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2 years ago

Kite EFGH is inscribed in a rectangle such that F and H are midpoints and EG is parallel to the side of the rectangle.

Mathematics

1 answer:

Radda [10]2 years ago

7 0

Answer:

Step-by-step explanation:

EFGH is a kite, so EF ≅ FG and EH ≅ HG.

The area of the kite consists of two area of triangles EFG and EHG.

1. Area of triangle EFG:

$A_{\trangle EFG}=\dfrac{1}{2}\cdot EG\cdot h,$

where h is the height drawn from point F to the side EG.

1. Area of triangle EHG:

$A_{\trangle EHG}=\dfrac{1}{2}\cdot EG\cdot H,$

where H is the height drawn from point H to the side EG.

3. Note that

$EG \cong \text{rectangle's length}$

$h+H\cong \text{rectangle's width}$

So,

$A_{\text{kite }EFGH}\\ \\=A_{\triangle EFG}+A_{\triangle EHG}\\ \\=\dfrac{1}{2}\cdot EG\cdot (h+H)\\ \\=\dfrac{1}{2}\cdot \text{rectangle's length}\cdot \text{rectangle's width}\\ \\=\dfrac{1}{2}A_{\text{rectangle}}$

Thus, option C is true (the area of the kite doesn't depend on ratio in which points E and G divide the side)

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At a certain airport, 75% of the flights arrive on time. A sample of 10 flights is studied. Assume each flight is independent of

mars1129 [50]

Answer:

52.56% probability that eight or more of the flights will arrive on time.

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it is on time, or it is not. The probability of a flight being on time is independent from other flights. So we use the binomial probability distribution to solve this question.]

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

At a certain airport, 75% of the flights arrive on time.

This means that $p = 0.75$

A sample of 10 flights is studied.

This means that $n = 10$

Find the probability that eight or more of the flights will arrive on time.

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.75)^{8}.(0.25)^{2} = 0.2816$

$P(X = 9) = C_{10,9}.(0.75)^{9}.(0.25)^{1} = 0.1877$

$P(X = 10) = C_{10,10}.(0.75)^{10}.(0.25)^{0} = 0.0563$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.2816 + 0.1877 + 0.0563 = 0.5256$

52.56% probability that eight or more of the flights will arrive on time.

4 0

2 years ago

Read 2 more answers

There is an integer n such that 2n2 − 5n + 2 is prime. To prove the statement it suffices to find a value of n such that (n, 2n2

fomenos

Answer:

n = 0 or 3

Step-by-step explanation:

2n² - 5n + 2

2n² - 4n - n + 2

2n(n - 2) -1(n - 2)

(n - 2)(2n - 1)

Prime number is one which is divisible by itself and 1

n-2 = 1 n = 3

2n-1 = 1 n = 0

7 0

2 years ago

Colin invests £2350 into a savings account. The bank gives 4.2% compound interest for the first 4 years and 4.9% thereafter. How

mixer [17]

To solve this, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years
$P$ is the initial investment
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year

For the first 4 years we know that: $P=2350$ , $r= \frac{4.2}{100} =0.042$ , $t=4$ , and since the problem is not specifying how often the interest is communed, we are going to assume it is compounded annually; therefore, $n=1$ . Lest replace those values in our formula:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=2350(1+ \frac{0.042}{1} )^{(1)(4)}$
$A=2350(1+0.042)^{4}$
$A=2770.38$

Now, for the next 6 years the intial investment will be the final amount from our previous step, so $P=2770.38$ . We also know that: $r= \frac{4.9}{100} =0.049$ , $t=6$ , and $n=1$ . Lets replace those values in our formula one more time:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=2770.38(1+ \frac{0.049}{1})^{(1)(6)$
$A=2770.38(1+0.049)^6$
$A=3691.41$

We can conclude that Collin will have <span>£3691.41 in his account after 10 years.</span>

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2 years ago

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zaharov [31]

I chose the number 155