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Andre45 [30]
3 years ago
10

The diagonals of a parallelogram ABCD intersect at O.find ar(AOB),if ar(ABCD)is 44cm^2

Mathematics
1 answer:
prisoha [69]3 years ago
5 0

Answer:

11\text{ cm}^2

Step-by-step explanation:

Given: In a parallelogram ABCD, diagonals intersect at O and ar(ABCD) is 44 \text{cm}^2.

We need to find the area of triangle AOB.

We know that each diagonal divide the parallelogram in two equal parts and diagonals bisect each other.

It means both diagonals divide the parallelogram in 4 equal parts.

ar(AOB)=\dfrac{1}{4}\times ar(ABCD)

ar(AOB)=\dfrac{1}{4}\times 44

ar(AOB)=11

Hence, the values of ar(AOB) is 11\text{ cm}^2.

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Answer with explanation:

Let us assume that the 2 functions are:

1) f(x)

2) g(x)

Now by definition of concave function we have the first derivative of the function should be strictly decreasing thus for the above 2 function we conclude that

\frac{d}{dx}\cdot f(x)

Now the sum of the 2 functions is shown below

y=f(x)+g(x)

Diffrentiating both sides with respect to 'x' we get

\frac{dy}{dx}=\frac{d}{dx}\cdot f(x)+\frac{d}{dx}\cdot g(x)\\\\

Since each term in the right of the above equation is negative thus we conclude that their sum is also negative thus

\frac{dy}{dx}

Thus the sum of the 2 functions is also a concave function.

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The product of the 2 functions is shown below

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Diffrentiating both sides with respect to 'x' we get

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Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

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95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

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