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Yuliya22 [10]
4 years ago
6

Will y’all help me on 16 thank you

Mathematics
2 answers:
kap26 [50]4 years ago
3 0

Answer:R

Step-by-step explanation:

Calculate 15% of 8000:

8000*15/100 =1200

Add this to the total population of 2010 to get the total poulation in 2014:

8000+1200= 9200

To find the number of people who were age 12 or under multuply 9200 by 2/5:

9200*2/5= 3680

nekit [7.7K]4 years ago
3 0
The answer is for your question is R.
All you do is Multiply 8000x.15 then add that answer to 8000 then divide by 5 and multiply that answer by 2
You might be interested in
A game of chance involves spinning a wheel with 4 number on it. The wheel is designed so that the result of each spin Xhas the f
Kipish [7]

Distribution table :

X : ___ 1 _____ 2 _____ 3 ______ 4

P(x) __0.50 __0.25 __ 0.15 ____ 0.10

Answer:

1.85 ; 1.014 ;` 0.55 ; 3.042

Step-by-step explanation:

Probability distribution :

X : ___ 1 _____ 2 _____ 3 ______ 4

P(x) __0.50 __0.25 __ 0.15 ____ 0.10

The mean: E(x) = Σ(X * p(x))

(1*0.5) + (2*0.25) + (3*0.15) + (4 *0.10)

= 1.85

Standard deviation = sqrt(Var(x))

Var(x) = Σ(x²*p(x)) - E(x)²

Var(x) = ((1^2*0.5) + (2^2*0.25) + (3^2*0.15) + (4^2 *0.10)) - 1.85^2

= 4.45 - 3.4225

= 1.0275

Standard deviation = sqrt(1.0275)

Standard deviation = 1.0136567

Standard deviation(X) = 1.014

3.)

Cost of spin = $5

Amount, y to be received = 3 times the number that appears

y = 3x - cost of playing

y = 3x - 5

E(y) = E(3x - 5)

E(y) = E(3x) - 5

Recall :E(x) = 1.85

E(y) = 3(1.85) - 5

E(y) = 0.55

Standard deviation :Sd(y) =

Sd(3x - 5)

3(1.014)

= 3.042

5 0
3 years ago
If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1<br><br><br> help meeeee plessss ​
Trava [24]

A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}

where det(<em>A</em>) = 1×1 - 2×1 = -1.

B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}

where det(<em>B</em>) = 0×2 - (-1)×1 = 1. Then

B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

On the other side, we have

AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}

and det(<em>AB</em>) = det(<em>A</em>) det(<em>B</em>) = (-1)×1 = -1. So

(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

and both matrices are clearly the same.

More generally, we have by definition of inverse,

(AB)(AB)^{-1} = I

where I is the identity matrix. Multiply on the left by <em>A </em>⁻¹ to get

A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}

Multiplication of matrices is associative, so we can regroup terms as

(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}

Now multiply again on the left by <em>B</em> ⁻¹ and do the same thing:

B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}

7 0
3 years ago
7. Given: AB CD, BD =DE
MAVERICK [17]

Answer:

Step-by-step explanation:

Given that Given: AB = CD, BD =DE

The following addition postulates are true;

AB+BD = AD......  1

CD+DE = CE...... 2

Substituting AB = CD, BD =DEinto equation 1, the equations will become;

CD+DE = AD......  3

Subtract equation 2 from 3;

CD -CD+DE-DE = CE - AD

0+0 = CE-AD

0 = CE-AD

CE-AD = 0

Add AD to both sides

CE-AD+AD = 0+AD

CE = AD

AD = CE (Proved)

5 0
3 years ago
Omae wa sinderou nani (questions in the image btw)
Korvikt [17]

Let a be the number of hours worked at Job A and b the number of hours at Job B. Then

a+b=30

and

7.5a+8b=234.50

From the first equation,

b=30-a

and substituting this into the second gives

7.5a+8(30-a)=234.50\implies-0.5a+240=234.50

\implies0.5a=5.50

\implies\boxed{a=11}

5 0
4 years ago
Sal and three friends plan to bowl one or two games each. Each game costs $2.50. Identify the independent and dependent quantiti
GrogVix [38]

Answer:

C. Number of games; total cost; 4 to 8 games; $10 to $20.

Step-by-step explanation:

Since the number of games played affects the total cost, # of games is the independent variable (domain) and total cost is the dependent variable (range). There are 4 people in total and they plan to bowl 1 to 2 each, resulting in a total of 4 to 8 games. 2.50 dollars times 4 is $10 and 2.50 dollars times 8 is $20.

5 0
3 years ago
Read 2 more answers
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