¹⁶/₅ or 3 ¹/₅ or 3.2 ounces.
Given:
Question:
How much more chocolate did Mrs. Harrison use in the brownies than in the frosting?
The Process:
From the information above, we made a suitable diagram as follows:
Hence, one part equal to .
She used of the chocolate to make some frosting, meaning two of five parts of the chocolate.
And she used the rest, that is, of the chocolate to make brownies, meaning three of five parts of the chocolate.
Finally, let us find out how much more chocolate did Mrs. Harrison use in the brownies than in the frosting. Before that, prepare in the form of a fraction of the part in question.
or one of five parts of the chocolate.
We declare this fraction into ounces by multiplying it by 16 ounces.
Thus Mrs. Harrison used 3.2 ounces more chocolate in brownies than in the frosting.
Keywords: Mrs. Harrison, used 16 ounces, dark chocolate, while baking, she used 2/5, to make some frosting, and used the rest to make brownies. How much more chocolate did Mrs. Harrison use in the brownies than in the frosting
choose a second tile. Find the probability of the dependent event(s) both
occurring.
1/2/3/4/5/6/7
-
Choosing a 6 and then a prime number
Choosing two odd numbers
Answer:
Choosing a 6 and then a prime number:2/21
Choosing two odd numbers: 2/7
Step-by-step explanation:
Choosing a 6 and then a prime number:
The probability of choosing 6 out of 1, 2, 3, 4, 5, 6 and 7 is one event divided by number of total events (which is equal to 7). That results in 1/7. Once 6 is chosen, the probability of choosing a prime number (prime number is a number that can only be divided by 1 and itself) out of 1, 2, 3, 4, 5 and 7 is 4/6 (prime numbers are 2, 3, 5, 7 in total there are 4 number and total number of events are 6). Finally, the probability of choosing a 6 and then a prime number is (1/7)*(4/6)=2/21.
Choosing two odd numbers:
The probability of choosing 1st odd number is 4/7 (number of odd numbers is 4 which includes 1, 3, 5, 7 and the number of total events is 7). Once 1st odd number is chosen, the probability of choosing 2nd odd number is 3/6 (number of odd numbers is 3 - because 1 odd number is already chosen and the number of total events is 6). Finally, the probability of choosing two odd numbers in a sequence is (4/7)*(3/6)=2/7.
Answer:
112 and 12/49
Step-by-step explanation:
I'm not really sure if this answer is accurate but hope it helps somewhat. This is the answer for 35 5/7 x 22/7