Lets say that J is for Jennys messages, B is for Boris’ and E is for Erics (or Ericas?).
79 = J + B + E
J = E - 6
B = 3E
Since both Jenny’s and Boris’ number of text messages has some kind of relation to Eric’s number of messages, we can say that:
79 = E - 6 (Jenny) + 3E (Boris) + E (Eric), so:
79 = E - 6 + 3E + E /combine like terms
79 = 5E - 6 /add 6 to both sides
79 + 6 = 5E /switch sides + combine like terms
5E = 85 /divide both sides by 5
E = 85/5
E = 17
So, Eric sent 17 messages. Lets get back to those equations we made for Jenny and Boris:
J = E - 6 = 17 - 6 = 11
B = 3E = 3 * 17 = 51
And we can check if it adds up: 51 + 11 + 17 = 79, which is what we were supposed to get.
Answer:
1/4
Step-by-step explanation:
there are only 4 suits so there is a 1/4 chance of getting a card from the same suit as the other card was added back into the pack
To get the loss or gain, think about the conveying estimation of the bonds with the sum we pay to recover the bonds. In the event that the conveying esteem is more greater than the money paid, there is a gain on the security retirement. In the event that the carrying value is not as much as the money paid, there is a loss on the security retirement.
So plugging in:Par value - 109,000Carrying Value - 102,700
There is a loss of $6,300.
It is A because it can’t be B and it can’t be C and it can’t be D so my answer is A
Answer:
In one Pb&j it is 385 and in three it has 1,155 (385×3=1,155 )
