Answer:
<u>II. Second table</u>
A B Total
C 0.25 0.75 1.00
D 0.35 0.65 1.00
Total 0.30 0.70 1.00
Explanation:
<h2>Tables</h2>
<u>I. First table </u>
A B Total
C 0.25 0.25 0.50
D 0.25 0.25 0.50
Total 0.50 0.25 1.00
<u>II. Second table</u>
A B Total
C 0.25 0.75 1.00
D 0.35 0.65 1.00
Total 0.30 0.70 1.00
<u>III. Third table</u>
<u></u>
A B Total
C 0.75 0.25 0.50
D 0.25 0.75 0.50
Total 0.50 0.50 1.00
<u>IV. Fourth table</u>
A B Total
C 0.65 0.35 1.00
D 0.35 0.65 1.00
Total 1.00 1.00 1.00
<h2>Solution</h2>
A <em>conditional relative frequency table</em> shows the relative frequencies determined upon a row or column.
There are two types of relative conditional frequency table: 1) row conditional relative frequency, and 2) column conditional relative frequency.
When you divide the joint frequency by the marginal frequency of the column total you have the row conditional frequency table. When you dividethe joint frequency by the row total you have the colum conditional frequency table.
In a row conditional relative frequency each total of the right hand column equals 1. This is the case of the second table.
In a column conditional relative frequency each total of the bottom row equals 1. This is not happening with any of the shown tables.
Hence, only the second table could be a conditional relative frequency table.
Answer is B. Have a nice day.
Answer:
Step-by-step explanation:
Your cousin wants to buy a coat and can spend at most $95. This means that the total amount of money that he can spend in buying the coat would be lesser than or equal to $95.
He has a coupon for $20 off any coat. Let x represent the cost of the coat. The amount that he can pay for the coat will be x - 20.
The inequality to find the original prices of coats he can afford would be
x - 20 lesser than or equal to 95
$2.50+3x=$9.25, if you need to work it out, then you do $9.25-$2.50. You take your answer and divide by three... whatever your solution is, equals X. There you go!
Answer:
No, its not equivalent
1/2 is equivalent though
Please mark as brainliest
