Please answer this question

Mike deposited $6500 into two saving accounts bearing simple interest. One of the accounts has an interest rate of 3% while the

ololo11 [35]

0.03x+0.06 (6500-x)=225
Solve for x
0.03x+390-0.06x=225
0.03x-0.06x=225-390
-0.03x=-165
X=165÷0.03
X=5500 invested at 3%

6500-5500=1000 invested at 6%

7 0

3 years ago

Use the graph to determine which statement describes f(x).

Lerok [7]

Answer: B

Step-by-step explanation:

7 0

2 years ago

(6z 2 - 4z + 1)(8 - 3z).

Yakvenalex [24]

Distributive property
a(b+c)+ab+ac
a(b-c)=ab-ac

(6z^2-4z+1)(8-3z)
move for nicety
(8-3z)(6z^2-4z+1)
distribute
8(6z^2-4z+1)-3z(6z^2-4z+1)=
48z^2-32z+8-18z^3+12z^2-3z=
-18z^3+48z^2+12z^2-32z-3z+8=
-18z^3+60z-35z+8

7 0

3 years ago

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$