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spayn [35]
3 years ago
5

2(4x + 2) = 4x - 12(x - 1)

Mathematics
2 answers:
natta225 [31]3 years ago
7 0

Let's solve this problem step-by-step.

2(4x+2)=4x−12(x−1)

Step 1: Simplify both sides of the equation.

2(4x+2)=4x−12(x−1)

(2)(4x)+(2)(2)=4x+(−12)(x)+(−12)(−1)(Distribute)

8x+4=4x+−12x+12

8x+4=(4x+−12x)+(12)(Combine Like Terms)

8x+4=−8x+12

8x+4=−8x+12

Step 2: Add 8x to both sides.

8x+4+8x=−8x+12+8x

16x+4=12

Step 3: Subtract 4 from both sides.

16x+4−4=12−4

16x=8

Step 4: Divide both sides by 16.

16x/16=8/16

So, the answer for this problem is x=1/2.

lisabon 2012 [21]3 years ago
5 0

Answer:

1/2

Step-by-step explanation:

8x+4=4x-12x+12

8x+4=-8x+12

16x+4=12

16x=8

x=8/16=1/2

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Find X, last two anwsers are 4 and 8
IgorLugansk [536]

Answer:

9

Step-by-step explanation:

8X+4=360-108-80-96

8X+4=76

8X=76-4

8X=72

X=72/8

X=9

8 0
3 years ago
The function C (t) = StartFraction 4 t Over t squared + 2 EndFraction models the cost per student of a field trip when x student
choli [55]

Answer:

C. It is vertically stretched by a factor of 200 and shifted 10 units up.

Step-by-step explanation:

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7 0
3 years ago
As part of a research project on student debt at TWU, a researcher interviewed a sample of 35 students that were chosen at rando
Inga [223]

Answer:

Option a is right

Step-by-step explanation:

Given that as part of a research project on student debt at TWU, a researcher interviewed a sample of 35 students that were chosen at random concerning their monthly credit card balance.

Sample average = 2573

Variance = 4252

Sample size = 35

STd deviation of X = \sqrt{4252} \\=65.21

Score of student selected at random X=1700

Corresponding Z score = \frac{1700-2573}{65.201} \\=-13.38

Rounding of we get Z score = -13.4

option a is right

6 0
3 years ago
. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting
SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, \overline x_1 = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, \overline x_2 = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20  - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; t_{\alpha /2} = ±2.026192

The t-confidence interval is given as follows;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore;

\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

6 0
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