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2 years ago

How do you solve 3x+7=x

2 answers:

8 0

3x + 7 = x

First, subtract 3x from both sides. / Your problem should look like: 7 = x - 3x
Second, simplify x - 3x to -2x. / Your problem should look like: 7 = -2x
Third, divide both sides by -2. / Your problem should look like: $\frac{7}{-2}$ = x
Fourth, simplify $\frac{7}{-2}$ to $-\frac{7}{2}$ / Your problem should look like: $- \frac{7}{2}$ = x
Fifth, switch sides. / Your problem should look like: x = $-\frac{7}{2}$

Answer as fraction: $-\frac{7}{2}$
Answer as decimal: -3.5

Marizza181 [45]2 years ago

7 0

3x + 7 = x
3x - x = -7
2x = -7
x = - 7/2
x = -3.5

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Please help ASAP!

Oliga [24]

The age of the students is about 0.67 years away from the 13.3 years.

The Mean Absolute Deviation is the average distance away from the mean. For the last choice, you should select more variable. Since the average is higher, there is a greater distance between the points and the mean.

3 0

3 years ago

Q4.

goblinko [34]

The coding of the statistic is used to make it easier to work with the large sunshine data set

The mean of the sunshine is 3.05 $\overline 6$

The standard deviation is approximately 18.184

Reason:

The given parameters are;

The sample size, n = 3.

∑x = 947

Sample corrected sum of squares, Sₓₓ = 33,065.37

The mean and standard deviation = Required

Solution:

$Mean, \ \overline x = \dfrac{\sum x_i}{n}$

The mean of the daily total sunshine is therefore;

$Mean, \ \overline x = \dfrac{947}{30} \approx 31.5 \overline 6$

$s = \dfrac{x}{10 } - \dfrac{1}{10}$

$E(s) = \dfrac{Ex}{10 } - \dfrac{1}{10}$

$E(s) = \dfrac{31.5 \overline 6}{10 } - \dfrac{1}{10} = 3.05 \overline 6$

The mean ≈ 3.05 $\overline 6$

Alternatively

, $The \ mean \ of \ the \ daily \ total \ sunshine, \, s = \dfrac{31.5 \overline 6 - 1}{10 } = 3.05\overline 6$

The mean of the daily total sunshine, $\overline s$ ≈3.05 $\overline 6$

$Var(s) = Var \left(\dfrac{x}{10 } - \dfrac{1}{10} \right)$

$Var(s) = \left(\dfrac{1}{10}\right)^2 \times Var \left(x \right)$

Therefore;

$Var(s) = \left(\dfrac{1}{10}\right)^2 \times 33,065.37 = 330,6537$

Therefore;

$s = \sqrt{330.6537} \approx 18.184$

The standard deviation, $s_s$ ≈ 18.184

Learn more about coding of statistic data here:

brainly.com/question/14837870

3 0

2 years ago

What is the equation of the line that has a slope of -3 and passes through the point (4,6)?

garri49 [273]

Answer:

y = -3x + 18

Step-by-step explanation:

y = mx + b

m = -3

6 = (-3) * 4 +b

b = 6 - (-3) * 4

b = 6 +12

b = 18

y = -3x + 18

6 0

2 years ago

70.49 divided by 19 pls help

Masja [62]

Hello!

The answer is

3.71.

Hope this helps!! :)

5 0

2 years ago

In a recent year, a poll asked 2362 random adult citizens of a large country how they rated economic conditions. In the poll,

Harman [31]

Answer:

a) The 99% confidence interval is given by (0.198;0.242).

b) Based on the p value obtained and using the significance level assumed $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

c) $\alpha=0.01$

Step-by-step explanation:

Data given and notation

n=2362 represent the random sample taken

X represent the people who says that they would watch one of the television shows.

$\hat p=\frac{X}{n}=0.22$ estimated proportion of people rated as Excellent/Good economic conditions.

$p_o=0.24$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that 24% of people are rated with good economic conditions:

Null hypothesis: $p=0.24$

Alternative hypothesis: $p \neq 0.24$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Part a: Test the hypothesis

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np = 2362x0.22=519.64>10 and n(1-p)=2364*(1-0.22)=1843.92>10

Condition satisfied.

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.22 -0.24}{\sqrt{\frac{0.24(1-0.24)}{2362}}}=-2.28$

The confidence interval would be given by:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The critical value using $\alpha=0.01$ and $\alpha/2 =0.005$ would be $z_{\alpha/2}=2.58$ . Replacing the values given we have:

$0.22 - (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.198$

$0.22 + (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.242$

So the 99% confidence interval is given by (0.198;0.242).

Part b

Statistical decision 

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided is $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

So based on the p value obtained and using the significance level assumed $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

Part c

The confidence level assumed was 99%, so then the signficance is given by $\alpha=1-confidence=1-0.99=0.01$

6 0

2 years ago