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svp [43]
3 years ago
5

Exhibit 12-4 In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Li

beral Arts College, and 30% of the students were in the Education College. To see whether the proportions have changed, a random sample of 300 students from ABC University was selected. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. Refer to Exhibit 12-4. This problem is an example of a _____. Select one: a. normally distributed variable b. test for independence c. multinomial population d. uniformly distributed variable
Mathematics
1 answer:
Brums [2.3K]3 years ago
8 0

Answer:d. uniformly distributed variable

Step-by-step explanation:Uniform Distribution Definition- in statistics , this mean an outcome is equally likely each variable has an equal or same probability to appear in an outcome.

For example in a deck of cards ,they are uniformly distributed which means the likelihood of drawing a club, a diamond , and a heart is the same.

When we see the percentage of how these students are distributed in art college and business college it is likely that one obtains a uniform distributed results.

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In the diagram below, R is located at (24,0), N is located at (12,18), T is located at (12,6), and E is located at (18,15). Assu
julsineya [31]

The midpoint of a segment divides the segment into equal halves

  • The coordinates of K are: \mathbf{K = (12,12)}
  • The coordinates of I are: \mathbf{I = (24,24)}
  • The coordinates of B are: \mathbf{B = ( 30,33 )}

The given parameters are:

\mathbf{R =(24,0)}

\mathbf{N =(12,18)}

\mathbf{T =(12,6)}

\mathbf{E =(18,15)}

K is the midpoint of N and T.

So, we have:

\mathbf{K = (\frac{N_x + T_x}{2},\frac{N_y + T_y}{2})}

This gives

\mathbf{K = (\frac{12 + 12}{2},\frac{18+ 6}{2})}

\mathbf{K = (12,12)}

E is the midpoint of T and I.

So, we have:

\mathbf{E = (\frac{I_x + T_x}{2},\frac{I_y + T_y}{2})}

This gives

\mathbf{(18,15) = (\frac{I_x + 12}{2},\frac{I_y+ 6}{2})}

Multiply through by 2

\mathbf{(36,30) = (I_x + 12,I_y+ 6)}

By comparison

\mathbf{I_x + 12 = 36.\ I_y + 6 =30}

So, we have:

\mathbf{I_x= 24.\ I_y  =24}

Hence, the coordinates of I are:

\mathbf{I = (24,24)}

I is the midpoint of E and B.

So, we have:

\mathbf{I = (\frac{E_x + B_x}{2},\frac{E_y + B_y}{2})}

This gives

\mathbf{(24,24) = (\frac{18 + B_x}{2},\frac{15 + B_y}{2})}

Multiply through by 2

\mathbf{(48,48) = (18 + B_x,15 + B_y)}

By comparison

\mathbf{18 + B_x = 48,\ 15 + B_y = 48 }

So, we have:

\mathbf{B_x = 30,\  B_y = 33 }

Hence, the coordinates of B are:

\mathbf{B = ( 30,33 )}

Read more about midpoints at:

brainly.com/question/18068617

7 0
2 years ago
HELP please!!!!!!!!!!!
Valentin [98]

Answer:

Step-by-step explanation:

5 0
3 years ago
What is the true solution to the equation below?
Naddik [55]
Yeeee

assuming your equaiton is
2ln(e^{ln(2x)})-ln(e^{ln(10x)})=ln(30)


remember some nice log rules
log_a(b)=c translates to a^c=b
and
a^{log_a(b)}=b
and
xlog_c(b)=log_c(b^x)
and
ln(x)=log_e(x)
and
log(a)-log(b)=log(\frac{a}{b})
and
if log(a)=log(b) then a=b

so

we can simplify a bit of stuff here

the e^{ln(2x)} \space\ and \space\ the \space\ e^{ln(10x)} can be simplified to 2x \space\ and \space\ 10x

so we gots now

2ln(2x)-ln(10x)=ln(30)
ln((2x)^2)-ln(10x)=ln(30)
ln(4x^2)-ln(10x)=ln(30)
ln(\frac{4x^2}{10x})=ln(30)
same base so
\frac{4x^2}{10x}=30
\frac{2x}{5}=30
times both sides by 5
2x=150
divide both sides by 2
x=75
answer is x=75
5 0
3 years ago
I need help please...
telo118 [61]
The answer is b
Area=[pi]r2
= [pi](12 in)2
=452.16 in2
7 0
3 years ago
I can't solve this equation pleas help!x^2+2x+8=0
tiny-mole [99]
x^2+2x+8=0\\
x^2+2x+1+7=0\\
(x+1)^2=-7\\
x+1=-\sqrt{-7} \vee x+1=\sqrt{-7}\\
x=-1-\sqrt7 i \vee x=-1+\sqrt7 i
8 0
3 years ago
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