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3 years ago

What is the correct method to label a segment with endpoint E and endpoint F?

Mathematics

2 answers:

Nuetrik [128]3 years ago

7 0

Answer:

The correct option is B.

Step-by-step explanation:

Given information: End points of a line segments are E and F.

Line segment : Line segment is a part of a line. It is represented by end points on each end of the line segment.

If end points of a line segments are E and F, then it is labeled as $\overline {EF}$ and $\overline {FE}$ .

Line: A line goes on forever. It is represented by a line with arrows at each end. $\overleftrightarrow {EF}$ and $\overleftrightarrow {FE}$ .

We need to find the the correct method to label a segment with endpoint E and endpoint F. The label for line segment are

$\overline {EF}$ and $\overline {FE}$

Therefore the correct option is B.

Degger [83]3 years ago

4 0

Answer : option B

the correct method to label a segment with endpoint E and endpoint F

The endpoints of the line are E and F

The Line starts at E and ends at F. The line will not extend after F

The line having end points are called as line segment

We represent a line segment by putting a line at the top

to label a segment with endpoint E and endpoint F, we use EF or FE and a line at the top

So answer is option B

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There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=

yulyashka [42]

Answer:

S = {0,2,3,4}

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Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

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P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = $\frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }$

P(X=2) = 0.401

P(X=3) = $\frac{4C3 * 12P2}{12^{4} }$

P(x=3) = 0.025

P(X=4) = $\frac{12}{12^{4} }$

P(X=4) = 0.001

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Standard deviation, $SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}$

SD = 1.033

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3 years ago

You received a gift in the mail. The length of the box is 12 inches, the width is 8

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Answer: 296 or 592

Step-by-step explanation:

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3 years ago

Help plz!!!

lara [203]

Answer with explanation:

→→→Function 1

f(x)= - x²+ 8 x -15

Differentiating once , to obtain Maximum or minimum of the function

f'(x)= - 2 x + 8

Put,f'(x)=0

-2 x+ 8=0

2 x=8

Dividing both sides by , 2, we get

x=4

Double differentiating the function

f"(x)= -2, which is negative.

Showing that function attains maximum at ,x=4.

Now,f(4)=-4²+ 8× 4-15

= -16 +32 -15

= -31 +32

→→→Function 2:

f(x) = −x² + 2 x − 3

Differentiating once , to obtain Maximum or minimum of the function

f'(x)= -2 x +2

Put,f'(x)=0

-2 x +2=0

2 x=2

Dividing both sides by , 2, we get

x=1

Double differentiating the function,gives

f"(x)= -2 ,which is negative.

Showing that function attains maximum at ,x=1.

f(1)= -1²+2 ×1 -3

= -1 +2 -3

= -4 +2

= -2

⇒⇒⇒Function 1 has the larger maximum.

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