Put this in order from least to greatest<br> 210,700,000 1,210,707,000 12,007,000

Lillian is going to invest in an account paying an interest rate of 5.7% compounded continuously. How much would Lillian need to

maksim [4K]

Answer:

The amount Lilian needs to invest is $5,160

Step-by-step explanation:

The parameters of the amount Lilian is going to invest in an account are;

The interest rate the account pays = 5.7%

The final amount the investment is expected to reach after the given time period = $7,600

The number of years of the investment = 7 years

The formula for compound interest is presented as follows;

$A = P \cdot \left(1 + \dfrac{r}{n} \right)^{n\cdot t}$

Where;

A = The final amount = $7,600

P = The amount Lilian is to invest

r = The interest rate = 5.7% = 0.057

n = The number of times the interest is applied per time period = 1

t = The number of period that passes = 7 (years)

Therefore, we get;

$7,600 = P \cdot \left(1 + 0.057 \right)^{7}$

∴ P = 7,600/(1 + 0.057)⁷ = 5,155.7123754

The amount Lilian needs to invest (rounded to the nearest ten dollars) for the value of the account to reach $7,600 in 7 years, P = $5,160

6 0

3 years ago

Haley mixes 2 1/2 a cup of water with 1/3 of grape juice concentrate. Michael mixes 1 1/3 cups of water with1/4 of grape juice c

Olegator [25]

Answer:

2 4/6 is what Haley has and 1 5/12 is what Michael has. I was unsure what the question was but I hope this answered it!

Step-by-step explanation:

4 0

3 years ago

Sam intends to roll a six sided cube 100 times. What is the probability model he can use to predict whether or not each roll wil

bekas [8.4K]

The outcomes of rolling fair six sided die is $\{1,2,3,4,5,6\}$ . The prime numbers in the above outcomes are $\{2,3,5\}$ . The probability of not rolling a prime number is $p=\frac{3}{6}= 0.5$ .

Each trial is independent.

In the given experiment of 100 trials, the binomial probability model can be used to predict the number of occurrence of non-prime numbers (X). The probability mass function is

$P(X=x)=C(100,x)p^x(1-p)^{100-x},x=0,1,2,3,...,100$ . Since $p=0.5,1-p=0.5$

$P(X=x)=C(100,x)0.5^x(0.5)^{100-x},x=0,1,2,3,...,100 \\ P(X=x)=C(100,x)(0.5)^{100},x=0,1,2,3,...,100$

8 0

3 years ago

Read 2 more answers

Solve for the zeros of the quadratic function f(x) = 9x² + 6x + 1.

Nezavi [6.7K]

Answer:

$\large\boxed{x=-\dfrac{1}{3}}$

Step-by-step explanation:

$f(x)=9x^2+6x+1\\\\\text{The zeros are for}\ f(x)=0\\\\9x^2+6x+1=0\\\\9x^2+3x+3x+1=0\\\\3x(3x+1)+1(3x+1)=0\\\\(3x+1)(3x+1)=0\\\\(3x+1)^2=0\iff3x+1=0\qquad\text{subtract 1 from both sides}\\\\3x=-1\qquad\text{divide both sides by 3}\\\\x=-\dfrac{1}{3}$

8 0

3 years ago

A light bulb has a lifetime X that is exponentially distributed with a mean 340 hours. Find the probability that the bulb lifeti

LUCKY_DIMON [66]

Answer:

0.7026

Step-by-step explanation:

Let X denote the lifetime of light bulb. Given $X \sim Exp(\lambda)$ where the mean is $E(X) = 340 =\frac{1}{\lambda} \implies \lambda = \frac{1}{340} = 0.00294$ .

Recall that,

$\displaystyle P(X>x) = 1 - \int_0^x e^{-\lambda x} = 1 - (1 - e^{-\lambda x} = e^{-\lambda x}$

$\displaystyle P(X>220 | X>100) = \frac{P(X>220,X>100)}{P(X>100)} = \frac{e^{-\lambda \times 220}}{e^{-\lambda \times 100}} = \frac{0.5236}{0.7452} = 0.7026$

7 0

3 years ago

Put this in order from least to greatest 210,700,000 1,210,707,000 12,007,000