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2 years ago

In parallelogram ABCD , diagonals AC⎯⎯⎯⎯⎯ and BD⎯⎯⎯⎯⎯ intersect at point E, AE=x2−16 , and CE=6x .

Mathematics

1 answer:

MAXImum [283]2 years ago

5 0

Answer:

<h3>AC=96 units.</h3>

Step-by-step explanation:

We are given a parallelogram ABCD with diagonals AC and BD intersect at point E.

$AE=x^2-16.$ , and CE=6x .

<em>Note: The diagonals of a parallelogram intersects at mid-point.</em>

Therefore, AE = EC.

Plugging expressions for AE and EC, we get

$x^2-16=6x.$

Subtracting 6x from both sides, we get

$x^2-16-6x=6x-6x$

$x^2-6x-16=0$

Factoriong quadratic by product sum rule.

We need to find the factors of -16 that add upto -6.

-16 has factors -8 and +2 that add upto -6.

Therefore, factor of $x^2-6x-16=0$ quadratic is (x-8)(x+2)=0

Setting each factor equal to 0 and solve for x.

x-8=0 => x=8

x+2=0 => x=-2.

We can't take x=-2 as it's a negative number.

Therefore, plugging x=8 in EC =6x, we get

EC = 6(8) = 48.

<h3>AC = AE + EC = 48+48 =96 units.</h3>

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Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.22, n = 425$

$\mu = E(X) = np = 425*0.22 = 93.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54$

What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

This probability is the pvalue of Z when X = 82. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{82 - 93.5}{8.54}$

$Z = -1.35$

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8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

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1 year ago

Darren invests $4,500 into an account that earns 5% annual interests. How much will be in the account after 10 years if the inte

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We have been given that Darren invests $4,500 into an account that earns 5% annual interests. We are asked to find the amount in his account after 10 years, if the interest rate is compounded annually, quarterly, monthly, or daily.

We will use compound interest formula to solve our given problem.

$A=P(1+\frac{r}{n})^{nt}$ , where,

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

$5\%=\frac{5}{100}=0.05$

When compounded annually, $n=1$ :

$A=4500(1+\frac{0.05}{1})^{1\cdot 10}$

$A=4500(1.05)^{10}$

$A=4500(1.6288946267774414)$

$A=7330.025820498\approx 7330.03$

When compounded quarterly, $n=4$ :

$A=4500(1+\frac{0.05}{4})^{4\cdot 10}$

$A=4500(1.0125)^{40}$

$A=4500(1.6436194634870132)$

$A=7396.28758569\approx 7396.29$

When compounded monthly, $n=12$ :

$A=4500(1+\frac{0.05}{12})^{12\cdot 10}$

$A=4500(1.00416666)^{120}$

$A=4500(1.64700949769)$

$A=7411.542739605\approx 7411.54$

When compounded daily, $n=365$ :

$A=4500(1+\frac{0.05}{365})^{365\cdot 10}$

$A=4500(1.0001369863013699)^{3650}$

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Since amount earned will be maximum, when interest is compounded daily, therefore, Darren should use compounded daily interest rate.

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Your equation transforms into:

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Now, you have to apply the definition of a logarithm to express the equation in exponential form:

$\dfrac{x^2+3x}{x+5} = 4^1$

In case you don't remember, this is the definition of a logarithm:

$\log_b x = y \iff b^y = x$

The log is the exponent (y) you have to raise the base (b) to in order to get the power (x).

Finally, solve the rational equation:

$\dfrac{x^2+3x}{x+5} =4 \iff x^2+3x=4(x+5)$
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Assoli18 [71]

Answer:

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Step-by-step explanation:

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