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2 years ago

14

How much greater is the median number of pairs of shoes in the inventory at the Golf pro Shop than the median number of pairs of

shoes in the inventory at the Tennis Pro Shop?

1 answer:

lana66690 [7]2 years ago

4 0

Answer: 20

To get this answer, you find the median of the golf pro shop data (a = 40) and the median of the tennis pro shop data (b = 20). Then you subtract to get a-b = 40-20 = 20. The median of any box plot is the vertical line inside the box plot.

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Solve the equation 5x + 2(2x -23) = -154

Anon25 [30]

$\boxed{\large{\bold{\blue{ANSWER~:) }}}}$

5x+2(2x-23)=-154

5x+4x-46=-154

9x-46=-154

9x=-154+46

9x=-108

x= $\sf{\dfrac{-108}{9} }$

x=-12

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2 years ago

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It is a best practice to analyze data directly on an infected machine. t or f?

Alex Ar [27]

I would say your answer is FALSE

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2 years ago

Exponential Function assignments help please! ❤️

Sladkaya [172]

Answer:

Initial value is 4,000

Growth is -480

Step-by-step explanation:

Initial value is the value that you started with: hence 4000.

Growth would be how much it grows, it grows downward because it is 12% down. Next you need to find what 13% of 4000 is 100%-13%=88% .88*4000=480 or 13% of 4000.

Hope this helps

7 0

2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

2 years ago

What is the value of p so that the expession 6x+3-(8x-5) is equivalent to (px+8)?

kiruha [24]

Answer:

-2

Step-by-step explanation:

The first equation simplifies to -2x+8 so p should be replaced with -2

4 0

2 years ago