I think it’s b y axis. I believe this because 0 is going to be on either the y or c axis therefore c and d are eliminated and you go up three on the graph so B y axis.
Answer:
151+8x>_367
Step-by-step explanation:
x=sets of glasses.
151+8x>_367
^is greater than equal to i dont
have a sign for that.
Answer:
Mrs. Etercsid
Step-by-step explanation:
One thing we can do is just multiply the averages by the percents in each class and compare from there.
To get from percentages to decimals, we can divide by 100.
For Mr. Tats:
Homework: 25 % -> 0.25
Participation: 10% -> 0.1
Test: 40% -> 0.4
Final: 0.25 -> 0.25
For Mrs. Etercsid:
Homework: 15% -> 0.15
Participation: 10% -> 0.1
Test: 60% -> 0.6
Final: 15% -> 0.15
We can then multiply the averages by the decimals for each teacher and add them up.
Mr. Tats:
0.25 * 81 + 0.1 * 57 + 0.4 * 93 + 0.25 * 87 = 84.9
Mrs. Etercsid:
0.15 * 81 + 0.1 * 57 + 0.6 * 93 + 0.15 * 87 = 86.7
You would get the higher grade with Mrs. Etercsid
Answer:
0.1225
Step-by-step explanation:
Given
Number of Machines = 20
Defective Machines = 7
Required
Probability that two selected (with replacement) are defective.
The first step is to define an event that a machine will be defective.
Let M represent the selected machine sis defective.
P(M) = 7/20
Provided that the two selected machines are replaced;
The probability is calculated as thus
P(Both) = P(First Defect) * P(Second Defect)
From tge question, we understand that each selection is replaced before another selection is made.
This means that the probability of first selection and the probability of second selection are independent.
And as such;
P(First Defect) = P (Second Defect) = P(M) = 7/20
So;
P(Both) = P(First Defect) * P(Second Defect)
PBoth) = 7/20 * 7/20
P(Both) = 49/400
P(Both) = 0.1225
Hence, the probability that both choices will be defective machines is 0.1225
