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2 years ago

A symmetrical binomial distribution with large sample size can be approximated by a __________

Mathematics

1 answer:

sweet [91]2 years ago

7 0

Typical distribution?

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A square is constructed on side AD of quadrilateral ABCD such that FA lies on AB, as shown in the figure.

Elza [17]

Answer: The co-ordinates of F are (4.4, -4.6) and the co-ordinates of D are (4.4, -1.4).

Step-by-step explanation: Given that a square is constructed on side AD of quadrilateral ABCD such that FA lies on AB as shown in the figure. The co-ordinates of A are (6, -3) and the co-ordinates of B are (10, 1).

Also, AD : AB = 2 : 5 and the co-ordinates of the point E are (2.8, -3).

We are to select the correct co-ordinates of the points F and D.

Let, (a, b) are the co-ordinates of F and (c, d) are the co-ordinates of D.

Since ADEF is a square, so we have

AD = DE = EF = FA.

Given that

AD : AB = 2 : 5, so FA : AB = 2 : 5.

That is, \left(\dfrac{c+4.4}{2},\dfrac{d-4.6}{2}\right)=\left(\dfrac{2.8+6}{2},\dfrac{-3-3}{2}\right)

We have, after applying the internal division formula that

$\left(\dfrac{2\times 10+5\times a}{2+5},\dfrac{2\times 1+5\times b}{2+5}\right)=(6,-3)\\\\\\\Rightarrow \left(\dfrac{20+5a}{7},\dfrac{2+5b}{7}\right)=(6,-3)\\\\\\\Rightarrow \dfrac{20+5a}{7}=6,~~~~~\dfrac{2+5b}{7}=-3\\\\\\\Rightarrow 20+5a=42,~~~~\Rightarrow 2+5b=-21\\\\\\\Rightarrow 5a=22,~~~~~~~~~~\Rightarrow 5b=-23\\\\\\\Rightarrow a=4.4,~~~~~~~~~~~\Rightarrow b=-4.6.$

So, the co-ordinates of F are (4.4, -4.6).

Now, since ADEF is a square, and the diagonals of a square bisect each other.

So, the mid-points of both the diagonals are same.

That is,

$\textup{mid-point of DF}=\textup{mid-point of AE}\\\\\\\Rightarrow \left(\dfrac{c+4.4}{2},\dfrac{d-4.6}{2}\right)=\left(\dfrac{2.8+6}{2},\dfrac{-3-3}{2}\right)\\\\\\\Rightarrow \left(\dfrac{c+4.4}{2},\dfrac{d-4.6}{2}\right)=\left(\dfrac{8.8}{2},\dfrac{-6}{2}\right)\\\\\\\Rightarrow \dfrac{c+4.4}{2}=\dfrac{8.8}{2},~~~~~~\dfrac{d-4.6}{2}=-\dfrac{6}{2}\\\\\\\Rightarrow c+4.4=8.8,~~~~~\Rightarrow d-4.6=-6\\\\\Rightarrow c=4.4,~~~~~~~~~~~~\Rightarrow d=-1.4.$

So, the co-ordinates of D are (4.4, -1.4).

Thus, the co-ordinates of F are (4.4, -4.6) and the co-ordinates of D are (4.4, -1.4).

7 0

2 years ago

Read 2 more answers

A stadium asked for $760,000 for naming rights. During a bidding war for the stadium naming rights, one company bid 7% more than

alex41 [277]

Answer:

$\$\ 829464$

Step-by-step explanation:

Bid of first company:

Bid is $7\%$ more

$7\%$ of $760000=\frac{7}{100}\times760000=53200$

Bid of first $=760000+53200=813200$

Bid of second company:

Bid of second is $2\%$ greater than that of first

$2\%$ of first bid $=\frac{2}{100}\times 813200=16264$

Hence Bid of second company $=813200+16264=829464$

8 0

2 years ago

David made a class banner out of a large rectangular piece of paper. He cut a

Irina18 [472]

Answer:

100 in²

Step-by-step explanation:

The area of the banner is equal to the area of the initial rectangle minus the area of the cutout triangle.

The rectangle has a height of 8 inches and width of 14 inches, so its area is:

A = (8 in) (14 in) = 112 in²

The triangle has a base of 8 inches and a height of 3 inches, so its area is:

A = ½ (8 in) (3 in) = 12 in²

So the area of the banner is 112 in² − 12 in² = 100 in².

6 0

2 years ago

On this graph, 4:00 p.m. occurs at 16 hours after midnight, and 6:00 p.m. occurs at 18 hours after midnight. Which statements ar

KatRina [158]

Answer:

TRUE OPTIONS ARE:

"The temperature increased until 4:00 p.m. "

"The temperature decreased after 6:00 p.m. "

"The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m."

Step-by-step explanation:

graph attached and complete question below:

Which statements are true about the temperatures Luis recorded on the graph? Check all that apply.

The temperature increased until 4:00 p.m.
The temperature was not recorded between 4:00 p.m. and 6:00 p.m.
The temperature decreased after 6:00 p.m.
The temperature increased and then decreased before holding constant.
The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m.

Until 1600 hours, the graph increases, so we can say temperature increased until 4.00 pm (FIRST OPTION TRUE).

From 1600 to 1800 hours (4 - 6pm), the temperature stayed same (horizontal line). This doesn't mean the temperature wasn't recorded. (2nd OPTION FALSE).

After 1800 hours (6pm), the line goes downward, so temperature decreased after 6 pm. (3rd OPTION TRUE).

If you look at the temperature graph, we can see temperature increased, then increased more, then constant, then decreased. Thus the 4th option isnt true. (4th OPTION FALSE).

Before 12, the increase isn't as sharp as after 12. After 12 temperature increase has more slope, so this increase is more. (5th OPTION TRUE).