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3 years ago

14

1. The range of guest on New Year's Eve was 15>

2 answers:

leonid [27]3 years ago

7 0

Answer:

1.False

2. True

3.) True

4. False

5.) False

6.) True

Step-by-step explanation:

1.)

Range = maximum - minimum

New year's eve :

Maximum = 22 ; minimum = 3

Range = 22 - 3 = 19

2.)

Median for Thanksgiving = 17

50% up to 17 ;hence the remaining 50% is also 17 and above

4.)

IQR for Thanksgiving

Q3 - Q1

23 - 12 = 11

5.)

The median number of people at new year's eve for dinner was 7 (line in between the box)

6.)

3/4th of new year eve's dinner = Q3 = 15

1/4 = Q1 ; have up to 5 people

Therefore ; the remaining (1 - 1/4) = 3/4 have 5 and above

Marizza181 [45]3 years ago

3 0

Answer:

1.False

2. True

3.) True

4. False

5.) False

6.) True

Step-by-step explanation:

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Can please someone help me? The question is in the picture. Thank you

maxonik [38]

Answer:

C. $9\pi$

Step-by-step explanation:

We proceed to solve for $x$ by the Pythagorean Theorem:

$(x+1)^{2} = (x-1)^{2} + (2\cdot x - 4)^{2}$ (1)

$x^{2} + 2\cdot x + 1 = (x^{2}-2\cdot x + 1) + (4\cdot x^{2}-16\cdot x + 16)$

$x^{2} + 2\cdot x + 1 = 5\cdot x^{2}-18\cdot x + 17$

$4\cdot x^{2} -20\cdot x + 16 = 0$

$(2\cdot x)^{2} -10\cdot (2\cdot x) + 16 = 0$

$(2\cdot x -8)\cdot (2\cdot x -2) = 0$

There are two different solutions:

$x_{1} = 4$ , $x_{2} = 1$

A solution of $x$ is considered realistic when the length of every side of the triangle is a not negative number.

$x_{1} = 4$

$UV = 2\cdot (4) - 4$

$UV = 4$

$WU = 4 - 1$

$WU = 3$

$WV = 4 + 1$

$WV = 5$

$x_{2} = 1$

$UV = 2\cdot (1) - 4$

$UV = -2$

$WU = 1 - 1$

$WU = 0$

$WV = 1 + 1$

$WV = 2$

The only realistic set of solutions is for $x = 4$ , and the radius of the circle is represented by the line segment $WU$ : $r = 3$

And the area of the circle ( $A$ ) is calculated from the following formula:

$A = \pi\cdot r^{2}$

$A = \pi\cdot (9)$

$A = 9\pi$

The correct answer is C.

6 0

3 years ago