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4 years ago

Why would a bank charge a fee to noncustomers who use it’s ATM?

Mathematics

2 answers:

Naily [24]4 years ago

5 0

Answer:

Bank charges fee to non customers while using ATM. ATM's frequently charge fees to users who are not account holders from that same bank whose ATM you are operating.

This fee goes towards the compensation for the costs associated with owning and operating the machine.

Same bank persons pay many types of fee yearly to their banks, so same bank ATM do not charge any fee.

When using other banks, your bank has to pay some interchange amount to that bank, this is also the reason why banks charge fee, basically they recoup the money that they will have to pay to other banks for interchange.

serg [7]4 years ago

3 0

So what exactly is going on here?

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2 years ago

Select the correct answer from each drop-down menu. Julie invests $200 per month in an account that earns 6% interest per year,

Alona [7]

Answer:

After 10 years, Julie's account balance will be $ 363.88 and Leah's account balance will be $ 411.75, thus Leah will have more money in her account.

Step-by-step explanation:

Since Julie invests $ 200 per month in an account that earns 6% interest per year, compounded monthly, and Leah invests $ 250 per month in an account that earns 5% interest per year, compounded monthly, to determine the amount of each after 10 years, the following calculations must be performed:

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3 years ago

N-8+n=1-4n solve for n

MrRissso [65]

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Step-by-step explanation:

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4 years ago

If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta + (pi) / (6) ) (c) cos (the

EleoNora [17]

Answer:

a) $\cos(\theta) = \frac{\sqrt[]{33}}{7}$

b) $\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}$

c) $\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}$

d) $\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

Step-by-step explanation:

We will use the following trigonometric identities

$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ $\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ .

Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that

$x^2+4^2 = 7 ^2$

which implies that $x=-\sqrt[]{49-16} = -\sqrt[]{33}$ . Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then

$\cos(\theta) = \frac{-\sqrt[]{33}}{7}$

b)Recall that $\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}$ , then using the identity from above, we have that

$\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}$

c) Recall that $\sin(\pi)=0, \cos(\pi)=-1$ . Then,

$\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}$

d) Recall that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}$ . Then

$\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

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3 years ago