Ask question

Ask question

All categories

3 years ago

6

Gia opened two savings accounts at two diffrent banks. One account earns an annual 3.4% simple interest, and the other earns hal

f as much. If she deposited $509 in each account, how much total interests will sh have earned in 3 years? A:$153.00 B:$38.25 C:$76.50 D:$51.00

1 answer:

dsp733 years ago

5 0

Interest from first account is :

$I_1 = \dfrac{p\times r \times t}{100}\\\\I_1 = \dfrac{509\times 3.4\times 3}{100}\\\\I_1 =\$ 51.918$

Interest form second account is :

$I_2 = \dfrac{p\times r\times t}{100}\\\\I_2 = \dfrac{509\times 1.7\times 3}{100}\\\\I_2 = \$25.959$

So, total interest is :

$I = I_1 + I_2 \\\\I = 51.918 + 25.959\\\\I = \$77.877$

Therefore, total interest is $77.877 .

You might be interested in

Let X be a Bernoulli rv with pmf as in Example 3.18. a. Compute E(X2 ). b. Show that V(X) 5 p(1 2 p). c. Compute E(X79).

spayn [35]

The Bernoulli distribution is a distribution whose random variable can only take 0 or 1

The value of E(x2) is p
The value of V(x) is p(1 - p)
The value of E(x79) is p

<h3>How to compute E(x2)</h3>

The distribution is given as:

p(0) = 1 - p

p(1) = p

The expected value of x2, E(x2) is calculated as:

$E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * (1- p) + 1^2 * p$

Evaluate the exponents

$E(x^2) = 0 * (1- p) + 1 * p$

Multiply

$E(x^2) = 0 +p$

Add

$E(x^2) = p$

Hence, the value of E(x2) is p

<h3>How to compute V(x)</h3>

This is calculated as:

$V(x) = E(x^2) - (E(x))^2$

Start by calculating E(x) using:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * (1- p) + 1 * p$

$E(x) = p$

Recall that:

$V(x) = E(x^2) - (E(x))^2$

So, we have:

$V(x) = p - p^2$

Factor out p

$V(x) = p(1 - p)$

Hence, the value of V(x) is p(1 - p)

<h3>How to compute E(x79)</h3>

The expected value of x79, E(x79) is calculated as:

$E(x^{79}) = \sum x^{79} * P(x)$

So, we have:

$E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p$

Evaluate the exponents

$E(x^{79}) = 0 * (1- p) + 1 * p$

Multiply

$E(x^{79}) = 0 + p$

Add

$E(x^{79}) = p$

Hence, the value of E(x79) is p

Read more about probability distribution at:

brainly.com/question/15246027

5 0

2 years ago

One number is four times another number. If their sum is 95, what are the numbers?

nikdorinn [45]

Let x be one number and y be the other.

#1 - Set up a system of equations.
4x = y
x + y = 45

#2 - Solve the system by substitution.
x + (4x) = 45
5x = 45
x = 9

(9) + y = 45
y = 36

8 0

3 years ago

Read 2 more answers

To get the value of an expression you add

frez [133]

When we substitute a specific valuefor each variable, and then perform the operations, it's called evaluating theexpression. Let's evaluate theexpression 3y + 2y when 5 = y. Click on the steps to see how it's done. Perform the operations, to find the value of the expression.

6 0

3 years ago

Read 2 more answers

A practice law exam has 100 questions, each with 5 possible choices. A student took the exam and received 13 out of 100.If the s

Cloud [144]

Answer:

$z=\frac{13-20}{4}=-1.75$

Assuming:

H0: $\mu \geq 20$

H1: $\mu$

$p_v = P(Z$

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest (number of correct answers in the test), on this case we now that:

$X \sim Binom(n=100, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We need to check the conditions in order to use the normal approximation.

$np=100*0.2=20 \geq 10$

$n(1-p)=20*(1-0.2)=16 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=100*0.2=20$

$\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2(1-0.2)}=4$

So we can approximate the random variable X like this:

$X\sim N(\mu =20, \sigma=4)$

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean". The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$

The z score is given by this formula:

$z=\frac{x-\mu}{\sigma}$

If we replace we got:

$z=\frac{13-20}{4}=-1.75$

Let's assume that we conduct the following test:

H0: $\mu \geq 20$

H1: $\mu$

We want to check is the score for the student is significantly less than the expected value using random guessing.

So on this case since we have the statistic we can calculate the p value on this way:

$p_v = P(Z$

5 0

2 years ago

(Easy 30 points) Solve the equation to find a positive value of c: 3^2 + 4^ = c^2

shusha [124]

Answer:

C= 5, -5

Step-by-step explanation:

8 0

3 years ago