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3 years ago

Solve the equation. −2 1/4=r−4/5 will give brainiest for the first correct answer!!!

Mathematics

1 answer:

STatiana [176]3 years ago

6 0

Answer:

r = -1 9/20

Step-by-step explanation:

-2 1/4 = r - 4/5

Switch sides:

r - 4/5 = -2 1/4

Add 4/5 to both sides.

r = -2 1/4 + 4/5

Change the mixed numeral into a fraction:

r = -9/4 + 4/5

The LCD is 20.

r = -45/20 + 16/20

r = -29/20

r = -1 9/20

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Jim saw that other bank offered the same rates but compounded the interest more often. Consider if he still put $15,000 into a s

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Compound interest is the addition of interest on the interest of the principal amount.

<h3>What is compound interest?</h3>

Compound interest is the addition of interest on the interest of the principal amount. It is given by the formula,

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

As it is given that the principal amount is $15,000; while the rate of interest is 2.8%, and the amount is invested for a period of 5 years.

A.) When the interest is charged weekly,

As we know that there are 52 weeks in a year, therefore, n = 52, substitute the values,

$A = P(1+ \dfrac{r}{n})^{nt}\\\\A = 15000(1+ \dfrac{0.028}{52})^{52 \times 5}\\\\A = \$17,253.46$

B.) When the interest is charged daily,

As we know that there are 365 days in a year, therefore, n = 365, substitute the values,

$A = P(1+ \dfrac{r}{n})^{nt}\\\\A = 15000(1+ \dfrac{0.028}{365})^{365 \times 5}\\\\A = \$19,554.55$

C.) When the interest is charged Continuously,

As we know that the formula for continuous compounding is given as,

$A = Pe^{rt}$

Substitute the values, we will get,

$A = 15000 \times (e^{0.024 \times 5})\\\\A= \$16,912.45$

D.) When the interest is charged Monthly,

As we know that there are 12 months in a year, therefore, n = 12, substitute the values,

$A = P(1+ \dfrac{r}{n})^{nt}\\\\A = 15000(1+ \dfrac{0.028}{12})^{12\times 5}\\\\A = \$15211.23$

E.) When the interest is charged Semi-annually,

As we know that the interest is charged Semi-annually, therefore, n = 2, substitute the values,

$A = P(1+ \dfrac{r}{n})^{nt}\\\\A = 15000(1+ \dfrac{0.028}{2})^{2\times 5}\\\\A = \$17,237.36$

F.) When the interest is charged Quarterly,

As we know that the interest is charged Quarterly, therefore, n = 4, substitute the values,

$A = P(1+ \dfrac{r}{n})^{nt}\\\\A = 15000(1+ \dfrac{0.028}{4})^{4\times 5}\\\\A = \$17,245.7$

Learn more about Compound Interest:

brainly.com/question/25857212

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3 years ago

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A radioactive substance is decaying exponentially. If there are initially 15 grams of the substance and after 3 hours there are

sukhopar [10]

Answer:

Ok, an exponential decay is written as:

P(t) = A*(1 - r)^t

Where A is the initial population, r is the rate of decay and t is the unit of time.

We know that the initial population is 15g

then:

P(t) = 15g*(1 - r)^t

And at t = 3hs, the populations is 5g

P(5h) = 5g = 15g*(1 - r)^5

5/15 = 1/3 = (1 - r)^5

(1/3)^(1/5) = (1 - r) = 0.8

Now, the half life time of the sustance is t = x, such that the population reduces to it's half:

P(x) = A/2 = 15g/2 = 7.5g

Then:

7.5g = 15g*0.8^x

7.5g/15g = 1/2 = 0.8^x

Now, remember that if we have:

a = b^x

then

x = ln(a)/ln(b)

ln(1/2)/ln(0.8) = x = 3,11 hours

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-3,7) yes I am pretty sure that’s the answer

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