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3 years ago

PLZ HELP FAST I'm pretty sure that the stuff i put in the boxes is correct

Mathematics

1 answer:

arlik [135]3 years ago

6 0

Answer:

3(x + 4)- X= 12

Step-by-step explanation:

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How do you work out the rate, time and principal from the compound interest formula?

wlad13 [49]

Answer:

Part 1) $P=A/(1+\frac{r}{n})^{nt}$ (see the explanation)

Part 2) $r=n[\frac{A}{P}^{1/(nt)}-1]$ (see the explanation)

Part 3) $t=log(\frac{A}{P})/[(n)log(1+\frac{r}{n})]$ (see the explanation)

Step-by-step explanation:

we know that

The compound interest formula is equal to

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

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part 1) Find the Principal P

The values of A,r,n and t are given

Isolate the variable P

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

Divide both side by $(1+\frac{r}{n})^{nt}$

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

Part 2) Find the rate r

The values of A,P,n and t are given

Isolate the variable r

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

Divide both sides by P

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

Elevated both sides to 1/(nt)

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

subtract 1 both sides

$\frac{A}{P}^{1/(nt)}-1 =\frac{r}{n}$

Multiply by n both sides

$r=n[\frac{A}{P}^{1/(nt)}-1]$

Part 3) Find the time t

The values of A,P,r and n are given

Isolate the variable t

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

Divide both sides by P

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

Apply log both sides

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

Apply property of exponents

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

Divide both side by $(n)log(1+\frac{r}{n})$

$t=log(\frac{A}{P})/[(n)log(1+\frac{r}{n})]$

3 0

3 years ago

An urn contains 6 red marbles and 4 black marbles. Two marbles are randomly drawn (one by one) from the urn without replacement.

strojnjashka [21]

Answer: The required probability is $\dfrac{2}{15}.$

Step-by-step explanation: Given that an urn contains 6 red marbles and 4 black marbles. Two marbles are randomly drawn one by one from the urn without replacement.

We are to find the probability that both drawn marbles are black.

Let E and F denote the events of two marbles one by one without replacement and let S and S' denote the corresponding sample spaces.

Then, we have

$n(E)=^4C_1=4,\\\\n(F)=^3C_1=3,\\\\n(S)=^{10}C_1=10,\\\\n(S')=^9C_1=9.$

Therefore, the probability that both marbles are red is given by

$p=P(E)\times P(F)=\dfrac{n(E)}{n(S)}\times\dfrac{n(F)}{n(S')}=\dfrac{4}{10}\times\dfrac{3}{9}=\dfrac{2}{15}.$

Thus, the required probability is $\dfrac{2}{15}.$

4 0

3 years ago

Please help i am stuck please

anzhelika [568]

Answer:

it is D

Step-by-step explanation:

3 0

4 years ago

What is (3 x 10^3) / (12 x 10^6)

Vinil7 [7]

Answer:

(3 x 10^3) / (12 x 10^6) = 0.00025

4 0

3 years ago

Read 2 more answers

What is the answer to<br> 1. b+4=2b-5<br> 2. -6-29=5x-7<br> 3. 10h+12=8h+4<br> 4. 7a-17=4a+1

RoseWind [281]

Answer:

b=9

(assuming you meant -6x) x = -2

h = -4

a = 6

Step-by-step explanation:

6 0

3 years ago

PLZ HELP FAST I'm pretty sure that the stuff i put in the boxes is correct​

PLZ HELP FAST I'm pretty sure that the stuff i put in the boxes is correct