13.23 divided by 1.89

How much money, as a one-time deposit, PV, would you need to deposit into an account that earns 1.5% compounded monthly to earn

Karolina [17]

Answer:

The amount needed as a one-time deposit to earn $7,500 in 3 years is <em>$4388.17</em>

Step-by-step explanation:

<u>Basic Finance Formulas </u>

One of the most-used formulas to compute present and future values is

$FV=PV(1+r)^{n}$

Where FV is the future value, PV is the present value, r is the interest rate and n is the number of periods. It's vital to keep in mind that r and n must be referred to the same compounded time, e.g. r is compounded monthly and n is expressed in months

The question requires to compute the PV needed as a one-time deposit to earn a future value of $7,500 in 3 years at a 1.5% rate compounded monthly.

FV=7,500

r=1.5%=0.015

n=3*12=36 months

We converted n to months because r is compounded monthly . The formula

$FV=PV(1+r)^{n}$

must be managed to make PV isolated

$PV=FV(1+r)^{-n}$

$PV=7,500(1+0.015)^{-36}$

$PV=\$4388.17$

Answer: The amount needed as a one-time deposit to earn $7,500 in 3 years is $4388.17

3 0

2 years ago

Law of sines: StartFraction sine (uppercase A) Over a EndFraction = StartFraction sine (uppercase B) Over b EndFraction = StartF

vladimir2022 [97]

Answer:

30°

Step-by-step explanation:

Law of Sines = $\frac{a}{Sin A} = \frac{b}{Sin B} = \frac{c}{Sin C}$

$\frac{2}{Sin 30} = \frac{2}{Sin B}$

$\frac{2}{Sin 30} = \frac{2}{Sin 30}$

Therefore, m∠B = 30°

Hope that's right and helps

6 0

2 years ago

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$