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3 years ago

12=3x-6

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

8 0

Answer:

12=3x-6 is 6

16=4x+8 is 2

4z+8=40 is 8

5x-7=38 is 9

-3x-9=24 is -11

Step-by-step explanation:

for 12=3x-6 you would get x by itself so first you add the 6 to the 12

you get 18 then you try to get the x by itself so you divide 3x by 3 and divide 18 by 3 as well. So you end up with 6=x. Repeat with every problem

Montano1993 [528]3 years ago

7 0

Answer: 1 is b

2 is c

3 is c

4 is d

5 is b

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Help me please, the question is in the picture^^^!

Nat2105 [25]

Answer:

79.3

Step-by-step explanation:

2,037,634= 100 percent

1,614,840=x

x=1,614,840×100÷2,037,634

x=79.25

7 0

3 years ago

You have just opened a new dance club, Swing Haven, but are unsure of how high to set the cover charge (entrance fee). One week

bonufazy [111]

Answer:

a) The demand function is

$q(p) = -4 p + 107$

b) The nightly revenue is

$R(p) = -4 p^2 + 107 p$

c) The profit function is

$P(p) = -4 p^2 + 133.75 p - 939$

d) The entrance fees that allow Swing Haven to break even are between 10.03 and 23.41 dollars per guest.

Step-by-step explanation:

a) Lets find the slope s of the demand:

$s = \frac{79-43}{7-16} = \frac{36}{-9} = -4$

Since the demand takes the value 79 in 7, then

$q(p) = -4 (p-7) + 79 = -4 p + 107$

b) The nightly revenue can be found by multiplying q by p

$R(p) = p*q(p) = p*( -4 p + 107) = -4 p^2 + 107 p$

c) The profit function is obtained from substracting the const function C(p) from the revenue function R(p)

$P(p) = R(p) - C(p) = p*q(p) = -4 p^2 + 107 p - (-26.75p + 939) = \\\\-4 p^2 + 133.75 p - 939$

d) Lets find out the zeros and positive interval of P. Since P is a quadratic function with negative main coefficient, then it should have a maximum at the vertex, and between the roots (if any), the function should be positive. Therefore, we just need to find the zeros of P

$r_1, r_2 = \frac{-133.75 \,^+_-\, \sqrt{133.75^2-4*(-4)*(-939)} }{-8} = \frac{-133.75 \,^+_-\, 53.526}{-8} \\r_1 = 10.03\\r_2 = 23.41$

Therefore, the entrance fees that allow Swing Haven to break even are between 10.03 and 23.41 dollars per guest.

7 0

3 years ago

Solve 7x^2–20x = 3. ;

sasho [114]

Answer:

x=3, -1/7

Step-by-step explanation:

7x^2-20x-3=0

Factor.

(x-3)(7x+1)=0

x-3=0 ⇒ x=3

7x+1=0 ⇒7x=-1 ⇒x= -1/7

3 0

3 years ago

Read 2 more answers

Esosa pays $43.19 for 3.7 pounds of salmon. What is the price per pound of the salmon?

joja [24]

Answer:

$11.67

Step-by-step explanation:

$43.19÷3.7=11.672972973, which rounds to $11.67.

3 0

3 years ago

Read 2 more answers

Lenovo uses the zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as follows:

Stella [2.4K]

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

$\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}$

The forcast for a period $F_{t+1}$ is given by the formular

$F_{t+1}=\alpha A_t+(1-\alpha)F_t$

where $A_t$ is the actual value for the preceding period and $F_t$ is the forcast for the preceding period.

Part 1A:
Given α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804$

Therefore, the foreast for period 11 is $1.80

Part 1B:

Given α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Therefore, the foreast for period 12 is $1.78

Part 2A:

Given α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70


Part 2B:

Given α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68


Part 3A:

Given α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65


Part 3B:

Given α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

Therefore, the forecast for period 12 is $1.64

Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

$\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157

Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = 
\frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = 
\frac{0.32}{3} \approx0.107$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107


Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = 
\frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = 
\frac{29}{3} \approx0.097$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097

5 0

3 years ago