees or 0.67 radians
5 0
2 years ago
What is <br> 4^ -3<br> 4^ -2<br><br> 2^ -3 + 3<br> 2^ -2 +3<br> 2^ -1 + 3
Scrat [10]
1/4³ or 1/64
1/4² or 1/16
1/2³ + 3 or 3 1/8 or 3.125
1/2²+3 or 1/4 +3 or 3 1/4 or 3.25
1/2¹ +3 or 3 1/2 or 3.5
6 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:

<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
3 years ago
Sort these in either Combination or Permutation PLEASE HELP!!!! 100 POINTS
balandron [24]
A permutation is where there is a difference in positions where the things will be selected into (order matters). Combination is where order doesn’t matter. I have sorted them but I am not sure about the starting line up one. If the players are selected into specific positions then it would be a permutation.
3 0
2 years ago
Read 2 more answers
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