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ad-work [718]
3 years ago
14

In the figure, is divided into equal parts. The coordinates of point A are (2, 4), and the coordinates of point B are (10, 6). M

atch each pair of coordinates to the corresponding point on .

Mathematics
1 answer:
timofeeve [1]3 years ago
8 0

Answer:

Step-by-step explanation:

We will use mid point formula:

According to the attached picture:

F is the mid point of A and B.

Thus by using formula we have,

F = (2+10)/2 , (4+6)/2

= 12/2 , 10,2

= 6,5

F should be (6,5)

Like wise the matched coordinates are:

D is the midpoint of A and F and is (4,4.5)

C is the midpoint of A and D and is (3,4.25)

E is the mid point of D and F and is (5,4.75)

H is the midpoint of F and B and is (8,5.5)

G is the midpoint of F and H and is (7, 5.25)

I is the midpoint of H and B and is (9,5.75) ....

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The weights of adult male beagles are normally distributed, with a mean of 25 pounds and a standard deviation of 4 pounds.
Andrews [41]

Answer:

A) 8

Step-by-step explanation:

Mean weight = u = 25 pounds

Standard Deviation = \sigma = 4 pounds

We have to find how many beagles out of 75 will weigh more than 30. Since the data is normally distributed, we can use z score to find this value. First we will what is the percentage(probability) of a randomly selected beagle to weigh more than x = 30 pounds, using this percentage we can then find our answer.

The formula for z score is:

z=\frac{x-u}{\sigma}

Using the values, we get:

z=\frac{30-25}{4}=1.25

So, P(Weight > 30) is equivalent to P(z > 1.25). Using the z table, we can write:

P(z > 1.25) = P(Weight > 30) = 0.1056

This, 0.1056 or 10.56% of the beagles are expected to weigh more than 30 pounds.

So,out of 75 beagles, 10.56% of 75 are expected to weigh more than 30 pounds.

10.56% of 75 = 0.1056 x 75 = 7.92 = 8 (rounding to nearest integer)

Therefore, out of 75 beagles 8 are expected to weigh more than 30 pounds.

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IceJOKER [234]
The answers is A !! No problem
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Answer:

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