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kipiarov [429]
4 years ago
9

Ron and Annie have $1,349.85 in their checking account. During the week, Annie goes to an ATM and withdraws $80. The following w

eek Ron deposits his paycheck of $699.65. Annie then pays bills online in the amounts of: $215.70, $53, $49.76, and $100.35. What is the current balance in their checking account

Mathematics
1 answer:
scZoUnD [109]4 years ago
3 0

Answer:

  $1550.69

Step-by-step explanation:

Deposits get added and withdrawals and bill payments get subtracted from the balance. The new balance is ...

  $1349.85 -80 +699.65 -215.70 -53 -49.76 -100.35 = $1550.69

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What is the value of x?
stiks02 [169]

Answer:

Step-by-step explanation:

Use Pythagoras Theorem

x² + 16² = 34²

x² + 256 =  1156

x² = 1156 - 256

x² = 900

x= √3*3*10*10

x= 3*10

x = 30

4 0
3 years ago
Read 2 more answers
If one end of a line segment is (4, 4) and midpoint is (-2,2), find the co-ordinates of the other end. ​
Soloha48 [4]

Answer:

-8,0

Step-by-step explanation:

From one end of a segment to the midpoint, it takes -6x to get to the midpoint. Based on that, you can go another 6 over the y graph and get -8 for x.

For y, the segment goes from 4 to 2 (a -2 over the x graph). you can infer then that the other end will be 0.

6 0
2 years ago
The statement 6<8 conveys the same information as 8 > 6
yan [13]

Answer:

yes 6<8 it shows 8 is larger the 6 and 8>6 is the same thing

7 0
3 years ago
8. For many computer tablets, the owner can set a 4-digit pass code to lock the device.
Anton [14]

Answer:

a. There 5040 different pass codes if the digits cannot be repeated

b. The probability that the pass code is 1234 is ≈ 0.0002

c.  The probability that two people both have a pass code of 1234 is 4×10^{-8}

Step-by-step explanation:

a. How many different 4-digit pass codes are possible if the digits cannot be repeated?

There are

  • 10 possibilities for the first digit
  • 9 possibilities for the second digit
  • 8 possibilities for the third digit
  • 7 possibilities for the fourth digit

Thus there are  10×9×8×7=5040 different pass codes

b. If the digits of a pass code are chosen at random and without replacement from the digits, what is the probability that the pass code is 1234

The probability that the first digit is 1 is \frac{1}{10}

The probability that the second digit is 2 is \frac{1}{9}

The probability that the second digit is 3 is \frac{1}{8}

The probability that the fourth digit is 4 is \frac{1}{7}

Thus  the probability that the pass code is 1234 is \frac{1}{10} * \frac{1}{9} *\frac{1}{8} *\frac{1}{7} =\frac{1}{5040} ≈ 0.0002

c. The probability that two people, who both chose a pass code by selecting digits at random and without replacement, both have a pass code of 1234?

The probability that the pass code of the one person is 1234 is 0.0002

The probability that the pass code of the other person is 1234 is 0.0002

Thus, the probability that two people have both pass code of 1234 is

0.0002×0.0002 = 4×10^{-8}

3 0
4 years ago
-8z+(4.5)+3.5z+7y-1.5
bogdanovich [222]

Answer:

  • \boxed{\sf{7y+3-4.5z}}

Step-by-step explanation:

In order to combine like terms, you have to isolate x and y from one side of the equation.

\sf{-8z+\left(4.5\right)+3.5z+7y-1.5}

<u>First, thing you do is remove parentheses.</u>

\Longrightarrow: \sf{-8z+4.5+3.5z+7y-1.5}

<u>Solve.</u>

<u>Then, you combine like terms.</u>

\Longrightarrow:\sf{-8z+3.5z+7y+4.5-1.5}

<u>Add/subtract the numbers from left to right.</u>

-8z+3.5z=-4.5z

<u>Rewrite the problem down.</u>

\Longrightarrow: \sf{-4.5z+7y+4.5-1.5}

<u>Solve.</u>

4.5-1.5=3

\Longrightarrow: \boxed{\sf{7y+3-4.5z}}

  • <u>Therefore, the correct answer is 7y+3-4.5z.</u>

I hope this helps, let me know if you have any questions.

6 0
2 years ago
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