Answer:
GCF < 2 : 2 and 7, 12 and 13
GCF = 2 : 2 and 6
GCF = 3 : 3 and 6
GCF > 3 : 4 and 12, 6 and 18
Step-by-step explanation:
The greatest common Factor (GCF) of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers.
(a) 2 and 7: Both 2 and 7 are prime numbers, therefore the GCF of 2 and 7 is 1 which is less than 2.
(b) 2 and 6: Both 2 and 6 are even numbers, therefore the GCF of 2 and 6 is 2.
(c) 4 and 12: Both 4 and 12 are even numbers and multiple of 4, therefore the GCF of 4 and 12 is 4 which is greater than 3.
(d) 12 and 13: 13 is a prime number, therefore the GCF of 12 and 13 is 1 which is less than 2.
(e) 6 and 18: Both 6 and 18 are even numbers and multiple of 6, therefore the GCF of 6 and 18 is 6 which is greater than 2 and greater than 3.
(f) 3 and 6: Both are multiple of 3, therefore the GCF of 3 and 6 is 3.
To determine the median, we need to set up our numbers from least to greatest, and then place T in later to figure out what T is.
8, 9, 9, 9, 10, 11, 12, 15. Cross out the smallest number with the largest number.
9, 9, 9, 10, 11, 12.
9, 9, 10, 11.
9, 10.
9.5 is our median currently.
Since we need to get a number after 10 to make 10 the median, let's use 12.
8, 9, 9, 9, 10, 11, 12, 12, 15.
9, 9, 9, 10, 11, 12 ,12.
9, 9, 10, 11, 12.
9, 10, 11.
10 is now our median since we inserted 12 into our list.
Your answer is 12.
I hope this helps!
Answer:
<u>After four games, a player can lose up to $ 16 to win up to $ 26. These are the probabilities for every game:</u>
<u>1/8 or 12.5% of landing three "heads"</u>
<u>3/8 or 37.5% of landing two "heads"</u>
<u>4/8 or 50% of landing no or only one "head".</u>
Step-by-step explanation:
1. Let's review the information given to us to answer the question correctly:
If three coins land "heads" the player wins $ 10
If two coins land "heads" the player wins $ 5
Cost of playing = $ 4
2. What is the player's expected outcome after four games?
Probability of two coins out of three lands "heads" = 3/8
Probability of three coins out of three lands "heads" = 1/8
Now, let's calculate the player's expected outcome, as follows:
Four games:
Cost = 4 * 4 = $ 16
Worst-case scenario: No wins
Best-case scenario: 4 out of 4 of $ 10 win
Worst-case scenario profit or loss = 0 - 16 = Loss of $ 16
Best-case scenario profit or loss = 40 - 16 = Profit of $ 24
After four games, a player can lose up to $ 16 to win up to $ 26. These are the probabilities for every game:
1/8 or 12.5% of landing three "heads"
3/8 or 37.5% of landing two "heads"
4/8 or 50% of landing no or only one "head".
Answer:
0.12 ± 1.96 * √(0.12(0.88) / 100)
Step-by-step explanation:
Confidence interval :
Phat ± Zcritical * √(phat(1 -phat) / n)
Phat = 12/100 = 0.12
1 - phat = 0.88
Zcritical at 95% = 1.96
Hence, we have :
0.12 ± 1.96 * √(0.12(0.88) / 100)
0.12 ± 1.96 * 0.0324961
0.12 ± 0.0636924
Lower boundary = (0.12 - 0.0636924) = 0.0563
Upper boundary = 0.12 + 0.0636924 = 0.1837
7 is the correct answer
because a coefficient is the number next to the variable which is k. And the second term is 7k.
