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tester [92]
3 years ago
14

How many shares of Lombard Incorporated traded on the 17th?

Mathematics
1 answer:
Naily [24]3 years ago
8 0

Answer:

Step-by-step explanation:

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I think true?? i think! im really not sure but yea
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2 years ago
Which of the following could not be considered a polynomial
Aleksandr [31]

Answer:

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Step-by-step explanation:

I cant solve it without knowing the choices

7 0
3 years ago
Determine if this is a function.
Kruka [31]

Answer: This is <u>not</u> a function

Why not?

The input x = 2 leads to more than one output. In this case, it leads to y = 20 and y = 40 simultaneously. This is the reason why we don't have a function. A function is only possible if each input x leads to exactly one y output.

Put another way: we cannot have x repeating any values if we want a function.

Side note: We could have y repeat however. So it's possible to have a function with the two points (1,5) and (3,5).

6 0
2 years ago
A passbook saving account has a rate of 6%. find the effective annual yield, rounded to the nearest tenth of a percent, if the i
irinina [24]

Answer:

a,b,c,d,e,f) The effective annual yield is 1.1P-P = 0.1P = 10%P.

Step-by-step explanation:

This is a compound interest problem

Compound interest formula:

The compound interest formula is given by:

A = P(1 + \frac{r}{n})^{nt}

A: Amount of money(Balance)

P: Principal(Initial deposit)

r: interest rate(as a decimal value)

n: number of times that interest is compounded per unit t

t: time the money is invested or borrowed for.

a)

r = 0.06

n: 2(semianually means that the interest is compounded twice a year).

t = 1.

A = P(1 + \frac{0.06}{2})^{2*1}

A = 1.1P

The acount started the year with P, and it ended with 1.1P, so the effective annual yield is 1.1P-P = 0.1P = 10%P.

b)

Now we have n = 3, since if the interest is compounded quarterly, is is compounded three times a year(a year has 3 quarters). So:

A = P(1 + \frac{0.06}{3})^{3*1}

A = 1.1P

The effective annual yield is 1.1P-P = 0.1P = 10%P.

c) Now we have n = 12, since the interest is compounded monthly, and there are 12 months a year. So:

A = P(1 + \frac{0.06}{12})^{12*1}

A = 1.1P

The effective annual yield is 1.1P-P = 0.1P = 10%P.

d) Since the interest is compounded daily, and we assume 360 days in a year, n = 360. So:

A = P(1 + \frac{0.06}{360})^{360*1}

A = 1.1P

The effective annual yield is 1.1P-P = 0.1P = 10%P.

e) The interest is compounded 1000 times a year, so n = 1000

A = P(1 + \frac{0.06}{1000})^{1000*1}

A = 1.1P

The effective annual yield is 1.1P-P = 0.1P = 10%P.

f) The interest is compounded 100000 times a year, so n = 100000

A = P(1 + \frac{0.06}{100000})^{100000*1}

A = 1.1P

The effective annual yield is 1.1P-P = 0.1P = 10%P.

6 0
3 years ago
A computer costs $800. It loses ¼ of its value every 2 years after it is purchased.
Rufina [12.5K]

The computer will be worth $200.

Step-by-step explanation:

A quarter of $800 is $200, so the computer loses $200 dollars in value every two years. Since 6÷2 is 3, you multiple $200 by 3 to get $600. You then subtract $600 from $800 and get $200.

5 0
2 years ago
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