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3 years ago

Solve for x. 2(x + 14) + x = 70a. 11b. 12c. 14d. 17

Mathematics

1 answer:

Mariulka [41]3 years ago

6 0

Hello!

To solve for the variable x, we need to follow the rules of SADMEP. SADMEP is ONLY used when solving algebraic equations. SADMEP is an acronym for subtraction, addition, division, multiplication, exponents, and parentheses. Also, when solving for x, our main goal is to isolate x to one side of the equation.

2(x + 14) + x = 70 (use the distributive property)

2x + 28 + x = 70 (simplify - add)

3x + 28 = 70 (subtract 28 from both sides)

3x = 42 (divide both sides by 3)

x = 14

When solving ANY algebraic equation, we need check our answer. To check your answer, you substitute the x-value, which is 14 in this case, into the original equation and solve using PEMDAS.

2(x + 14) + x = 70 (substitute x = 14)

2(14 + 14) + 14 = 70 (simplify the terms inside the parentheses)

2(28) + 14 = 70 (simplify - multiply)

56 + 14 = 70 (simplify - add alike term)

70 = 70 | This is correct!

Therefore, your final answer is x = 14, which is choice C.

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Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is

$\sigma = 12.6 \: hours$

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

$n = 250$

The sample mean is

$\bar{x} = 9 \: hours$

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

$\mu = \bar{x} = 9 \: hours$

The sample standard deviation would be

$s = {\frac{\sigma}{\sqrt{n} }$

Where is the population standard deviation and n is the sample size.

$s = {\frac{12.6}{\sqrt{250} }$

$s = 0.7969 \: hours$

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

$\text {confidence interval} = \mu \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sampling size, s is the sample standard deviation and is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

$MoE = t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\\\MoE = 1.9695\cdot \frac{12.6}{\sqrt{250} } \\\\MoE = 1.9695\cdot 0.7969\\\\MoE = 1.569\\\\$

So the required 95% confidence interval is

$\text {confidence interval} = \mu \pm MoE\\\\\text {confidence interval} = 9 \pm 1.569\\\\\text {LCI } = 9 - 1.569 = 7.431\\\\\text {UCI } = 9 + 1.569 = 10.569$

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

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