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Kay [80]
2 years ago
14

Is 122/48 in its simplest form or is it’s simplest form 61/24?

Mathematics
2 answers:
horsena [70]2 years ago
8 0

Answer:

61/24 is its simplest form! :)

Step-by-step explanation:

sergiy2304 [10]2 years ago
3 0
The simplest form is 61/24 as your able to divide both the numerator and denominator of 122/48 by 2 to simplify it
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Estimate 9.8% of 320
Jobisdone [24]
10% of 320 = 32

So 9.8% of 320 should be some number between 31 and 32
8 0
2 years ago
Pleaseee helpppp answer correctly !!!!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!
Anit [1.1K]

Answer:

540°

General Formulas and Concepts:

<u>Math</u>

  • Counting

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Geometry</u>

  • Sum of Angles: 180(n - 2)°

Step-by-step explanation:

<u>Step 1: Define</u>

We are given a 5-sided polygon (irregular pentagon)

n = 5

<u>Step 2: Find Sum</u>

  1. Substitute in <em>n</em> [Sum of Angles]:                                                                    180(5 - 2)°
  2. (Parenthesis) Subtract:                                                                                   180(3)°
  3. Multiply:                                                                                                           540°
5 0
2 years ago
Read 2 more answers
Evaluate the expression when x=12,y=-3 and z=-2
Degger [83]

Answer:

(12)(3)(-2)

Step-by-step explanation:

5 0
2 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
What is the image point of (1, -1) after a translation left 4 units and down 5 units?
seraphim [82]
Answer is A (-4;-6).
6 0
3 years ago
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