Which is the least valid way to stimulate how many boys and girls are in a random sample of 20 students from a school that is ha lf boys and half girls? A. Flip a coin 20 times, assigning one out come to boys and the other to girls
B. Drop 20 coin at once and count the number of each outcome
C. Count how many boys and girls are in your mat class and use a proportion
D. Have a calculator generate 20 random integers and count the number of even and odd integers
2 answers:
Total number of Student in the School in which half are boys and Half are girls =20
Probability of an event
Probability of Number of Boys as well as Probability of Number of Girls
The Most appropriate option is
Option C
Count how many boys and girls are in your math class and use a proportion.
In Option A, B and D
The probability of each outcome that is Success and Failure will be nearly equal to 0.50 .
But in Option C, you can't say precisely that Probability will be equal to , 0.50 in each case , that is Success and Failure.
You might be interested in
Answer:
8/9
Step-by-step explanation:
1 blue, 2 yellow, 7 red, 6 green, and 2 purple marbles = 18 marbles
The number that are not yellow = total - yellow
P( not yellow) = number that are not yellow / total
= (18-2) / 18
= 16/18
=8/9
The length of that semi circle is 6pi. So the perimeter should be 6pi + 24
Answer:
See picture below.
Ok -7-7=-14 because your subtracting a negative with a negative and then it will be a negative then if you multiply a negative and a negative it will be a positive
The answer is -.2.5 hope that help