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3 years ago

Seth just opened a checking account. His account includes an ATM card. What will he NOT be able to do with his card?

Mathematics

1 answer:

OverLord2011 [107]3 years ago

6 0

Answer:

Option a

Step-by-step explanation:

Given that Seth just opened a checking account. His account includes an ATM card.

With ATM card

d) he can withdraw money from any ATM centre provided sufficient balance is kept in his checking account

b) he can deposit money for fixed denominations normally in multiples of 100 in the ATM centre with ATM card

c) he can give the card with pin in any shop to purchase an item instead of paying cash.

Hence b,c,d are not right answers.

a) transfer money is not possible using ATM card.

Only with netbanking transfer of money is possible

Option a is right answer

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6 1/2+n=12. please explain for me I dunno how to do this.

Temka [501]

Hey there!

Let's first define our goal. Our goal here is to be able to isolate the variable, n, and solve for it by moving everything to the other side. We see that we have two sides of the equation, and by using the property of equality which states that whatever you do to one side you do to the other and the expression remains equal, we can solve this equation.

It looks like the only thing we have on the left side with n is that 6 1/2. In order to get rid of it, we must subtract it from both sides, because the property of equality states you must do the inverse operation.

We have:

n = 12 - 6 1/2

n = 5 1/2

Hope this helps!

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2 years ago

Which of the following is a step in simplifying the expression x multiplied by y to the power of 3 over x to the power of negati

Sveta_85 [38]

Answer:

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Step-by-step explanation:

4 0

2 years ago

Jovita divides $12,000 into three investments: a savings account paying 4% annual interest, a bond paying 6%, and a money market

xenn [34]

Answer:

$2080 was invested in the savings account, $4241 was invested in the bond account, & $6539 was invested in the money fund.

Step-by-step explanation:

Let a, b, and c be the fractions of the $12,000 that is invested in the savings, bond, & money fund respectively.

Let S1, S2, S3 be the total monies from the savings, bond, & money fund after the year with interest respectively.

We know that a+b+c=1

and

S1+S2+S3 = 12000 + 860

S1+S2+S3 = 12860

From, the question, we know that b = 2a, and c = 1 - 3a based on the condition of 2 times more in the bond account that the savings account.

Thus;

S1 = a(12000)(1.04) = 12480a

S2 = 2a(12000)(1.06) = 25,440a

S3 = (1 - 3a)(12000)(1.09) = 13080 - 39240a

We already know that;

S1+S2+S3 = 12860

Thus, Adding the 3 equations gives;

12480a + 25,440a + 13080 - 39240a = 12860

Thus, simplifying gives;

13080 - 12860 = 39240a - 12480a - 25,440a

220 = 1320a

a = 220/1320

a = 0.1667

So,b = 2 x 0.1667 = 0.3334

c = 1 - (3 x 0.1667)

c = 1 - 0.5

c = 0.5

Thus,

S1 = 12480a = 12480 x 0.1667 ≈ $2080

S2 = 25,440a = 25440 x 0.1667 ≈ $4241

S3 = 13080 - 39240(0.1667) ≈ 6539

So, $2080 was invested in the savings account, $4241 was invested in the bond account, & $6539 was invested in the money fund.

3 0

3 years ago

Read 2 more answers

Divide (18x^3 + 12x – 3x) ÷ 6x^2

andre [41]

First let's simplify it

$\frac{18x^3 + 9x}{6x^2}$

Now let's factor it
$\frac{9x(2x^2 + 1)}{6x^2}$

Lastly, let's simplify
$\frac{3(2x^2 + 1)}{2x}$

4 0

2 years ago

What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that t

Inga [223]

Answer:

f(2n)-f(n)=log2

b.lg(lg2+lgn)-lglgn

c. f(2n)/f(n)=2

d.2nlg2+nlgn

e.f(2n)/(n)=4

f.f(2n)/f(n)=8

g. f(2n)/f(n)=2

Step-by-step explanation:

What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos- sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]

from a

f(n)=logn

f(2n)=lg(2n)

f(2n)-f(n)=log2n-logn

lo(2*n)=lg2+lgn-lgn

f(2n)-f(n)=lg2+lgn-lgn

f(2n)-f(n)=log2

2.f(n)=lglgn

F(2n)=lglg2n

f(2n)-f(n)=lglg2n-lglgn

lg2n=lg2+lgn

lg(lg2+lgn)-lglgn

3.f(n)=100n

f(2n)=100(2n)

f(2n)/f(n)=200n/100n

f(2n)/f(n)=2

the time will double

4.f(n)=nlgn

f(2n)=2nlg2n

f(2n)-f(n)=2nlg2n-nlgn

f(2n)-f(n)=2n(lg2+lgn)-nlgn

2nLg2+2nlgn-nlgn

2nlg2+nlgn

5.we shall look for the ratio

f(n)=n^2

f(2n)=2n^2

f(2n)/(n)=2n^2/n^2

f(2n)/(n)=4n^2/n^2

f(2n)/(n)=4

the time will be times 4 the initial tiote tat ratio are used because it will be easier to calculate and compare

6.n^3

f(n)=n^3

f(2n)=(2n)^3

f(2n)/f(n)=(2n)^3/n^3

f(2n)/f(n)=8

the ratio will be times 8 the initial

7.2n

f(n)=2n

f(2n)=2(2n)

f(2n)/f(n)=2(2n)/2n

f(2n)/f(n)=2

5 0

2 years ago