Two events are occurring:
1) Rolling a die
Sample Space = {1,2,3,4,5,6}
Total number of outcomes in sample space = 6
Favorable outcomes = Odd number
Number of Favorable outcomes = 3
Probability of getting an odd number = 3/6
2) Tossing a coin
Sample Space = {H, T}
Probability of getting a head= 1/2
The probability of getting odd number and head will be the product of two probabilities, which will be = 3/6 x 1/2 = 3/12
Thus there is 3/12 = 1/4 (0.25 or 25%) probability of getting an odd number and a head in given scenario.
So correct answer is option C
Should be four times to get in between 9 and 10
Sorry i can’t quite see the image can you do another one
You have to solve this algebraicly.
To do this set up the proportion and add x to each term.
You get 4x and 7x
Set this equal to 33
4x+7x=33
Combine like terms and get 11x=33
Divide by 11
x=3
Now go back to 4x and 7x
Multiply them by x and get 12(4×3=12) and 21(7×3=21)
One group is 12 and the other is 21.
Let the total number of questions in the math test be defined by the variable x.
Now, we know that Parker correctly answered 35 questions. These 35 questions make 70% of the total number of questions on his math test.
The information we have above can be expressed as an equation given below:
70% of x=35
This can be rewritten as:
Thus, to find the total number of questions we will have to isolate x.
Therefore, there were a total of 50 questions in the math test.
