75÷500 = 0.15 multiply with a hundred to get percentage equals 15℅
<u>Answer:</u>
1/5
<u>Step-by-step explanation:</u>
To find this you would need to multiply the probability of pulling a white marble to the probability of pulling out a green marble.
1)First you would need the probability of pulling out a white marble. There are 10 marbles in total and out of those 2 are white. So the probability of pulling out a white marble would be 2/10. If you simplify that you would get 1/5 for the probability of pulling out a white marble.
2)Next, you would find the probability of pulling out a green marble. Using the same process that we used to find the probability of pulling out a white marble, we would find the answer to be 3/10. All that we did here was <em>green marbles/total marbles</em>. By filling that in we got 3/10 for the probability of pulling out a green marble.
3)Now all that is left is doing <em>probability of pulling a white marble × probability of pulling out a green marble</em>. This would be 1/5 × 3/10. After solving the answer would be 3/15 which we would simplify down to 1/5 as our final answer.
Answer:
The smallest value is 2) the opposite of 50 because it is farthest left of the zero (this makes it the smallest)
Step-by-step explanation:
1) = 100
2) = -50
3) = 0
4) = -25
5) = -5
Answer:
7.71
Step-by-step explanation:
The diameter is 3, so the arc length is 180/360 * 3pi or 3pi/2. Now you add the 3 in the base, so it is 3pi/2 + 3 or approximately 7.71 (I used a calculator for that).
I've attached the complete question.
Answer:
Only participant 1 is not cheating while the rest are cheating.
Because only participant 1 has a z-score that falls within the 95% confidence interval.
Step-by-step explanation:
We are given;
Mean; μ = 3.3
Standard deviation; s = 1
Participant 1: X = 4
Participant 2: X = 6
Participant 3: X = 7
Participant 4: X = 0
Z - score for participant 1:
z = (x - μ)/s
z = (4 - 3.3)/1
z = 0.7
Z-score for participant 2;
z = (6 - 3.3)/1
z = 2.7
Z-score for participant 3;
z = (7 - 3.3)/1
z = 3.7
Z-score for participant 4;
z = (0 - 3.3)/1
z = -3.3
Now from tables, the z-score value for confidence interval of 95% is between -1.96 and 1.96
Now, from all the participants z-score only participant 1 has a z-score that falls within the 95% confidence interval.
Thus, only participant 1 is not cheating while the rest are cheating.
