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patriot [66]
4 years ago
12

A team is being formed that includes eight different people. There are eight different positions on the teams. How many differen

t ways are there to assign the right people to the eight ​positions
Mathematics
1 answer:
abruzzese [7]4 years ago
6 0

Answer:

40320 different ways

Step-by-step explanation:

That problem is a permutation one

We have eight people to occupy  one position in a team, without any constraint at all

So

Total number of events   = P(8)

P (8)  =  8!

P (8)  =  8*7*6*5*4*3*2*1

P (8)  =  40320 different ways

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Find the missing side 5 and 6.​
Lady_Fox [76]

Answer:

Hypotenuse= 10, short leg= 5

Step-by-step explanation:

6 0
3 years ago
If 0.6 of a number is 120,what is 0.25 of the same number​
earnstyle [38]
0.6/1=120/x
0.6x=120
x=200

200*0.25=50
8 0
4 years ago
Function 1 is represented by the equation, y = 0.75x - 1. Function 2 is represented by the graph below. Which about the function
VARVARA [1.3K]

Answer: 1- function 1 has the greater rate of change but function 2 has the greater y-intercept

Step-by-step explanation:

We know that function 1 is:

y = f₁(x) = 0.75*x - 1

And for the second linear equation we can see that it passes through the points (-3, -1) and (0,1)

Then the slope is:

a = (1 - (-1))/(0 - (-3)) = 2/3

then the equation is something like:

y = f₂(x) = (2/3)*x + b

To find the value of b, we use the fact that when x = 0, we have y = 1.

Then:

1 = (2/3)*0 + b

1 = b

Then the equation for function 2 is:

f₂(x) = (2/3)*x + 1

So the two equations are:

f₁(x) = 0.75*x - 1

f₂(x) = (2/3)*x + 1

Where we can see that function 1 has a greater slope (rate of change) and function two has a greater y-intercept.

Then the correct option is

(1- function 1 has the greater rate of change but function 2 has the greater y-intercept)

3 0
3 years ago
How many times does 6 go into 133
garik1379 [7]

Answer:

22 times with 1 as the remainder.

Step-by-step explanation:

3 0
4 years ago
Read 2 more answers
Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).
andre [41]
Check the picture below.  So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\

\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}

7 0
4 years ago
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