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3 years ago

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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 .

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spayn [35]

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

The value of E(x2) is p
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<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:

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

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

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:

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2 years ago

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Answer:

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

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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:

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If we replace we got:

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

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

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