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Alona [7]
3 years ago
7

David has a checking account. His opening balance on 3/1 was $120. The bank charges a 30¢ fee for each debit card transaction an

d a 20¢ fee for each check. David made the following transactions:
Used debit card for $20 movie tickets on 3/5
Deposited $38 paycheck on 3/12
Transferred $25 to savings account on 3/16
Withdrew $20 cash from ATM on 3/20



What is David’s balance after the transaction on 3/20?
Mathematics
2 answers:
sukhopar [10]3 years ago
7 0
0! David's balance is $0 after the transaction.
Luden [163]3 years ago
6 0
David has zero balance in his account.
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U need question problem 2 to solve problem 3
Sergio [31]

Answer:

You are correct

Step-by-step explanation:

Start with 1 1/2. This can be made into an improper fraction which is 3/2

Now multiply both top and bottom of 3/2 by 5

(3*5)/(2 * 5) = 15 / 10

16/10 is just slightly bigger than 15/10

3 0
3 years ago
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Find the optimal solution for the following problem. (Round your answers to 3 decimal places.)
Sphinxa [80]

Answer:

x=2.125

y=0

C=19.125

Step-by-step explanation:

To solve this problem we can use a graphical method, we start first noticing the restrictions x\geq 0 and  y\geq 0, which restricts the solution to be in the positive quadrant. Then we plot the first restriction 8x+10y\leq 17 shown in purple, then we can plot the second one 11x+12y\leq 25 shown in the second plot in green.

The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.

So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.

3 0
3 years ago
CAN SOMEONE HELP ME FIND THE CIRCUMFERENCE
aleksandrvk [35]
Tbh idk all I got was 51.496 I’m dumb i just want points ^_^
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3 years ago
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A study involves a population of 400 tall women. This population has a mean height of 179.832 cm and a standard deviation of 12.
Fofino [41]

Answer:  0.0386

Step-by-step explanation:

Given: The population of 400 tall women has a mean height(\mu) of 179.832 cm and a standard deviation(\sigma) of 12.192 cm.

Let X be a random variable that represents the height of woman.

Sample size : n= 50

The probability that the mean for this sample group is above 182.88 will be :

P(\overline{X}>182.88)\\\\=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{182.88-179.832}{\dfrac{12.192}{\sqrt{50}}})\\\\ =P(Z>1.7678)\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-P(Z

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At Martin Nature Center, the park ranger was showing Mrs. Rhames' class a comparison of a tree's age and the diameter of the tre
sdas [7]
If x is the age and y is the diameter of the trunk, write the equation:
ax+b=y

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5 0
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