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1 answer:
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9514 1404 393
Answer:
$189.32
Step-by-step explanation:
Add all of the numbers shown to the beginning balance to find the ending balance.
$92.19 -39.35 +49.76 +87.35 +9.97 -10.60 = $189.32
Tristan's balance at the end of the month was $189.32.
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There should be no confusion. Money deposited into the account increases the balance by the amount deposited.
Money spent from the account decreases the balance by the amount spent.
Here, amounts deposited are given a plus sign (+); amounts spent are given a minus sign (-).
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<em>Additional comment</em>
It can be tedious to enter these numbers correctly into a calculator. Many folks prefer a calculator that provides a "tape", a record of the amounts and operations the calculator uses in the computation. Alternatively, you can put the numbers in a spreadsheet and use the SUM function to do the addition for you. That, too, makes it easy to check that numbers have been entered correctly.
The attachment shows the calculation done in a spreadsheet.
Hello!
Let's write some important information contained in the exercise:
• hotdogs: $2.59 (,x,)
,• hamburgers: $5.29 (,y,)
He needs 11 packages. Let's write it:
• x + y = 11
He spent a total of $39.29. We can write it as:
• 2.59x + 5.29y = 39.29
Now, let's solve these two equations as a linear system:
First, let's isolate x in equation A:
Now, we will replace where's x by 11-y in equation B:
As I called the hamburgers as 'y', we know that he bought 4 packages of hamburgers.
Answer:
The probability that he ends up with a full house is 0.0083.
Step-by-step explanation:
We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.
We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).
We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);
- If he is given with two kings.
- If he is given one king and one ace.
Only in these two situations, he will end up with a full house.
Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.
So, the ways in which we can draw two kings from available three kings is given by = {∵ one king is already there}
= {∵
}
= = 0.0028
Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =
=
= = 0.0055
Now, probability that he ends up with a full house =
= = <u>0.0083</u>.